\displaystyle \textbf{Question 1: }~\text{There are }6\%\text{ defective items in a large bulk of items. Find the} \\ \text{probability that a sample of }8\text{ items will include not more than one defective item.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ denote the number of defective items in a sample of }8\text{ items.}
\displaystyle \text{Then }X\text{ follows a binomial distribution with }n=8.
\displaystyle p=0.06,\ q=1-p=0.94.
\displaystyle P(X=r)={}^{8}C_{r}(0.06)^r(0.94)^{8-r},\ r=0,1,2,3,\ldots,8.
\displaystyle \text{The required probability is the probability of not more than one defective item.}
\displaystyle =P(X\leq 1).
\displaystyle =P(X=0)+P(X=1).
\displaystyle ={}^{8}C_{0}(0.06)^0(0.94)^{8-0}+{}^{8}C_{1}(0.06)^1(0.94)^{8-1}.
\displaystyle =(0.94)^8+8(0.06)(0.94)^7.
\displaystyle =(0.94)^7(0.94+0.48).
\displaystyle =1.42(0.94)^7.

\displaystyle \textbf{Question 2: }~\text{A coin is tossed }5\text{ times. What is the probability of getting at} \\ \text{least }3\text{ heads?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ denote the number of heads in }5\text{ tosses.}
\displaystyle X\text{ follows a binomial distribution with }n=5.
\displaystyle p=\frac{1}{2},\ q=1-p=\frac{1}{2}.
\displaystyle P(X=r)={}^{5}C_{r}\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{5-r},\ r=0,1,2,\ldots,5.
\displaystyle \text{The required probability is }P(\text{getting at least }3\text{ heads}).
\displaystyle =P(X\geq 3).
\displaystyle =P(X=3)+P(X=4)+P(X=5).
\displaystyle ={}^{5}C_{3}\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^{5-3}+{}^{5}C_{4}\left(\frac{1}{2}\right)^4\left(\frac{1}{2}\right)^{5-4}+{}^{5}C_{5}\left(\frac{1}{2}\right)^5\left(\frac{1}{2}\right)^{5-5}.
\displaystyle =10\left(\frac{1}{2}\right)^5+5\left(\frac{1}{2}\right)^5+1\left(\frac{1}{2}\right)^5.
\displaystyle =\left(\frac{1}{2}\right)^5(10+5+1).
\displaystyle =\left(\frac{1}{2}\right)^5\times 16.
\displaystyle =\frac{1}{2}.

\displaystyle \textbf{Question 3: }~\text{A coin is tossed }5\text{ times. What is the probability that tail appears} \\ \text{an odd number of times?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ denote the number of tails when a coin is tossed }5\text{ times.}
\displaystyle X\text{ follows a binomial distribution with }n=5.
\displaystyle p=\frac{1}{2},\ q=1-p=\frac{1}{2}.
\displaystyle P(X=r)={}^{5}C_{r}\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{5-r}={}^{5}C_{r}\left(\frac{1}{2}\right)^5.
\displaystyle \text{The required probability is }P(X\text{ is odd}).
\displaystyle =P(X=1)+P(X=3)+P(X=5).
\displaystyle ={}^{5}C_{1}\left(\frac{1}{2}\right)^5+{}^{5}C_{3}\left(\frac{1}{2}\right)^5+{}^{5}C_{5}\left(\frac{1}{2}\right)^5.
\displaystyle =\left(\frac{1}{2}\right)^5(5+10+1).
\displaystyle =\frac{16}{32}.
\displaystyle =\frac{1}{2}.

\displaystyle \textbf{Question 4: }~\text{A pair of dice is thrown }6\text{ times. If getting a total of }9\text{ is considered} \\ \text{a success, what is the probability of at least }5\text{ successes?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ be the number of successes in }6\text{ throws of two dice.}
\displaystyle \text{Probability of success }=\text{ probability of getting a total of }9.
\displaystyle =\text{ probability of getting }(3,6),(4,5),(5,4),(6,3)\text{ out of }36\text{ outcomes.}
\displaystyle p=\frac{4}{36}=\frac{1}{9},\ q=1-p=\frac{8}{9},\ n=6.
\displaystyle X\text{ follows a binomial distribution with }n=6,\ p=\frac{1}{9},\ q=\frac{8}{9}.
\displaystyle P(X=r)={}^{6}C_{r}\left(\frac{1}{9}\right)^r\left(\frac{8}{9}\right)^{6-r}.
\displaystyle \text{The required probability is the probability of at least }5\text{ successes.}
\displaystyle =P(X\geq 5).
\displaystyle =P(X=5)+P(X=6).
\displaystyle ={}^{6}C_{5}\left(\frac{1}{9}\right)^5\left(\frac{8}{9}\right)^{6-5}+{}^{6}C_{6}\left(\frac{1}{9}\right)^6\left(\frac{8}{9}\right)^{6-6}.
\displaystyle =6\left(\frac{1}{9}\right)^5\left(\frac{8}{9}\right)^1+1\left(\frac{1}{9}\right)^6\left(\frac{8}{9}\right)^0.
\displaystyle =6\cdot\frac{1}{9^5}\cdot\frac{8}{9}+1\cdot\frac{1}{9^6}\cdot 1.
\displaystyle =\frac{48}{9^6}+\frac{1}{9^6}.
\displaystyle =\frac{49}{9^6}.

\displaystyle \textbf{Question 5: }~\text{A fair coin is tossed }8\text{ times, find the probability of (i) exactly } \\ 5\text{ heads (ii) at least six heads (iii) at most six heads.}
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)  }
\displaystyle \text{Let }X\text{ denote the number of heads obtained when a fair coin is tossed }8\text{ times.}
\displaystyle \text{Now, }X\text{ follows a binomial distribution with }n=8.
\displaystyle p=\frac{1}{2},\ q=1-p=\frac{1}{2}.
\displaystyle P(X=r)={}^{8}C_{r}\left(\frac{1}{2}\right)^{8-r}\left(\frac{1}{2}\right)^r={}^{8}C_{r}\left(\frac{1}{2}\right)^8,\ r=0,1,2,\ldots,8.
\displaystyle \text{Probability of getting exactly }5\text{ heads}=P(X=5).
\displaystyle ={}^{8}C_{5}\left(\frac{1}{2}\right)^8.
\displaystyle =\frac{56}{256}.
\displaystyle =\frac{7}{32}.
\displaystyle \textbf{(ii)  }
\displaystyle \text{Let }X\text{ denote the number of heads obtained when a fair coin is tossed }8\text{ times.}
\displaystyle \text{Now, }X\text{ follows a binomial distribution with }n=8.
\displaystyle p=\frac{1}{2},\ q=1-p=\frac{1}{2}.
\displaystyle P(X=r)={}^{8}C_{r}\left(\frac{1}{2}\right)^{8-r}\left(\frac{1}{2}\right)^r={}^{8}C_{r}\left(\frac{1}{2}\right)^8,\ r=0,1,2,\ldots,8.
\displaystyle \text{Probability of getting at least }6\text{ heads.}
\displaystyle =P(X\geq 6).
\displaystyle =P(X=6)+P(X=7)+P(X=8).
\displaystyle ={}^{8}C_{6}\left(\frac{1}{2}\right)^8+{}^{8}C_{7}\left(\frac{1}{2}\right)^8+{}^{8}C_{8}\left(\frac{1}{2}\right)^8.
\displaystyle =(28+8+1)\cdot\frac{1}{256}.
\displaystyle =\frac{37}{256}.
\displaystyle \textbf{(iii)  }
\displaystyle \text{Let }X\text{ denote the number of heads obtained when a fair coin is tossed }8\text{ times.}
\displaystyle \text{Now, }X\text{ follows a binomial distribution with }n=8.
\displaystyle p=\frac{1}{2},\ q=1-p=\frac{1}{2}.
\displaystyle P(X=r)={}^{8}C_{r}\left(\frac{1}{2}\right)^{8-r}\left(\frac{1}{2}\right)^r={}^{8}C_{r}\left(\frac{1}{2}\right)^8,\ r=0,1,2,\ldots,8.
\displaystyle \text{Probability of getting at most }6\text{ heads.}
\displaystyle =P(X\leq 6).
\displaystyle =1-[P(X=7)+P(X=8)].
\displaystyle =1-\left[{}^{8}C_{7}\left(\frac{1}{2}\right)^8+{}^{8}C_{8}\left(\frac{1}{2}\right)^8\right].
\displaystyle =1-\left(\frac{8}{256}+\frac{1}{256}\right).
\displaystyle =1-\frac{9}{256}.
\displaystyle =\frac{247}{256}.

\displaystyle \textbf{Question 6: }~\text{Find the probability of }4\text{ turning up at least once in two tosses of a} \\ \text{fair die.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ be the probability of getting }4\text{ in two tosses of a fair die.}
\displaystyle X\text{ follows a binomial distribution with }n=2.
\displaystyle p=\frac{1}{6},\ q=\frac{5}{6}.
\displaystyle P(X=r)={}^{2}C_{r}\left(\frac{1}{6}\right)^r\left(\frac{5}{6}\right)^{2-r}.
\displaystyle \text{Probability of getting }4\text{ at least once }=P(X\geq 1).
\displaystyle =1-P(X=0).
\displaystyle =1-{}^{2}C_{0}\left(\frac{1}{6}\right)^0\left(\frac{5}{6}\right)^{2-0}.
\displaystyle =1-\frac{25}{36}.
\displaystyle =\frac{11}{36}.

\displaystyle \textbf{Question 7: }~\text{A coin is tossed }5\text{ times. What is the probability that head appears an} \\ \text{even number of times?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ be the number of heads that appear when a coin is tossed }5\text{ times.}
\displaystyle X\text{ follows a binomial distribution with }n=5\text{ and }p=q=\frac{1}{2}.
\displaystyle P(X=r)={}^{5}C_{r}\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{5-r}.
\displaystyle ={}^{5}C_{r}\left(\frac{1}{2}\right)^5.
\displaystyle P(\text{head appears an even number of times})=P(X=0)+P(X=2)+P(X=4).
\displaystyle ={}^{5}C_{0}\left(\frac{1}{2}\right)^5+{}^{5}C_{2}\left(\frac{1}{2}\right)^5+{}^{5}C_{4}\left(\frac{1}{2}\right)^5.
\displaystyle =\frac{1+10+5}{2^5}.
\displaystyle =\frac{16}{32}.
\displaystyle =\frac{1}{2}.

\displaystyle \textbf{Question 8: }~\text{The probability of a man hitting a target is }1/4.\text{ If he fires times, what is} \\ \text{the probability of his hitting the target at least twice?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ be the number of times the target is hit.}
\displaystyle X\text{ follows a binomial distribution with }n=7.
\displaystyle p=\frac{1}{4},\ q=\frac{3}{4}.
\displaystyle P(X=r)={}^{7}C_{r}\left(\frac{1}{4}\right)^r\left(\frac{3}{4}\right)^{7-r}.
\displaystyle P(\text{hitting the target at least twice}).
\displaystyle =P(X\geq 2).
\displaystyle =1-\{P(X=0)+P(X=1)\}.
\displaystyle =1-\left[{}^{7}C_{0}\left(\frac{1}{4}\right)^0\left(\frac{3}{4}\right)^{7-0}+{}^{7}C_{1}\left(\frac{1}{4}\right)^1\left(\frac{3}{4}\right)^{7-1}\right].
\displaystyle =1-\left[\left(\frac{3}{4}\right)^7+7\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^6\right].
\displaystyle =1-\frac{2187+5103}{16384}.
\displaystyle =1-\frac{3645}{8192}.
\displaystyle =\frac{4547}{8192}.

\displaystyle \textbf{Question 9: }~\text{Assume that on an average one telephone number out of }15\text{ called between } \\ 2\text{ P.M. and }3\text{ P.M. on week days is busy. What is the probability that if six randomly} \\ \text{selected telephone numbers are called, at least }3\text{ of them will be busy?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ be the number of busy calls for }6\text{ randomly selected telephone numbers.}
\displaystyle X\text{ follows a binomial distribution with }n=6.
\displaystyle p=\frac{1}{15},\ q=\frac{14}{15}.
\displaystyle P(X=r)={}^{6}C_{r}\left(\frac{1}{15}\right)^r\left(\frac{14}{15}\right)^{6-r}.
\displaystyle \text{Probability that at least }3\text{ of them are busy }=P(X\geq 3).
\displaystyle =1-[P(X=0)+P(X=1)+P(X=2)].
\displaystyle =1-\left\{{}^{6}C_{0}\left(\frac{1}{15}\right)^0\left(\frac{14}{15}\right)^{6-0}+{}^{6}C_{1}\left(\frac{1}{15}\right)^1\left(\frac{14}{15}\right)^{6-1}+{}^{6}C_{2}\left(\frac{1}{15}\right)^2\left(\frac{14}{15}\right)^{6-2}\right\}.
\displaystyle =1-\left\{\left(\frac{14}{15}\right)^6+6\left(\frac{1}{15}\right)\left(\frac{14}{15}\right)^5+15\left(\frac{1}{15}\right)^2\left(\frac{14}{15}\right)^4\right\}.

\displaystyle \textbf{Question 10: }~\text{If getting }5\text{ or }6\text{ in a throw of an unbiased die is a success and the random} \\ \text{variable }X\text{ denotes the number of successes in six throws of the die, find } \\ P(X\ge4).
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ denote the number of successes, i.e. of getting }5\text{ or }6\text{ in a throw of a die in }6\text{ throws.}
\displaystyle \text{Then }X\text{ follows a binomial distribution with }n=6.
\displaystyle p=\frac{1}{6}+\frac{1}{6}=\frac{1}{3},\ q=1-p=\frac{2}{3}.
\displaystyle P(X=r)={}^{6}C_{r}\left(\frac{1}{3}\right)^r\left(\frac{2}{3}\right)^{6-r}.
\displaystyle P(X\geq 4)=P(X=4)+P(X=5)+P(X=6).
\displaystyle ={}^{6}C_{4}\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)^{6-4}+{}^{6}C_{5}\left(\frac{1}{3}\right)^5\left(\frac{2}{3}\right)^{6-5}+{}^{6}C_{6}\left(\frac{1}{3}\right)^6\left(\frac{2}{3}\right)^{6-6}.
\displaystyle =\frac{1}{3^6}(60+12+1).
\displaystyle =\frac{73}{729}.

\displaystyle \textbf{Question 11: }~\text{Eight coins are thrown simultaneously. Find the chance of obtaining at least} \\ \text{six heads.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ be the number of heads in tossing }8\text{ coins.}
\displaystyle X\text{ follows a binomial distribution with }n=8.
\displaystyle p=\frac{1}{2},\ q=\frac{1}{2}.
\displaystyle P(X=r)={}^{8}C_{r}\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{8-r}={}^{8}C_{r}\left(\frac{1}{2}\right)^8.
\displaystyle \text{Probability of obtaining at least }6\text{ heads }=P(X\geq 6).
\displaystyle =P(X=6)+P(X=7)+P(X=8).
\displaystyle ={}^{8}C_{6}\left(\frac{1}{2}\right)^8+{}^{8}C_{7}\left(\frac{1}{2}\right)^8+{}^{8}C_{8}\left(\frac{1}{2}\right)^8.
\displaystyle =\frac{1}{2^8}(28+8+1).
\displaystyle =\frac{37}{256}.

\displaystyle \textbf{Question 12: }~\text{Five cards are drawn successively with replacement from a well shuffled} \\ \text{pack of }52\text{ cards. What is the probability that (i) all the five cards are spades (ii) only }3 \\ \text{ cards are spades (iii) none is spade?}
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)  }
\displaystyle \text{Let }X\text{ denote the number of spade cards when }5\text{ cards are drawn with replacement.}
\displaystyle \text{Because it is with replacement,}
\displaystyle X\text{ follows a binomial distribution with }n=5.
\displaystyle p=\frac{13}{52}=\frac{1}{4},\ q=1-p=\frac{3}{4}.
\displaystyle P(X=r)={}^{5}C_{r}\left(\frac{1}{4}\right)^r\left(\frac{3}{4}\right)^{5-r}.
\displaystyle P(\text{all cards are spades})=P(X=5).
\displaystyle ={}^{5}C_{5}\left(\frac{1}{4}\right)^5\left(\frac{3}{4}\right)^0.
\displaystyle =\frac{1}{1024}.
\displaystyle \textbf{(ii)  }
\displaystyle \text{Let }X\text{ denote the number of spade cards when }5\text{ cards are drawn with replacement.}
\displaystyle \text{Because it is with replacement,}
\displaystyle X\text{ follows a binomial distribution with }n=5.
\displaystyle p=\frac{13}{52}=\frac{1}{4},\ q=1-p=\frac{3}{4}.
\displaystyle P(X=r)={}^{5}C_{r}\left(\frac{1}{4}\right)^r\left(\frac{3}{4}\right)^{5-r}.
\displaystyle P(\text{only }3\text{ cards are spades})=P(X=3).
\displaystyle ={}^{5}C_{3}\left(\frac{1}{4}\right)^3\left(\frac{3}{4}\right)^2.
\displaystyle =\frac{90}{1024}.
\displaystyle =\frac{45}{512}.
\displaystyle \textbf{(iii)  }
\displaystyle \text{Let }X\text{ denote the number of spade cards when }5\text{ cards are drawn with replacement.}
\displaystyle \text{Because it is with replacement,}
\displaystyle X\text{ follows a binomial distribution with }n=5.
\displaystyle p=\frac{13}{52}=\frac{1}{4},\ q=1-p=\frac{3}{4}.
\displaystyle P(X=r)={}^{5}C_{r}\left(\frac{1}{4}\right)^r\left(\frac{3}{4}\right)^{5-r}.
\displaystyle P(\text{none is a spade})=P(X=0).
\displaystyle ={}^{5}C_{0}\left(\frac{1}{4}\right)^0\left(\frac{3}{4}\right)^5.
\displaystyle =\frac{243}{1024}.

\displaystyle \textbf{Question 13: }~\text{A bag contains }7\text{ red, }5\text{ white and }8\text{ black balls. If four balls are drawn} \\ \text{one by one with replacement, what is the probability that (i) none is white (ii) all are} \\ \text{white (iii) any two are white?}
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)  }
\displaystyle \text{Let }X\text{ be the number of white balls drawn when }4\text{ balls are drawn with replacement.}
\displaystyle X\text{ follows a binomial distribution with }n=4.
\displaystyle p=\text{Probability for a white ball}=\frac{\text{No of white balls}}{\text{Total number of balls}}.
\displaystyle =\frac{5}{20}.
\displaystyle =\frac{1}{4}.
\displaystyle \text{and }q=1-p=\frac{3}{4}.
\displaystyle P(X=r)={}^{4}C_{r}\left(\frac{1}{4}\right)^r\left(\frac{3}{4}\right)^{4-r}.
\displaystyle \text{Prob that none is white }=P(X=0).
\displaystyle ={}^{4}C_{0}\left(\frac{1}{4}\right)^0\left(\frac{3}{4}\right)^{4-0}.
\displaystyle =\frac{81}{256}.
\displaystyle \textbf{(ii)  }
\displaystyle \text{Let }X\text{ be the number of white balls drawn when }4\text{ balls are drawn with replacement.}
\displaystyle X\text{ follows a binomial distribution with }n=4.
\displaystyle p=\text{Probability for a white ball}=\frac{\text{No of white balls}}{\text{Total number of balls}}.
\displaystyle =\frac{5}{20}.
\displaystyle =\frac{1}{4}.
\displaystyle \text{and }q=1-p=\frac{3}{4}.
\displaystyle P(X=r)={}^{4}C_{r}\left(\frac{1}{4}\right)^r\left(\frac{3}{4}\right)^{4-r}.
\displaystyle \text{Prob that all are white }=P(X=4).
\displaystyle ={}^{4}C_{4}\left(\frac{1}{4}\right)^4\left(\frac{3}{4}\right)^{4-4}.
\displaystyle =\frac{1}{256}.
\displaystyle \textbf{(iii)  }
\displaystyle \text{Let }X\text{ be the number of white balls drawn when }4\text{ balls are drawn with replacement.}
\displaystyle X\text{ follows a binomial distribution with }n=4.
\displaystyle p=\text{Probability for a white ball}=\frac{\text{No of white balls}}{\text{Total number of balls}}.
\displaystyle =\frac{5}{20}.
\displaystyle =\frac{1}{4}.
\displaystyle \text{and }q=1-p=\frac{3}{4}.
\displaystyle P(X=r)={}^{4}C_{r}\left(\frac{1}{4}\right)^r\left(\frac{3}{4}\right)^{4-r}.
\displaystyle \text{Prob that any two are white }=P(X=2).
\displaystyle ={}^{4}C_{2}\left(\frac{1}{4}\right)^2\left(\frac{3}{4}\right)^{4-2}.
\displaystyle =\frac{54}{256}.
\displaystyle =\frac{27}{128}.

\displaystyle \textbf{Question 14: }~\text{A box contains }100\text{ tickets each bearing one of the numbers from }1\text{ to } \\ 100.\text{ If }5\text{ tickets are drawn successively with replacement from the box, find the} \\ \text{probability that all the tickets bear numbers divisible by }10.
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ be the variable representing the number on the ticket bearing a number divisible by }\\ 10\text{ out of the }5\text{ tickets drawn.}
\displaystyle \text{Then }X\text{ follows a binomial distribution with }n=5.
\displaystyle p=\text{Probability of getting a ticket bearing a number divisible by }10.
\displaystyle p=\frac{1}{100}(10)=\frac{1}{10},\ q=\frac{9}{10}.
\displaystyle P(X=r)={}^{5}C_{r}\left(\frac{1}{10}\right)^r\left(\frac{9}{10}\right)^{5-r}.
\displaystyle \text{Probability that all the tickets bear numbers divisible by }10.
\displaystyle =P(X=5).
\displaystyle ={}^{5}C_{5}\left(\frac{1}{10}\right)^5\left(\frac{9}{10}\right)^{5-5}.
\displaystyle =\left(\frac{1}{10}\right)^5.
\displaystyle \text{Hence, required probability is }\left(\frac{1}{10}\right)^5.

\displaystyle \textbf{Question 15: }~\text{A bag contains }10\text{ balls each marked with one of the digits } 0\text{ to }9.\text{ If} \\ \text{fourballs are drawn successively with replacement from the bag, what is the probability} \\ \text{that none is marked with the digit }0?
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ be the number of balls marked with the digit }0\text{ when }4\text{ balls are drawn} \\ \text{successively with replacement.}
\displaystyle \text{As this is with replacement, }X\text{ follows a binomial distribution with }n=4.
\displaystyle p=\text{probability that a ball randomly drawn bears digit }0=\frac{1}{10},\ q=1-p=\frac{9}{10}.
\displaystyle P(X=r)={}^{4}C_{r}\left(\frac{1}{10}\right)^r\left(\frac{9}{10}\right)^{4-r}.
\displaystyle P(\text{none bears the digit }0)=P(X=0).
\displaystyle ={}^{4}C_{0}\left(\frac{1}{10}\right)^0\left(\frac{9}{10}\right)^{4-0}.
\displaystyle =\left(\frac{9}{10}\right)^4.

\displaystyle \textbf{Question 16: }~\text{There are }5\%\text{ defective items in a large bulk of items. What is the} \\ \text{probability that a sample of }10\text{ items will include not more than one defective item?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ denote the number of defective items in a sample of }10\text{ items.}
\displaystyle X\text{ follows a binomial distribution with }n=10.
\displaystyle p=\text{probability of defective items}=5\%=0.05,\ q=1-p=0.95.
\displaystyle P(X=r)={}^{10}C_{r}(0.05)^r(0.95)^{10-r}.
\displaystyle \text{Probability that the sample of }10\text{ items will include not more than one defective item }=P(X\leq 1).
\displaystyle =P(X=0)+P(X=1).
\displaystyle ={}^{10}C_{0}(0.05)^0(0.95)^{10-0}+{}^{10}C_{1}(0.05)^1(0.95)^{10-1}.
\displaystyle =(0.95)^{10}+10(0.05)(0.95)^9.
\displaystyle =(0.95)^9(0.95+0.5).
\displaystyle =1.45(0.95)^9.
\displaystyle =\frac{29}{20}\left(\frac{19}{20}\right)^9.

\displaystyle \textbf{Question 17: }~\text{The probability that a bulb produced by a factory will fuse after }150\\ \text{ days of use is }0.05.\text{ Find the probability that out of }5\text{ such bulbs (i) none} \\ \text{(ii) not more than one (iii) more than one (iv) at least one will fuse after }150\text{ days of use.}
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)  }
\displaystyle \text{Let }X\text{ be the number of bulbs that fuse after }150\text{ days.}
\displaystyle X\text{ follows a binomial distribution with }n=5.
\displaystyle p=0.05\text{ and }q=0.95.
\displaystyle \text{Or }p=\frac{1}{20}\text{ and }q=\frac{19}{20}.
\displaystyle P(X=r)={}^{5}C_{r}\left(\frac{1}{20}\right)^r\left(\frac{19}{20}\right)^{5-r}.
\displaystyle \text{Probability (none will fuse after }150\text{ days of use})=P(X=0).
\displaystyle ={}^{5}C_{0}\left(\frac{1}{20}\right)^0\left(\frac{19}{20}\right)^{5-0}.
\displaystyle =\left(\frac{19}{20}\right)^5.
\displaystyle \textbf{(ii)  }
\displaystyle \text{Let }X\text{ be the number of bulbs that fuse after }150\text{ days.}
\displaystyle X\text{ follows a binomial distribution with }n=5.
\displaystyle p=0.05\text{ and }q=0.95.
\displaystyle \text{Or }p=\frac{1}{20}\text{ and }q=\frac{19}{20}.
\displaystyle P(X=r)={}^{5}C_{r}\left(\frac{1}{20}\right)^r\left(\frac{19}{20}\right)^{5-r}.
\displaystyle \text{Probability (not more than }1\text{ will fuse after }150\text{ days of use})=P(X\leq 1).
\displaystyle =P(X=0)+P(X=1).
\displaystyle =\left(\frac{19}{20}\right)^5+{}^{5}C_{1}\left(\frac{1}{20}\right)^1\left(\frac{19}{20}\right)^{5-1}.
\displaystyle =\left(\frac{19}{20}\right)^4\left(\frac{1}{20}+\frac{5}{20}\right).
\displaystyle =\frac{6}{5}\left(\frac{19}{20}\right)^4.
\displaystyle \textbf{(iii)  }
\displaystyle \text{Let }X\text{ be the number of bulbs that fuse after }150\text{ days.}
\displaystyle X\text{ follows a binomial distribution with }n=5.
\displaystyle p=0.05\text{ and }q=0.95.
\displaystyle \text{Or }p=\frac{1}{20}\text{ and }q=\frac{19}{20}.
\displaystyle P(X=r)={}^{5}C_{r}\left(\frac{1}{20}\right)^r\left(\frac{19}{20}\right)^{5-r}.
\displaystyle \text{Probability (more than }1\text{ will fuse after }150\text{ days of use})=P(X>1).
\displaystyle =1-P(X\leq 1).
\displaystyle =1-\frac{6}{5}\left(\frac{19}{20}\right)^4.
\displaystyle \textbf{(iv)  }
\displaystyle \text{Let }X\text{ be the number of bulbs that fuse after }150\text{ days.}
\displaystyle X\text{ follows a binomial distribution with }n=5.
\displaystyle p=0.05\text{ and }q=0.95.
\displaystyle \text{Or }p=\frac{1}{20}\text{ and }q=\frac{19}{20}.
\displaystyle P(X=r)={}^{5}C_{r}\left(\frac{1}{20}\right)^r\left(\frac{19}{20}\right)^{5-r}.
\displaystyle \text{Probability (at least one will fuse after }150\text{ days of use})=P(X\geq 1).
\displaystyle =1-P(X=0).
\displaystyle =1-\left(\frac{19}{20}\right)^5.

\displaystyle \textbf{Question 18: }~\text{Suppose that }90\%\text{ of people are right-handed. What is the} \\ \text{probability that at most }6\text{ of a random sample of }10\text{ people are right-handed?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ be the number of people that are right-handed in the sample of }10\text{ people.}
\displaystyle X\text{ follows a binomial distribution with }n=10.
\displaystyle p=0.9,\ q=0.1.
\displaystyle P(X=r)={}^{10}C_{r}(0.9)^r(0.1)^{10-r}.
\displaystyle \text{Probability that at most }6\text{ are right-handed }=P(X\leq 6).
\displaystyle =P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6).
\displaystyle =1-[P(X=7)+P(X=8)+P(X=9)+P(X=10)].
\displaystyle =1-\sum_{r=7}^{10}{}^{10}C_{r}(0.9)^r(0.1)^{10-r}.

\displaystyle \textbf{Question 19: }~\text{A bag contains }7\text{ green, }4\text{ white and }5\text{ red balls. If four balls are} \\ \text{drawn one by one with replacement, what is the probability that one is red?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ denote the number of red balls drawn from }16\text{ balls with replacement.}
\displaystyle X\text{ follows a binomial distribution with }n=4.
\displaystyle p=\frac{5}{16},\ q=1-p=\frac{11}{16}.
\displaystyle P(X=r)={}^{4}C_{r}\left(\frac{5}{16}\right)^r\left(\frac{11}{16}\right)^{4-r}.
\displaystyle P(\text{one ball is red})=P(X=1).
\displaystyle ={}^{4}C_{1}\left(\frac{5}{16}\right)^1\left(\frac{11}{16}\right)^{4-1}.
\displaystyle =4\left(\frac{5}{16}\right)\left(\frac{11}{16}\right)^3.
\displaystyle =\frac{5}{4}\left(\frac{11}{16}\right)^3.

\displaystyle \textbf{Question 20: }~\text{A bag contains }2\text{ white, }3\text{ red and }4\text{ blue balls. Two balls are drawn} \\ \text{at random from the bag. If }X\text{ denotes the number of white balls among the two balls drawn,} \\ \text{describe the probability distribution of }X.
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ denote the number of white balls when }2\text{ balls are drawn from the bag.}
\displaystyle X\text{ follows a distribution with values }0,1\text{ or }2.
\displaystyle P(X=0)=P(\text{all balls non-white})=\frac{{}^{7}C_{2}}{{}^{9}C_{2}}=\frac{42}{72}=\frac{21}{36}.
\displaystyle P(X=1)=P(\text{first ball white and second ball non-white}).
\displaystyle =\frac{{}^{7}C_{2}\,{}^{2}C_{1}}{{}^{9}C_{2}}=\frac{14}{36}.
\displaystyle P(X=2)=P(\text{both balls white})=\frac{{}^{2}C_{2}}{{}^{9}C_{2}}=\frac{1}{36}.
\displaystyle \text{It can be shown in tabular form as follows.}
\displaystyle \begin{array}{c|ccc} X & 0 & 1 & 2\\ \hline P(X) & \frac{21}{36} & \frac{14}{36} & \frac{1}{36} \end{array}

\displaystyle \textbf{Question 21: }~\text{An urn contains }4\text{ white and }3\text{ red balls. Find the probability} \\ \text{distribution of the number of red balls in three draws, with replacement from the urn.}
\displaystyle \text{Answer:}
\displaystyle \text{As three balls are drawn with replacement, the number of white balls, say }X,\text{ follows a} \\ \text{binomial distribution with }n=3.
\displaystyle p=\frac{3}{7}\text{ and }q=\frac{4}{7}.
\displaystyle P(X=r)={}^{3}C_{r}\left(\frac{3}{7}\right)^r\left(\frac{4}{7}\right)^{3-r},\ r=0,1,2,3.
\displaystyle P(X=0)={}^{3}C_{0}\left(\frac{3}{7}\right)^0\left(\frac{4}{7}\right)^{3-0}.
\displaystyle P(X=1)={}^{3}C_{1}\left(\frac{3}{7}\right)^1\left(\frac{4}{7}\right)^{3-1}.
\displaystyle P(X=2)={}^{3}C_{2}\left(\frac{3}{7}\right)^2\left(\frac{4}{7}\right)^{3-2}.
\displaystyle P(X=3)={}^{3}C_{3}\left(\frac{3}{7}\right)^3\left(\frac{4}{7}\right)^{3-3}.
\displaystyle \begin{array}{c|cccc} X & 0 & 1 & 2 & 3\\ \hline P(X) & \frac{64}{343} & \frac{144}{343} & \frac{108}{343} & \frac{27}{343} \end{array}

\displaystyle \textbf{Question 22: }~\text{Find the probability distribution of the number of doublets in }4\text{ throws} \\ \text{of a pair of dice. Also, find the mean and variance of this distribution.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ be the number of doublets in }4\text{ throws of a pair of dice.}
\displaystyle X\text{ follows a binomial distribution with }n=4.
\displaystyle p=\text{number of getting }(1,1),(2,2),\ldots,(6,6)=\frac{6}{36}=\frac{1}{6}.
\displaystyle q=1-p=\frac{5}{6}.
\displaystyle P(X=r)={}^{4}C_{r}\left(\frac{1}{6}\right)^r\left(\frac{5}{6}\right)^{4-r},\ r=0,1,2,3,4.
\displaystyle P(X=0)={}^{4}C_{0}\left(\frac{1}{6}\right)^0\left(\frac{5}{6}\right)^{4-0}.
\displaystyle P(X=1)={}^{4}C_{1}\left(\frac{1}{6}\right)^1\left(\frac{5}{6}\right)^{4-1}.
\displaystyle P(X=2)={}^{4}C_{2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^{4-2}.
\displaystyle P(X=3)={}^{4}C_{3}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^{4-3}.
\displaystyle P(X=4)={}^{4}C_{4}\left(\frac{1}{6}\right)^4\left(\frac{5}{6}\right)^{4-4}.
\displaystyle \text{The distribution is as follows.}
\displaystyle \begin{array}{c|ccccc} X & 0 & 1 & 2 & 3 & 4\\ \hline P(X) & \frac{625}{1296} & \frac{500}{1296} & \frac{150}{1296} & \frac{20}{1296} & \frac{1}{1296} \end{array}

\displaystyle \textbf{Question 23: }~\text{Find the probability distribution of the number of sixes in three tosses} \\ \text{of a die.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ be the number of }6\text{ in }3\text{ tosses of a die.}
\displaystyle \text{Then }X\text{ follows a binomial distribution with }n=3.
\displaystyle p=\frac{1}{6},\ q=1-p=\frac{5}{6}.
\displaystyle P(X=r)={}^{3}C_{r}\left(\frac{1}{6}\right)^r\left(\frac{5}{6}\right)^{3-r},\ r=0,1,2,3.
\displaystyle P(X=0)={}^{3}C_{0}\left(\frac{1}{6}\right)^0\left(\frac{5}{6}\right)^{3-0}.
\displaystyle P(X=1)={}^{3}C_{1}\left(\frac{1}{6}\right)^1\left(\frac{5}{6}\right)^{3-1}.
\displaystyle P(X=2)={}^{3}C_{2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^{3-2}.
\displaystyle P(X=3)={}^{3}C_{3}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^{3-3}.
\displaystyle \text{Hence, the distribution of }X\text{ is as follows.}
\displaystyle \begin{array}{c|cccc} X & 0 & 1 & 2 & 3\\ \hline P(X) & \frac{125}{216} & \frac{75}{216} & \frac{15}{216} & \frac{1}{216} \end{array}

\displaystyle \textbf{Question 24: }~\text{A coin is tossed }5\text{ times. If }X\text{ is the number of heads observed,} \\ \text{find the probability distribution of }X.
\displaystyle \text{Answer:}
\displaystyle \text{Let }X=\text{number of heads in }5\text{ tosses. Then the binomial distribution for }X\text{ has }n=5.
\displaystyle p=\frac{1}{2}\text{ and }q=\frac{1}{2}.
\displaystyle P(X=r)={}^{5}C_{r}\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{5-r},\ r=0,1,2,3,4,5.
\displaystyle =\frac{{}^{5}C_{r}}{2^5}.
\displaystyle \text{Substituting }r=0,1,2,3,4,5\text{ we get the following probability distribution.}
\displaystyle \begin{array}{c|cccccc} X & 0 & 1 & 2 & 3 & 4 & 5\\ \hline P(X) & \frac{1}{32} & \frac{5}{32} & \frac{10}{32} & \frac{10}{32} & \frac{5}{32} & \frac{1}{32} \end{array}

\displaystyle \textbf{Question 25: }~\text{An unbiased die is thrown twice. A success is getting a number greater} \\ \text{than }4.\text{ Find the probability distribution of the number of successes.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ denote getting a number greater than }4.
\displaystyle \text{Then }X\text{ follows a binomial distribution with }n=2.
\displaystyle p=P(X>4)=P(X=5\text{ or }6).
\displaystyle =\frac{1}{6}+\frac{1}{6}.
\displaystyle =\frac{1}{3}.
\displaystyle q=1-p=\frac{2}{3}.
\displaystyle P(X=r)={}^{2}C_{r}\left(\frac{1}{3}\right)^r\left(\frac{2}{3}\right)^{2-r},\ r=0,1,2.
\displaystyle \text{Substituting for }r\text{ we get the probability distribution of }X\text{ as follows.}
\displaystyle \begin{array}{c|ccc} X & 0 & 1 & 2\\ \hline P(X) & \frac{4}{9} & \frac{4}{9} & \frac{1}{9} \end{array}

\displaystyle \textbf{Question 26: }~\text{A man wins a rupee for head and loses a rupee for tail when a coin is} \\ \text{tossed. Suppose that he tosses once and quits if he wins but tries once more if he} \\ \text{loses on the first toss. Find the probability distribution of the number of rupees} \\ \text{the man wins.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ be the number of rupees the man wins.}
\displaystyle \text{First, let us assume that he gets head in the first toss.}
\displaystyle \text{Probability would be }\frac{1}{2}.\text{ Also, he wins Rs }1.
\displaystyle \text{The second possibility is that he gets a tail in the first toss.}
\displaystyle \text{Then he tosses again.}
\displaystyle \text{Suppose he gets a head in the second toss.}
\displaystyle \text{Then he wins Rs }1\text{ in the second toss but loses Rs }1\text{ in the first toss.}
\displaystyle \text{So, the money won }=\text{Rs }0.
\displaystyle \text{Probability for winning Rs }0=\text{getting tail in first toss}\times\text{getting head in second toss.}
\displaystyle =\frac{1}{2}\times\frac{1}{2}.
\displaystyle =\frac{1}{4}.
\displaystyle \text{The third possibility is getting tail in the first toss and also tail in the second toss.}
\displaystyle \text{Then the money that he would win }=-2\text{ (as he loses Rs }2).
\displaystyle \text{Probability for the third possibility }=\frac{1}{2}\times\frac{1}{2}.
\displaystyle =\frac{1}{4}.
\displaystyle \text{Probability distribution of }X\text{ is as follows.}
\displaystyle \begin{array}{c|ccc} X & 1 & 0 & -2\\ \hline P(X) & \frac{1}{2} & \frac{1}{4} & \frac{1}{4} \end{array}

\displaystyle \textbf{Question 27: }~\text{Five dice are thrown simultaneously. If the occurrence of }3,4\text{ or } \\ 5\text{ in a single die is considered a success, find the probability of at least }3\text{ successes.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ denote the occurrence of }3,4\text{ or }5\text{ in a single die. Then }X\text{ follows a binomial} \\ \text{distribution with }n=5.
\displaystyle \text{Let }p=\text{probability of getting }3,4\text{ or }5\text{ in a single die.}
\displaystyle p=\frac{3}{6}=\frac{1}{2}.
\displaystyle q=1-\frac{1}{2}=\frac{1}{2}.
\displaystyle P(X=r)={}^{5}C_{r}\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{5-r}.
\displaystyle P(\text{at least }3\text{ successes})=P(X\geq 3).
\displaystyle =P(X=3)+P(X=4)+P(X=5).
\displaystyle ={}^{5}C_{3}\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^{5-3}+{}^{5}C_{4}\left(\frac{1}{2}\right)^4\left(\frac{1}{2}\right)^{5-4}+{}^{5}C_{5}\left(\frac{1}{2}\right)^5\left(\frac{1}{2}\right)^{5-5}.
\displaystyle =\frac{{}^{5}C_{3}+{}^{5}C_{4}+{}^{5}C_{5}}{2^5}.
\displaystyle =\frac{1}{2}.

\displaystyle \textbf{Question 28: }~\text{The items produced by a company contain }10\%\text{ defective items.} \\ \text{Show that the probability of getting }2\text{ defective items in a sample of }8\text{ items is }\frac{28\times9^{6}}{10^{8}}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ denote the number of defective items in the items produced by the company.}
\displaystyle \text{Then }X\text{ follows a binomial distribution with }n=8.
\displaystyle p=10\%=\frac{1}{10}.
\displaystyle q=1-p=\frac{9}{10}.
\displaystyle \text{Hence, the distribution is given by}
\displaystyle P(X=r)={}^{8}C_{r}\left(\frac{1}{10}\right)^r\left(\frac{9}{10}\right)^{8-r}.
\displaystyle \text{Prob of getting }2\text{ defective items }=P(X=2).
\displaystyle ={}^{8}C_{2}\left(\frac{1}{10}\right)^2\left(\frac{9}{10}\right)^{8-2}.
\displaystyle =\frac{28\times 9^6}{10^8}.

\displaystyle \textbf{Question 29: }~\text{A card is drawn and replaced in an ordinary pack of }52\text{ cards. How} \\ \text{many times must a card be drawn so that (i) there is at least an even chance of drawing} \\ \text{a heart, (ii) the probability of drawing a heart is greater than }\frac34\text{ ?}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }p\text{ denote the probability of drawing a heart from a deck of }52\text{ cards.}
\displaystyle p=\frac{13}{52}=\frac{1}{4}.
\displaystyle q=1-p=\frac{3}{4}.
\displaystyle \text{Let the card be drawn }n\text{ times. So, the binomial distribution is given by }P(X=r)={}^{n}C_{r}p^rq^{\,n-r}.
\displaystyle \text{Let }X\text{ denote the number of hearts drawn from a pack of }52\text{ cards.}
\displaystyle \text{We have to find the smallest value of }n\text{ for which }P(X=0)<\frac{1}{4}.
\displaystyle {}^{n}C_{0}\left(\frac{1}{4}\right)^0\left(\frac{3}{4}\right)^{n-0}<\frac{1}{4}.
\displaystyle \left(\frac{3}{4}\right)^n<\frac{1}{4}.
\displaystyle \text{Put }n=1,\ \left(\frac{3}{4}\right)^1\text{ not less than }\frac{1}{4}.
\displaystyle n=2,\ \left(\frac{3}{4}\right)^2\text{ not less than }\frac{1}{4}.
\displaystyle n=3,\ \left(\frac{3}{4}\right)^3\text{ not less than }\frac{1}{4}.
\displaystyle \text{So, the smallest value of }n=3.
\displaystyle \text{Therefore, the card must be drawn three times.}
\displaystyle \text{(ii) Given the probability of drawing a heart }>\frac{3}{4}.
\displaystyle 1-P(X=0)>\frac{3}{4}.
\displaystyle 1-{}^{n}C_{0}\left(\frac{1}{4}\right)^0\left(\frac{3}{4}\right)^{n-0}>\frac{3}{4}.
\displaystyle 1-\left(\frac{3}{4}\right)^n>\frac{3}{4}.
\displaystyle \frac{1}{4}>\left(\frac{3}{4}\right)^n.
\displaystyle \text{For }n=1,\ \left(\frac{3}{4}\right)^1\text{ not less than }\frac{1}{4}.
\displaystyle n=2,\ \left(\frac{3}{4}\right)^2\text{ not less than }\frac{1}{4}.
\displaystyle n=3,\ \left(\frac{3}{4}\right)^3\text{ not less than }\frac{1}{4}.
\displaystyle n=4,\ \left(\frac{3}{4}\right)^4\text{ not less than }\frac{1}{4}.
\displaystyle n=5,\ \left(\frac{3}{4}\right)^5<\frac{1}{4}.
\displaystyle \text{So, the card must be drawn }5\text{ times.}

\displaystyle \textbf{Question 30: }~\text{The mathematics department has }8\text{ graduate assistants who are} \\ \text{assigned to the same office. Each assistant is just as likely to study at home as in the} \\ \text{office. How many desks must there be in the office so that each assistant has a desk} \\ \text{at least } 90%\text{ of the time?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }k\text{ be the number of desks and }X\text{ be the number of graduate assistants in the office.}
\displaystyle \text{Therefore, }X=8,\ p=\frac{1}{2},\ q=\frac{1}{2}.
\displaystyle \text{According to the given condition,}
\displaystyle P(X\leq k)>90\%.
\displaystyle \Rightarrow P(X\leq k)>0.90.
\displaystyle \Rightarrow P(X>k)<0.10.
\displaystyle \Rightarrow P(X=k+1,k+2,\ldots,8)<0.10.
\displaystyle \text{Therefore, }P(X>6)=P(X=7\text{ or }X=8).
\displaystyle ={}^{8}C_{7}\left(\frac{1}{2}\right)^8+{}^{8}C_{8}\left(\frac{1}{2}\right)^8.
\displaystyle =0.04.
\displaystyle \text{Now, }P(X>5)=P(X=6,X=7\text{ or }X=8)=0.15.
\displaystyle P(X>6)<0.10.
\displaystyle \text{So, if there are }6\text{ desks then there is at least }90\%\text{ chance for every graduate to get a desk.}

\displaystyle \textbf{Question 31: }~\text{An unbiased coin is tossed }8\text{ times. Find, by using binomial distribution,} \\ \text{the probability of getting at least }6\text{ heads.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ be the number of heads in tossing the coin }8\text{ times.}
\displaystyle X\text{ follows a binomial distribution with }n=8.
\displaystyle p=\frac{1}{2}\text{ and }q=\frac{1}{2}.
\displaystyle \text{Hence, the distribution is given by}
\displaystyle P(X=r)={}^{8}C_{r}\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{8-r},\ r=0,1,2,3,4,5,6,7,8.
\displaystyle \text{Required probability }=P(X\geq 6).
\displaystyle =P(X=6)+P(X=7)+P(X=8).
\displaystyle =\frac{{}^{8}C_{6}+{}^{8}C_{7}+{}^{8}C_{8}}{2^8}.
\displaystyle =\frac{28+8+1}{256}.
\displaystyle =\frac{37}{256}.

\displaystyle \textbf{Question 32: }~\text{Six coins are tossed simultaneously. Find the probability of getting} \\ \text{(i) }3\text{ heads (ii) no heads (iii) at least one head.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ denote the number of heads obtained in tossing }6\text{ coins.}
\displaystyle \text{Then }X\text{ follows a binomial distribution with }n=6.
\displaystyle p=\frac{1}{2}\text{ and }q=\frac{1}{2}.
\displaystyle \text{Hence, the distribution is given by}
\displaystyle P(X=r)={}^{6}C_{r}\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{6-r},\ r=0,1,2,3,4,5,6.
\displaystyle =\frac{{}^{6}C_{r}}{2^6}.
\displaystyle \text{(i) }P(\text{getting }3\text{ heads})=P(X=3).
\displaystyle =\frac{{}^{6}C_{3}}{2^6}.
\displaystyle =\frac{20}{64}.
\displaystyle =\frac{5}{16}.
\displaystyle \text{(ii) }P(\text{getting no head})=P(X=0).
\displaystyle =\frac{{}^{6}C_{0}}{2^6}.
\displaystyle =\left(\frac{1}{2}\right)^6.
\displaystyle =\frac{1}{64}.
\displaystyle \text{(iii) }P(\text{getting at least }1\text{ head})=P(X\geq 1).
\displaystyle =1-P(X=0).
\displaystyle =1-\frac{1}{64}.
\displaystyle =\frac{63}{64}.

\displaystyle \textbf{Question 33: }~\text{Suppose that a radio tube inserted into a certain type of set has probability }\\ 0.2\text{ of functioning more than }500\text{ hours. If we test }4\text{ tubes at random what is the probability} \\ \text{that exactly three of these tubes function for more than }500\text{ hours?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ denote the number of tubes that function for more than }500\text{ hours.}
\displaystyle \text{Then }X\text{ follows a binomial distribution with }n=4.
\displaystyle \text{Let }p\text{ be the probability that the tubes function more than }500\text{ hours.}
\displaystyle \text{Here, }p=0.2,\ q=0.8.
\displaystyle \text{Hence, the distribution is given by}
\displaystyle P(X=r)={}^{4}C_{r}(0.2)^r(0.8)^{4-r},\ r=0,1,2,3,4.
\displaystyle \text{Therefore, required probability }=P(X=3).
\displaystyle ={}^{4}C_{3}(0.2)^3(0.8).
\displaystyle =0.0256.

\displaystyle \textbf{Question 34: }~\text{The probability that a certain kind of component will survive a given shock} \\ \text{test is }\frac34.\text{ Find the probability that among }5\text{ components tested (i) exactly }2\text{ will survive} \\ \text{(ii) at most }3\text{ will survive.}
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)  }
\displaystyle \text{Let }X\text{ denote the number of components that survive shock.}
\displaystyle \text{Then }X\text{ follows a binomial distribution with }n=5.
\displaystyle \text{Let }p\text{ be the probability that a certain kind of component will survive a given shock test.}
\displaystyle p=\frac{3}{4}\text{ and }q=\frac{1}{4}.
\displaystyle \text{Hence, the distribution is given by}
\displaystyle P(X=r)={}^{5}C_{r}\left(\frac{3}{4}\right)^r\left(\frac{1}{4}\right)^{5-r},\ r=0,1,2,3,4,5.
\displaystyle P(\text{exactly }2\text{ will survive})=P(X=2).
\displaystyle ={}^{5}C_{2}\left(\frac{3}{4}\right)^2\left(\frac{1}{4}\right)^{5-2}.
\displaystyle =\frac{10\times 9}{1024}.
\displaystyle =0.0879.
\displaystyle \textbf{(ii)  }
\displaystyle \text{Let }X\text{ denote the number of components that survive shock.}
\displaystyle \text{Then }X\text{ follows a binomial distribution with }n=5.
\displaystyle \text{Let }p\text{ be the probability that a certain kind of component will survive a given shock test.}
\displaystyle p=\frac{3}{4}\text{ and }q=\frac{1}{4}.
\displaystyle \text{Hence, the distribution is given by}
\displaystyle P(X=r)={}^{5}C_{r}\left(\frac{3}{4}\right)^r\left(\frac{1}{4}\right)^{5-r},\ r=0,1,2,3,4,5.
\displaystyle P(\text{at most }3\text{ will survive})=P(X\leq 3).
\displaystyle =P(X=0)+P(X=1)+P(X=2)+P(X=3).
\displaystyle ={}^{5}C_{0}\left(\frac{3}{4}\right)^0\left(\frac{1}{4}\right)^{5-0}+{}^{5}C_{1}\left(\frac{3}{4}\right)^1\left(\frac{1}{4}\right)^{5-1}+{}^{5}C_{2}\left(\frac{3}{4}\right)^2\left(\frac{1}{4}\right)^{5-2}+{}^{5}C_{3}\left(\frac{3}{4}\right)^3\left(\frac{1}{4}\right)^{5-3}.
\displaystyle =\left(\frac{1}{4}\right)^5+5\left(\frac{3}{4}\right)\left(\frac{1}{4}\right)^4+10\left(\frac{3}{4}\right)^2\left(\frac{1}{4}\right)^3+10\left(\frac{3}{4}\right)^3\left(\frac{1}{4}\right)^2.
\displaystyle =\frac{1+15+90+270}{1024}.
\displaystyle =\frac{376}{1024}.
\displaystyle =0.3672.

\displaystyle \textbf{Question 35: }~\text{Assume that the probability that a bomb dropped from an aeroplane} \\ \text{will strike a certain target is }0.2.\text{ If }6\text{ bombs are dropped find the probability that (i) exactly } \\ 2\text{ will strike the target (ii) at least }2\text{ will strike the target.}
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)  }
\displaystyle \text{Let }X\text{ be the number of bombs that hit the target.}
\displaystyle \text{Then }X\text{ follows a binomial distribution with }n=6.
\displaystyle \text{Let }p\text{ be the probability that a bomb dropped from an aeroplane will strike the target.}
\displaystyle p=0.2\text{ and }q=0.8.
\displaystyle \text{Hence, the distribution is given by}
\displaystyle P(X=r)={}^{6}C_{r}(0.2)^r(0.8)^{6-r}.
\displaystyle P(\text{exactly }2\text{ will strike the target})=P(X=2).
\displaystyle ={}^{6}C_{2}(0.2)^2(0.8)^4.
\displaystyle =15(0.04)(0.4096).
\displaystyle =0.2458.
\displaystyle \textbf{(ii)  }
\displaystyle \text{Let }X\text{ be the number of bombs that hit the target.}
\displaystyle \text{Then }X\text{ follows a binomial distribution with }n=6.
\displaystyle \text{Let }p\text{ be the probability that a bomb dropped from an aeroplane will strike the target.}
\displaystyle p=0.2\text{ and }q=0.8.
\displaystyle \text{Hence, the distribution is given by}
\displaystyle P(X=r)={}^{6}C_{r}(0.2)^r(0.8)^{6-r}.
\displaystyle P(\text{at least }2\text{ will strike the target})=P(X\geq 2).
\displaystyle =1-[P(X=0)+P(X=1)].
\displaystyle =1-(0.8)^6-6(0.2)(0.8)^5.
\displaystyle =1-0.2621-0.3932.
\displaystyle =0.3447.

\displaystyle \textbf{Question 36: }~\text{It is known that }60\%\text{ of mice inoculated with a serum are protected from} \\ \text{a certain disease. If }5\text{ mice are inoculated, find the probability that (i) none contract} \\ \text{the disease (ii) more than }3\text{ contract the disease.}
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)  }
\displaystyle \text{Let }X\text{ be the number of mice that contract the disease.}
\displaystyle \text{Then }X\text{ follows a binomial distribution with }n=5.
\displaystyle \text{Let }p\text{ be the probability of mice that contract the disease.}
\displaystyle p=0.4\text{ and }q=0.6.
\displaystyle \text{Hence, the distribution is given by}
\displaystyle P(X=r)={}^{5}C_{r}(0.4)^r(0.6)^{5-r},\ r=0,1,2,3,4,5.
\displaystyle P(X=0)={}^{5}C_{0}(0.4)^0(0.6)^{5-0}.
\displaystyle =(0.6)^5.
\displaystyle =0.0778.
\displaystyle \textbf{(ii)  }
\displaystyle \text{Let }X\text{ be the number of mice that contract the disease.}
\displaystyle \text{Then }X\text{ follows a binomial distribution with }n=5.
\displaystyle \text{Let }p\text{ be the probability of mice that contract the disease.}
\displaystyle p=0.4\text{ and }q=0.6.
\displaystyle \text{Hence, the distribution is given by}
\displaystyle P(X=r)={}^{5}C_{r}(0.4)^r(0.6)^{5-r},\ r=0,1,2,3,4,5.
\displaystyle P(X>3)=P(X=4)+P(X=5).
\displaystyle ={}^{5}C_{4}(0.4)^4(0.6)^{5-4}+{}^{5}C_{5}(0.4)^5(0.6)^{5-5}.
\displaystyle =0.0768+0.01024.
\displaystyle =0.08704.

\displaystyle \textbf{Question 37: }~\text{An experiment succeeds twice as often as it fails. Find the probability} \\ \text{that in the next }6\text{ trials there will be at least }4\text{ successes.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ denote the number of successes in }6\text{ trials.}
\displaystyle \text{It is given that successes are twice the failures.}
\displaystyle \Rightarrow p=2q.
\displaystyle p+q=1.
\displaystyle \Rightarrow 3q=1.
\displaystyle \Rightarrow q=\frac{1}{3}.
\displaystyle \therefore p=1-\frac{1}{3}=\frac{2}{3}.
\displaystyle n=6.
\displaystyle \text{Hence, the distribution is given by}
\displaystyle P(X=r)={}^{6}C_{r}\left(\frac{2}{3}\right)^r\left(\frac{1}{3}\right)^{6-r},\ r=0,1,2,\ldots,6.
\displaystyle P(\text{at least }4\text{ successes})=P(X\geq 4).
\displaystyle =P(X=4)+P(X=5)+P(X=6).
\displaystyle ={}^{6}C_{4}\left(\frac{2}{3}\right)^4\left(\frac{1}{3}\right)^{6-4}+{}^{6}C_{5}\left(\frac{2}{3}\right)^5\left(\frac{1}{3}\right)^{6-5}+{}^{6}C_{6}\left(\frac{2}{3}\right)^6\left(\frac{1}{3}\right)^{6-6}.
\displaystyle =\frac{15(2^4)+6(2^5)+2^6}{3^6}.
\displaystyle =\frac{240+192+64}{729}.
\displaystyle =\frac{496}{729}.

\displaystyle \textbf{Question 38: }~\text{In a hospital, there are }20\text{ kidney dialysis machines and the chance of any} \\ \text{one of them to be out of service during a day is }0.02.\text{ Determine the probability that exactly } \\ 3\text{ machines will be out of service on the same day.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ denote the number of machines out of service during a day.}
\displaystyle \text{Then }X\text{ follows a binomial distribution with }n=20.
\displaystyle \text{Let }p\text{ be the probability of any machine being out of service during a day.}
\displaystyle p=0.02\text{ and }q=0.98.
\displaystyle \text{Hence, the distribution is given by}
\displaystyle P(X=r)={}^{20}C_{r}(0.02)^r(0.98)^{20-r},\ r=0,1,2,\ldots,20.
\displaystyle P(\text{exactly }3\text{ machines will be out of service on the same day})=P(X=3).
\displaystyle ={}^{20}C_{3}(0.02)^3(0.98)^{20-3}.
\displaystyle =1140(0.000008)(0.7093).
\displaystyle =0.006469.

\displaystyle \textbf{Question 39: }~\text{The probability that a student entering a university will graduate is } \\ 0.4.\text{ Find the probability that out of }3\text{ students (i) none will graduate (ii) only one will} \\ \text{graduate (iii) all will graduate.}
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)  }
\displaystyle \text{Let }X\text{ be the number of students that graduate from among }3\text{ students.}
\displaystyle \text{Let }p=\text{probability that a student entering a university will graduate.}
\displaystyle \text{Here, }n=3,\ p=0.4\text{ and }q=0.6.
\displaystyle \text{Hence, the distribution is given by}
\displaystyle P(X=r)={}^{3}C_{r}(0.4)^r(0.6)^{3-r},\ r=0,1,2,3.
\displaystyle P(X=0)=q^3=0.216.
\displaystyle \textbf{(ii)  }
\displaystyle \text{Let }X\text{ be the number of students that graduate from among }3\text{ students.}
\displaystyle \text{Let }p=\text{probability that a student entering a university will graduate.}
\displaystyle \text{Here, }n=3,\ p=0.4\text{ and }q=0.6.
\displaystyle \text{Hence, the distribution is given by}
\displaystyle P(X=r)={}^{3}C_{r}(0.4)^r(0.6)^{3-r},\ r=0,1,2,3.
\displaystyle P(X=1)=3(0.4)(0.36)=0.432.
\displaystyle \textbf{(iii)  }
\displaystyle \text{Let }X\text{ be the number of students that graduate from among }3\text{ students.}
\displaystyle \text{Let }p=\text{probability that a student entering a university will graduate.}
\displaystyle \text{Here, }n=3,\ p=0.4\text{ and }q=0.6.
\displaystyle \text{Hence, the distribution is given by}
\displaystyle P(X=r)={}^{3}C_{r}(0.4)^r(0.6)^{3-r},\ r=0,1,2,3.
\displaystyle P(X=3)=p^3=0.064.

\displaystyle \textbf{Question 40: }~\text{Ten eggs are drawn successively, with replacement, from a lot containing} \\ 10\%\text{ defective eggs. Find the probability that there is at least one defective egg.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ be the number of defective eggs drawn from }10\text{ eggs.}
\displaystyle \text{Then }X\text{ follows a binomial distribution with }n=10.
\displaystyle \text{Let }p\text{ be the probability that a drawn egg is defective.}
\displaystyle p=10\%=\frac{1}{10},\ q=\frac{9}{10}.
\displaystyle \text{Hence, the distribution is given by}
\displaystyle P(X=x)={}^{10}C_{x}p^xq^{10-x}.
\displaystyle P(\text{there is at least one defective egg})=P(X\geq 1).
\displaystyle =1-P(X=0).
\displaystyle =1-{}^{10}C_{0}\left(\frac{1}{10}\right)^0\left(\frac{9}{10}\right)^{10-0}.
\displaystyle =1-\left(\frac{9}{10}\right)^{10}.
\displaystyle =1-\frac{9^{10}}{10^{10}}.

\displaystyle \textbf{Question 41: }~\text{In a }20\text{-question true-false examination suppose a student tosses a fair} \\ \text{coin to determine his answer to each question. If the coin falls heads, he answers true;} \\ \text{if it falls tails, he answers false. Find the probability that he answers at least } \\ 12\text{ questions correctly.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ denote the number of correct answers.}
\displaystyle \text{Then }X\text{ follows a binomial distribution with }n=20.
\displaystyle \text{Let }p\text{ be the probability of a correct answer.}
\displaystyle \Rightarrow p=\text{getting a head and a right answer to be true or getting a tail and a right answer to be false.}
\displaystyle \Rightarrow p=\frac{1}{2}.
\displaystyle \therefore q=1-\frac{1}{2}=\frac{1}{2}.
\displaystyle \text{Hence, the distribution is given by}
\displaystyle P(X=r)={}^{20}C_{r}\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{20-r},\ r=0,1,2,3,\ldots,20.
\displaystyle =\frac{{}^{20}C_{r}}{2^{20}}.
\displaystyle \text{Probability that the student answers at least }12\text{ questions correctly }=P(X\geq 12).
\displaystyle =P(X=12)+P(X=13)+\cdots+P(X=20).
\displaystyle =\frac{{}^{20}C_{12}+{}^{20}C_{13}+\cdots+{}^{20}C_{20}}{2^{20}}.

\displaystyle \textbf{Question 42: }~\text{Suppose }X\text{ has a binomial distribution with }n=6\text{ and } p=\frac12.\text{ Show } \text{that }X=3\text{ is the most likely outcome.}
\displaystyle \text{Answer:}
\displaystyle \text{We have }n=6\text{ and }p=\frac{1}{2}.
\displaystyle \therefore q=1-p=\frac{1}{2}.
\displaystyle \text{Hence, the distribution is given by}
\displaystyle P(X=r)={}^{6}C_{r}\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{6-r},\ r=0,1,2,3,4,5,6.
\displaystyle ={}^{6}C_{r}\left(\frac{1}{2}\right)^6.
\displaystyle P(X=r)=\frac{{}^{6}C_{r}}{2^6}.
\displaystyle \text{By substituting }r=0,1,2,3,4,5\text{ and }6,\text{ we get the following distribution for }X.
\displaystyle \begin{array}{c|ccccccc} X & 0 & 1 & 2 & 3 & 4 & 5 & 6\\ \hline P(X) & \frac{1}{64} & \frac{6}{64} & \frac{15}{64} & \frac{20}{64} & \frac{15}{64} & \frac{6}{64} & \frac{1}{64} \end{array}
\displaystyle \text{Comparing the probabilities, we get that }X=3\text{ is the most likely outcome.}

\displaystyle \textbf{Question 43: }~\text{In a multiple choice examination with three possible answers for each} \\ \text{of the five questions out of which only one is correct what is the probability that} \\ \text{a candidate would get four or more correct answers just by guessing?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ be the number of right answers in the }5\text{ questions.}
\displaystyle X\text{ can take values }0,1,2,\ldots,5.
\displaystyle X\text{ follows a binomial distribution with }n=5.
\displaystyle p=\text{probability of guessing right answer}=\frac{1}{3}.
\displaystyle q=\text{probability of guessing wrong answer}=\frac{2}{3}.
\displaystyle \text{Hence, the distribution is given by}
\displaystyle P(X=r)={}^{5}C_{r}\left(\frac{1}{3}\right)^r\left(\frac{2}{3}\right)^{5-r},\ r=0,1,2,\ldots,5.
\displaystyle P(\text{the student guesses }4\text{ or more correct answers})=P(X\geq 4).
\displaystyle =P(X=4)+P(X=5).
\displaystyle ={}^{5}C_{4}\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)^1+{}^{5}C_{5}\left(\frac{1}{3}\right)^5\left(\frac{2}{3}\right)^0.
\displaystyle =\frac{10+1}{3^5}.
\displaystyle =\frac{11}{243}.

\displaystyle \textbf{Question 44: }~\text{A person buys a lottery ticket in }50\text{ lotteries, in each of which his chance} \\ \text{of winning a prize is }\frac1{100}.\text{ What is the probability that he will win a prize (i) at least once} \\ \text{(ii) exactly once (iii) at least twice?}
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)  }
\displaystyle \text{Let }X=\text{number of winning prizes.}
\displaystyle p=\text{probability of winning a prize.}
\displaystyle p=\frac{1}{100}.
\displaystyle q=1-p=1-\frac{1}{100}=\frac{99}{100}.
\displaystyle \text{Given }n=50.
\displaystyle X\sim B\left(50,\frac{1}{100}\right).
\displaystyle \text{The p.m.f. of }X\text{ is given by }P(X=x)={}^{n}C_x p^x q^{\,n-x}.
\displaystyle p(x)={}^{50}C_x\left(\frac{1}{100}\right)^x\left(\frac{99}{100}\right)^{50-x},\ x=0,1,2,\ldots,50.
\displaystyle P(\text{a person wins a prize at least once})=P(X\geq 1).
\displaystyle =1-P(X<1)=1-P(X=0).
\displaystyle =1-{}^{50}C_0\left(\frac{1}{100}\right)^0\left(\frac{99}{100}\right)^{50-0}.
\displaystyle =1-1\times1\times\left(\frac{99}{100}\right)^{50}.
\displaystyle =1-\left(\frac{99}{100}\right)^{50}.
\displaystyle \text{Hence, probability of winning a prize at least once}=1-\left(\frac{99}{100}\right)^{50}.
\displaystyle \textbf{(ii)  }
\displaystyle \text{Let }X\text{ denote the number of times the person wins the lottery.}
\displaystyle \text{Then, }X\text{ follows a binomial distribution with }n=50.
\displaystyle \text{Let }p\text{ be the probability of winning a prize.}
\displaystyle p=\frac{1}{100},\ q=1-\frac{1}{100}=\frac{99}{100}.
\displaystyle \text{Hence, the distribution is given by}
\displaystyle P(X=r)={}^{50}C_r\left(\frac{1}{100}\right)^r\left(\frac{99}{100}\right)^{50-r},\ r=0,1,2,\ldots,50.
\displaystyle P(\text{winning exactly once})=P(X=1).
\displaystyle ={}^{50}C_1\left(\frac{1}{100}\right)^1\left(\frac{99}{100}\right)^{50-1}.
\displaystyle =\frac{1}{2}\left(\frac{99}{100}\right)^{49}.
\displaystyle \text{Hence, probability of winning a prize exactly once}=\frac{1}{2}\left(\frac{99}{100}\right)^{49}.
\displaystyle \textbf{(iii)  }
\displaystyle \text{Let }X\text{ denote the number of times the person wins the lottery.}
\displaystyle \text{Then, }X\text{ follows a binomial distribution with }n=50.
\displaystyle \text{Let }p\text{ be the probability of winning a prize.}
\displaystyle p=\frac{1}{100},\ q=1-\frac{1}{100}=\frac{99}{100}.
\displaystyle \text{Hence, the distribution is given by}
\displaystyle P(X=r)={}^{50}C_r\left(\frac{1}{100}\right)^r\left(\frac{99}{100}\right)^{50-r},\ r=0,1,2,\ldots,50.
\displaystyle P(\text{winning at least twice})=P(X\ge 2).
\displaystyle =1-P(X=0)-P(X=1).
\displaystyle =1-\left(\frac{99}{100}\right)^{50}-{}^{50}C_1\left(\frac{1}{100}\right)\left(\frac{99}{100}\right)^{49}.
\displaystyle =1-\frac{99^{50}}{100^{50}}-\frac{50\cdot 99^{49}}{100^{50}}.
\displaystyle =1-\frac{99^{49}\times 149}{100^{50}}.
\displaystyle \text{Hence, the probability of winning the prize at least twice}=1-\frac{99^{49}\times 149}{100^{50}}.

\displaystyle \textbf{Question 45: }~\text{The probability of a shooter hitting a target is }\frac34.\text{ How many minimum} \\ \text{number of times must he/she fire so that the probability of hitting the target at least once is} \\ \text{more than }0.99\text{ ?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the shooter fire }n\text{ times and let }X\text{ denote the number of times the shooter hits the target.}
\displaystyle \text{Then, }X\text{ follows binomial distribution with }p=\frac{3}{4}\text{ and }q=\frac{1}{4}\text{ such that}
\displaystyle P(X=r)={}^{n}C_r\left(\frac{3}{4}\right)^r\left(\frac{1}{4}\right)^{\,n-r}.
\displaystyle \Rightarrow P(X=r)={}^{n}C_r\frac{3^r}{4^n}.
\displaystyle \text{It is given that }P(X\ge 1)>0.99.
\displaystyle \Rightarrow 1-P(X=0)>0.99.
\displaystyle \Rightarrow 1-\frac{1}{4^n}>0.99.
\displaystyle \Rightarrow \frac{1}{4^n}<0.01.
\displaystyle \Rightarrow 4^n>\frac{1}{0.01}.
\displaystyle \Rightarrow 4^n>100.
\displaystyle \text{The least value of }n\text{ satisfying this inequality is }4.
\displaystyle \text{Hence, the shooter must fire at least }4\text{ times.}

\displaystyle \textbf{Question 46: }~\text{How many times must a man toss a fair coin so that the probability of having} \\ \text{at least one head is more than }90\%\text{ ?}
\displaystyle \text{Answer:}
\displaystyle \text{Suppose the man tosses a fair coin }n\text{ times and }X\text{ denotes the number of heads in }n\text{ tosses.}
\displaystyle \text{As }p=\frac{1}{2}\text{ and }q=\frac{1}{2},
\displaystyle P(X=r)={}^{n}C_r\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{\,n-r},\ r=0,1,2,3,\ldots,n.
\displaystyle \text{It is given that }P(X\ge 1)>0.9.
\displaystyle \Rightarrow 1-P(X=0)>0.9.
\displaystyle \Rightarrow 1-{}^{n}C_0\left(\frac{1}{2}\right)^n>0.9.
\displaystyle \Rightarrow \left(\frac{1}{2}\right)^n<\frac{1}{10}.
\displaystyle \Rightarrow 2^n>10.
\displaystyle \Rightarrow n=4,5,6,\ldots
\displaystyle \text{Hence, the man must toss the coin at least }4\text{ times.}

\displaystyle \textbf{Question 47: }~\text{How many times must a man toss a fair coin so that the probability of having} \\ \text{at least one head is more than }80\%\text{ ?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ be the number of heads and }n\text{ be the minimum number of times that a man} \\ \text{must toss a fair coin so that probability of }X\ge 1\text{ is more than }80\%\text{ and }X\text{ follows} \\ \text{a binomial distribution with }p=\frac{1}{2},\ q=\frac{1}{2}.
\displaystyle P(X=r)={}^{n}C_r\left(\frac{1}{2}\right)^n.
\displaystyle \text{We have }P(X\ge 1)=1-P(X=0)=1-{}^{n}C_0\left(\frac{1}{2}\right)^n=1-\frac{1}{2^n}.
\displaystyle \text{and }P(X\ge 1)>80\%.
\displaystyle 1-\frac{1}{2^n}>80\%=0.80.
\displaystyle \frac{1}{2^n}<1-0.80=0.20.
\displaystyle 2^n>\frac{1}{0.2}=5.
\displaystyle \text{We know }2^2<5\text{ while }2^3>5.
\displaystyle \text{So, }n=3.
\displaystyle \text{So, }n\text{ should be at least }3.

\displaystyle \textbf{Question 48: }~\text{A pair of dice is thrown }4\text{ times. If getting a doublet is considered a} \\ \text{success, find the probability distribution of the number of successes.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }p\text{ denote the probability of getting a doublet in a single throw of a pair of dice. Then,}
\displaystyle p=\frac{6}{36}=\frac{1}{6}.
\displaystyle q=1-p=1-\frac{1}{6}=\frac{5}{6}.
\displaystyle \text{Let }X\text{ be the number of doublets in }4\text{ throws of a pair of dice. Then, }X\text{ follows} \\ \text{abinomial distribution with }n=4,
\displaystyle p=\frac{1}{6}\text{ and }q=\frac{5}{6}.
\displaystyle P(X=r)=\text{Probability of getting }r\text{ doublets.}
\displaystyle P(X=r)={}^{4}C_r\left(\frac{1}{6}\right)^r\left(\frac{5}{6}\right)^{4-r},\ r=0,1,2,3,4.
\displaystyle \text{If }X=0,\text{ then }P(X=0)={}^{4}C_0\left(\frac{1}{6}\right)^0\left(\frac{5}{6}\right)^4.
\displaystyle \Rightarrow P=\left(\frac{5}{6}\right)^4.
\displaystyle \text{If }X=1,\text{ then }P={}^{4}C_1\left(\frac{1}{6}\right)^1\left(\frac{5}{6}\right)^3=\frac{2}{3}\left(\frac{5}{6}\right)^3.
\displaystyle \text{If }X=2,\text{ then }P={}^{4}C_2\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^2=\frac{1}{6}\left(\frac{5}{6}\right)^2.
\displaystyle \text{If }X=3,\text{ then }P={}^{4}C_3\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^1=\frac{10}{3}\left(\frac{1}{6}\right)^3.
\displaystyle \text{If }X=4,\text{ then }P={}^{4}C_4\left(\frac{1}{6}\right)^4\left(\frac{5}{6}\right)^0=\left(\frac{1}{6}\right)^4=\frac{1}{1296}.
\displaystyle \begin{array}{c|ccccc} X & 0 & 1 & 2 & 3 & 4\\ \hline P(X) & \left(\frac{5}{6}\right)^4 & \frac{2}{3}\left(\frac{5}{6}\right)^3 & \frac{25}{216} & \frac{5}{324} & \frac{1}{1296} \end{array}

\displaystyle \textbf{Question 49: }~\text{From a lot of }30\text{ bulbs which include }6\text{ defectives, a sample of } 4\text{ bulbs} \\ \text{is drawn at random with replacement. Find the probability distribution of the number} \\ \text{of defective bulbs.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ denote the number of defective bulbs in a sample of }4\text{ bulbs drawn successively} \\ \text{with replacement.}
\displaystyle \text{Then, }X\text{ follows a binomial distribution with parameters }n=4.
\displaystyle p=\frac{6}{30}=\frac{1}{5}\text{ and }q=\frac{4}{5}.
\displaystyle \text{Then, the distribution is given by}
\displaystyle P(X=r)={}^{4}C_r\left(\frac{1}{5}\right)^r\left(\frac{4}{5}\right)^{4-r},\ r=0,1,2,3,4.
\displaystyle P(X=0)=\left(\frac{4}{5}\right)^4.
\displaystyle =\frac{256}{625}.
\displaystyle P(X=1)=4\left(\frac{1}{5}\right)^1\left(\frac{4}{5}\right)^3.
\displaystyle =\frac{256}{625}.
\displaystyle P(X=2)=6\left(\frac{1}{5}\right)^2\left(\frac{4}{5}\right)^2.
\displaystyle =\frac{96}{625}.
\displaystyle P(X=3)=4\left(\frac{1}{5}\right)^3\left(\frac{4}{5}\right)^1.
\displaystyle =\frac{16}{625}.
\displaystyle P(X=4)=\left(\frac{1}{5}\right)^4.
\displaystyle =\frac{1}{625}.
\displaystyle \begin{array}{c|ccccc} X & 0 & 1 & 2 & 3 & 4\\ \hline P(X) & \frac{256}{625} & \frac{256}{625} & \frac{96}{625} & \frac{16}{625} & \frac{1}{625} \end{array}

\displaystyle \textbf{Question 50: }~\text{Find the probability that in }10\text{ throws of a fair die a score which is a} \\ \text{multiple of }3\text{ will be obtained in at least }8\text{ of the throws.}
\displaystyle \text{Answer:}
\displaystyle \text{Let getting a multiple of }3\text{ be a success.}
\displaystyle \text{We have,}
\displaystyle p=\text{probability of getting a success}=\frac{2}{6}=\frac{1}{3}.
\displaystyle \text{So, }q=\text{probability of getting a failure}=1-p=1-\frac{1}{3}=\frac{2}{3}.
\displaystyle \text{Let }X\text{ denote the number of successes in a sample of }10\text{ trials. Then,}
\displaystyle X\text{ follows binomial distribution with parameters }n=10,\ p=\frac{1}{3}\text{ and }q=\frac{2}{3}.
\displaystyle P(X=r)={}^{10}C_r\left(\frac{1}{3}\right)^r\left(\frac{2}{3}\right)^{10-r},\ r=0,1,2,\ldots,10.
\displaystyle \text{Now,}
\displaystyle \text{Required probability}=P(X\ge 8).
\displaystyle =P(X=8)+P(X=9)+P(X=10).
\displaystyle =\frac{{}^{10}C_8\,2^2}{{3}^{10}}+\frac{{}^{10}C_9\,2}{{3}^{10}}+\frac{{}^{10}C_{10}}{{3}^{10}}.
\displaystyle =\frac{45\times 2^2}{3^{10}}+\frac{10\times 2}{3^{10}}+\frac{1}{3^{10}}.
\displaystyle =\frac{180+20+1}{3^{10}}.
\displaystyle =\frac{201}{3^{10}}.

\displaystyle \textbf{Question 51: }~\text{A die is thrown }5\text{ times. Find the probability that an odd number will} \\ \text{come up exactly three times.}
\displaystyle \text{Answer:}
\displaystyle \text{Let getting an odd number be a success in a trial.}
\displaystyle \text{We have,}
\displaystyle p=\text{probability of getting an odd number in a trial}=\frac{3}{6}=\frac{1}{2}.
\displaystyle \text{Also, }q=1-p=1-\frac{1}{2}=\frac{1}{2}.
\displaystyle \text{Let }X\text{ denote the number of successes in a sample of }5\text{ trials. Then,}
\displaystyle X\text{ follows binomial distribution with parameters }n=5\text{ and }p=q=\frac{1}{2}.
\displaystyle P(X=r)={}^{5}C_r p^r q^{5-r}={}^{5}C_r\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{5-r},\ r=0,1,2,3,4,5.
\displaystyle \text{Now,}
\displaystyle \text{Required probability}=P(X=3).
\displaystyle ={}^{5}C_3\left(\frac{1}{2}\right)^5.
\displaystyle =\frac{10}{32}.
\displaystyle =\frac{5}{16}.

\displaystyle \textbf{Question 52: }~\text{The probability of a man hitting a target is }0.25.\text{ He shoots }7\text{ times.} \\ \text{What is the probability of his hitting at least twice?}
\displaystyle \text{Answer:}
\displaystyle \text{Let hitting the target be a success in a shoot.}
\displaystyle \text{We have,}
\displaystyle p=\text{probability of hitting the target}=0.25=\frac{1}{4}.
\displaystyle \text{Also, }q=1-p=1-\frac{1}{4}=\frac{3}{4}.
\displaystyle \text{Let }X\text{ denote the number of successes in a sample of }7\text{ trials. Then,}
\displaystyle X\text{ follows a binomial distribution with parameters }n=7\text{ and }p=\frac{1}{4}.
\displaystyle P(X=r)={}^{7}C_r p^r q^{7-r}={}^{7}C_r\left(\frac{1}{4}\right)^r\left(\frac{3}{4}\right)^{7-r}=\frac{{}^{7}C_r\,3^{7-r}}{4^7},\ r=0,1,2,3,4,5,6,7.
\displaystyle \text{Now,}
\displaystyle \text{Required probability}=P(X\ge 2).
\displaystyle =1-\left[P(X=0)+P(X=1)\right].
\displaystyle =1-\left[\frac{{}^{7}C_0\,3^7}{4^7}+\frac{{}^{7}C_1\,3^6}{4^7}\right].
\displaystyle =1-\left[\frac{2187}{16384}+\frac{5103}{16384}\right].
\displaystyle =1-\frac{7290}{16384}.
\displaystyle =\frac{9094}{16384}.
\displaystyle =\frac{4547}{8192}.

\displaystyle \textbf{Question 53: }~\text{A factory produces bulbs. The probability that one bulb is defective is } \\ \frac1{50}\text{ and they are packed in boxes of }10.\text{ From a single box, find the probability that} \\ \text{(i) none of the bulbs is defective (ii) exactly two bulbs are defective (iii) more than }8\text{ bulbs} \\ \text{work properly.}
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)  }
\displaystyle \text{Let getting a defective bulb from a single box be a success.}
\displaystyle \text{We have}
\displaystyle p=\text{probability of getting a defective bulb}=\frac{1}{50}.
\displaystyle \text{Also, }q=1-p=1-\frac{1}{50}=\frac{49}{50}.
\displaystyle \text{Let }X\text{ denote the number of successes in a sample of }10\text{ trials. Then,}
\displaystyle X\text{ follows binomial distribution with parameters }n=10\text{ and }p=\frac{1}{50}.
\displaystyle P(X=r)={}^{10}C_r p^r q^{10-r}={}^{10}C_r\left(\frac{1}{50}\right)^r\left(\frac{49}{50}\right)^{10-r},\ r=0,1,2,\ldots,10.
\displaystyle \text{Now,}
\displaystyle \text{Required probability}=P(\text{none of the bulbs is defective})=P(X=0).
\displaystyle ={}^{10}C_0\left(\frac{1}{50}\right)^0\left(\frac{49}{50}\right)^{10}.
\displaystyle =\left(\frac{49}{50}\right)^{10}.
\displaystyle \textbf{(ii)  }
\displaystyle \text{Let getting a defective bulb from a single box be a success.}
\displaystyle \text{We have}
\displaystyle p=\text{probability of getting a defective bulb}=\frac{1}{50}.
\displaystyle \text{Also, }q=1-p=1-\frac{1}{50}=\frac{49}{50}.
\displaystyle \text{Let }X\text{ denote the number of successes in a sample of }10\text{ trials. Then,}
\displaystyle X\text{ follows binomial distribution with parameters }n=10\text{ and }p=\frac{1}{50}.
\displaystyle P(X=r)={}^{10}C_r p^r q^{10-r}={}^{10}C_r\left(\frac{1}{50}\right)^r\left(\frac{49}{50}\right)^{10-r}=\frac{{}^{10}C_r\,49^{10-r}}{50^{10}},\ r=0,1,2,3,\ldots,10.
\displaystyle \text{Now,}
\displaystyle \text{Required probability}=P(\text{exactly two bulbs are defective}).
\displaystyle =P(X=2).
\displaystyle =\frac{{}^{10}C_2\,49^8}{50^{10}}.
\displaystyle =\frac{45\times 49^8}{50^{10}}.
\displaystyle \textbf{(iii)  }
\displaystyle \text{Let getting a defective bulb from a single box be a success.}
\displaystyle \text{We have}
\displaystyle p=\text{probability of getting a defective bulb}=\frac{1}{50}.
\displaystyle \text{Also, }q=1-p=1-\frac{1}{50}=\frac{49}{50}.
\displaystyle \text{Let }X\text{ denote the number of successes in a sample of }10\text{ trials. Then,}
\displaystyle X\text{ follows binomial distribution with parameters }n=10\text{ and }p=\frac{1}{50}.
\displaystyle P(X=r)={}^{10}C_r p^r q^{10-r}={}^{10}C_r\left(\frac{1}{50}\right)^r\left(\frac{49}{50}\right)^{10-r}=\frac{{}^{10}C_r\,49^{10-r}}{50^{10}},\ r=0,1,2,3,\ldots,10.
\displaystyle \text{Now,}
\displaystyle \text{Required probability}=P(\text{more than }8\text{ bulbs work properly}).
\displaystyle =P(\text{at most one bulb is defective}).
\displaystyle =P(X\le 1).
\displaystyle =P(X=0)+P(X=1).
\displaystyle =\frac{{}^{10}C_0\,49^{10}}{50^{10}}+\frac{{}^{10}C_1\,49^9}{50^{10}}.
\displaystyle =\frac{49^{10}}{50^{10}}+\frac{10\times 49^9}{50^{10}}.
\displaystyle =\frac{49^9(49+10)}{50^{10}}.
\displaystyle =\frac{59\times 49^9}{50^{10}}.

\displaystyle \textbf{Question 54: }~\text{A box has }20\text{ pens of which }2\text{ are defective. Calculate probability} \\ \text{that out of }5\text{ pens drawn one by one with replacement, at most }2\text{ are defective.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }p\text{ denote the probability of drawing a defective pen. Then,}
\displaystyle p=\frac{2}{20}=\frac{1}{10}.
\displaystyle \Rightarrow q=1-p=1-\frac{1}{10}=\frac{9}{10}.
\displaystyle \text{Let }X\text{ denote the number of defective pens drawn. Then, }X\text{ is a binomial variate with parameters }n=5\text{ and }p=\frac{1}{10}.
\displaystyle \text{Now, }P(X=r)=\text{Probability of drawing }r\text{ defective pens}={}^{5}C_r\left(\frac{1}{10}\right)^r\left(\frac{9}{10}\right)^{5-r},\ r=0,1,2,3,4,5.
\displaystyle \text{Required probability of drawing at most }2\text{ defective pens}=P(X\le 2).
\displaystyle =P(X=0)+P(X=1)+P(X=2).
\displaystyle ={}^{5}C_0\left(\frac{1}{10}\right)^0\left(\frac{9}{10}\right)^5+{}^{5}C_1\left(\frac{1}{10}\right)^1\left(\frac{9}{10}\right)^4+{}^{5}C_2\left(\frac{1}{10}\right)^2\left(\frac{9}{10}\right)^3.
\displaystyle =\left(\frac{9}{10}\right)^3\left(\frac{81}{100}+5\times\frac{9}{100}+10\times\frac{1}{100}\right).
\displaystyle =\frac{729}{1000}\times\frac{136}{100}.
\displaystyle =0.99144.

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