\displaystyle \textbf{Question 1:}\ \ \text{A random variable }X\text{ has the following probability distribution:}
\displaystyle \begin{array}{c|cccccccc} X: & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline P(X): & 0 & k & 2k & 3k & k^{2} & 2k^{2} & 7k^{2} & k \end{array}
\displaystyle \text{Find each of the following:}
\displaystyle \text{(i) }k \qquad \text{(ii) }P(X<6) \qquad \text{(iii) }P(X\ge 6) \qquad \text{(iv) }P(0<X<5)\qquad [\text{CBSE 2011}]
\displaystyle \text{Answer:}
\displaystyle \text{(i) Since the sum of all the probabilities in a probability distribution is always unity.}
\displaystyle \text{Therefore,}
\displaystyle P(X=0)+P(X=1)+\cdots+P(X=7)=1
\displaystyle 0+k+2k+3k+k^{2}+2k^{2}+7k^{2}+k=1
\displaystyle \Rightarrow 10k^{2}+9k-1=0
\displaystyle \Rightarrow (10k-1)(k+1)=0
\displaystyle \Rightarrow k=\frac{1}{10}
\displaystyle [\because\ k\ge 0\ \therefore\ k+1\ne 0]
\displaystyle \text{(ii)}
\displaystyle P(X<6)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)
\displaystyle P(X<6)=0+k+2k+3k+k^{2}
\displaystyle \Rightarrow P(X<6)=k^{2}+8k
\displaystyle \Rightarrow P(X<6)=\left(\frac{1}{10}\right)^{2}+\frac{8}{10}
\displaystyle \Rightarrow P(X<6)=\frac{81}{100}
\displaystyle [\because\ k=\frac{1}{10}]
\displaystyle \text{(iii)}
\displaystyle P(X\ge 6)=P(X=6)+P(X=7)
\displaystyle P(X\ge 6)=2k^{2}+7k^{2}+k
\displaystyle \Rightarrow P(X\ge 6)=9k^{2}+k
\displaystyle \Rightarrow P(X\ge 6)=\frac{9}{100}+\frac{1}{10}
\displaystyle \Rightarrow P(X\ge 6)=\frac{19}{100}
\displaystyle [\because\ k=\frac{1}{10}]

\displaystyle \textbf{Question 2:}\ \ \text{Three cards are drawn from a pack of 52 playing cards. Find the probability} \\ \text{distribution of the number of aces.}\qquad [\text{CBSE 2001}]
\displaystyle \text{Answer:}
\displaystyle  \text{Let }X\text{ denote the number of aces in a sample of 3 cards drawn from a well shuffled pack} \\ \text{of 52 playing cards. Since there are four aces in the pack, therefore in the sample of} \\ \text{3 cards drawn either there can be no ace or there can be one ace or two aces or three aces.} \\ \text{Thus, }X\text{ can take values }0,1,2,\text{ and }3.
\displaystyle \text{Now, }\ P(X=0)=\text{Probability of getting no ace = Probability of getting 3 other cards}
\displaystyle =\frac{{}^{48}C_{3}}{{}^{52}C_{3}}=\frac{4324}{5525}
\displaystyle P(X=1)=\text{Probability of getting one ace and two other cards}=\frac{{}^{4}C_{1}\times{}^{48}C_{2}}{{}^{52}C_{3}}=\frac{1128}{5525}
\displaystyle P(X=2)=\text{Probability of getting two aces and one other card}=\frac{{}^{4}C_{2}\times{}^{48}C_{1}}{{}^{52}C_{3}}=\frac{72}{5525}
\displaystyle \text{and, }\ P(X=3)=\text{Probability of getting 3 aces}=\frac{{}^{4}C_{3}}{{}^{52}C_{3}}=\frac{1}{5525}
\displaystyle \text{Thus, the probability distribution of random variable }X\text{ is given by}
\displaystyle \begin{array}{c|cccc} X: & 0 & 1 & 2 & 3\\ \hline P(X): & \frac{4324}{5525} & \frac{1128}{5525} & \frac{72}{5525} & \frac{1}{5525} \end{array}
\displaystyle \text{It is to note here that the sum of the probabilities is 1 which is the condition for} \\ \text{a distribution to be a probability distribution.}

\displaystyle \textbf{Question 3:}\ \ \text{An urn contains 4 white and 6 red balls. Four balls are drawn at random} \\ \text{from the urn. Find the probability distribution of the number of white balls.}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }X\text{ denote the number of white balls drawn from the urn. Since there are 4} \\ \text{white balls, therefore }X\text{ can take values }0,1,2,3\text{ and }4.
\displaystyle \text{Now, }\ P(X=0)=\text{Probability of getting no white ball = Probability that 4 balls drawn are red}
\displaystyle =\frac{{}^{6}C_{4}}{{}^{10}C_{4}}=\frac{1}{14}
\displaystyle P(X=1)=\text{Probability of getting one white ball}=\frac{{}^{4}C_{1}\times{}^{6}C_{3}}{{}^{10}C_{4}}=\frac{8}{21}
\displaystyle P(X=2)=\text{Probability of getting two white balls}=\frac{{}^{4}C_{2}\times{}^{6}C_{2}}{{}^{10}C_{4}}=\frac{6}{14}
\displaystyle P(X=3)=\text{Probability of getting three white balls}=\frac{{}^{4}C_{3}\times{}^{6}C_{1}}{{}^{10}C_{4}}=\frac{4}{35}
\displaystyle \text{and, }\ P(X=4)=\text{Probability of getting 4 white balls}=\frac{{}^{4}C_{4}}{{}^{10}C_{4}}=\frac{1}{210}
\displaystyle \text{Thus, the probability distribution of }X\text{ is given by}
\displaystyle \begin{array}{c|ccccc} X: & 0 & 1 & 2 & 3 & 4\\ \hline P(X): & \frac{1}{14} & \frac{8}{21} & \frac{6}{14} & \frac{4}{35} & \frac{1}{210} \end{array}

\displaystyle \textbf{Question 4:}\ \ \text{Four bad oranges are mixed accidently with 16 good oranges. Find the} \\ \text{probability distribution of the number of bad oranges in a draw of two oranges.} \\ \qquad [\text{CBSE 2002C}]
\displaystyle \text{Answer:}
\displaystyle  \text{Let }X\text{ denote the number of bad oranges in a draw of 4 oranges drawn from group} \\ \text{of 16 good oranges and 4 bad oranges. Since there are 4 bad oranges in the group,} \\ \text{therefore } X\text{ can take values }0,1\text{ and }2.
\displaystyle \text{Now, }\ P(X=0)=\text{Probability of getting no bad orange = Probability of getting 2 good oranges}
\displaystyle =\frac{{}^{16}C_{2}}{{}^{20}C_{2}}=\frac{12}{19}
\displaystyle P(X=1)=\text{Probability of getting one bad orange}=\frac{{}^{4}C_{1}\times{}^{16}C_{1}}{{}^{20}C_{2}}=\frac{32}{95}
\displaystyle \text{and, }\ P(X=2)=\text{Probability of getting two bad oranges}=\frac{{}^{4}C_{2}}{{}^{20}C_{2}}=\frac{3}{95}
\displaystyle \text{Thus, the probability distribution of }X\text{ is given by}
\displaystyle \begin{array}{c|ccc} X: & 0 & 1 & 2\\ \hline P(X): & \frac{12}{19} & \frac{32}{95} & \frac{3}{95} \end{array}
\displaystyle \Rightarrow P(X=2)=\frac{3}{8}\times\frac{2}{7}\times\frac{5}{6}+\frac{5}{8}\times\frac{3}{7}\times\frac{2}{6}+\frac{3}{8}\times\frac{5}{7}\times\frac{2}{6}=\frac{15}{56}
\displaystyle \text{and,}
\displaystyle P(X=3)=P(G_{1}\cap G_{2}\cap G_{3})=P(G_{1})\,P(G_{2}/G_{1})\,P(G_{3}/G_{1}\cap G_{2})=\frac{3}{8}\times\frac{2}{7}\times\frac{1}{6}=\frac{1}{56}
\displaystyle \text{Thus, the probability distribution of the number of green balls is given by}
\displaystyle \begin{array}{c|cccc} X: & 0 & 1 & 2 & 3\\ \hline P(X): & \frac{5}{28} & \frac{15}{28} & \frac{15}{56} & \frac{1}{56} \end{array}

\displaystyle \textbf{Question 5:}\ \ \text{From a lot of 10 items containing 3 defectives, a sample of 4 items is drawn} \\ \text{at random. Let the random variable }X\text{ denote the number of defective items in the sample.} \\ \text{If thee sample is drawn randomly, find}
\displaystyle \text{(i) the probability distribution of }X  \qquad \text{(ii) }P(X\le 1)
\displaystyle \text{(iii) }P(X<1)\qquad \text{(iv) }P(0<X<2) \ \ \ \ \ \ \ \ \ \ \qquad [\text{CBSE 2010}] 
\displaystyle \text{Answer:}
\displaystyle  \text{(i) Clearly, }X\text{ can assume values }0,1,2,3\text{ such that}
\displaystyle P(X=0)=\text{(Probability of getting no defective item)}=\frac{{}^{7}C_{4}}{{}^{10}C_{4}}=\frac{1}{6}
\displaystyle P(X=1)=\text{(Probability of getting one defective item)}=\frac{{}^{3}C_{1}\times{}^{7}C_{3}}{{}^{10}C_{4}}=\frac{1}{2}
\displaystyle P(X=2)=\text{(Probability of getting two defective items)}=\frac{{}^{3}C_{2}\times{}^{7}C_{2}}{{}^{10}C_{4}}=\frac{3}{10}
\displaystyle \text{and,}\ P(X=3)=\text{(Probability of getting three defective items)}=\frac{{}^{3}C_{3}\times{}^{7}C_{1}}{{}^{10}C_{4}}=\frac{1}{30}
\displaystyle \text{Hence, the probability distribution of }X\text{ is}
\displaystyle \begin{array}{c|cccc} X: & 0 & 1 & 2 & 3\\ \hline P(X): & \frac{1}{6} & \frac{1}{2} & \frac{3}{10} & \frac{1}{30} \end{array}
\displaystyle \text{(ii)}\ \ P(X\le 1)=P(X=0)+P(X=1)=\frac{1}{6}+\frac{1}{2}=\frac{2}{3}
\displaystyle \text{(iii)}\ \ P(X<1)=P(X=0)=\frac{1}{6}
\displaystyle \text{(iv)}\ \ P(0<X<2)=P(X=1)=\frac{1}{2}

\displaystyle \textbf{Question 6:}\ \ \text{The random variable }X\text{ can take values }0,1,2,3.\text{ Given that } \\ P(X=0)=P(X=1)=p\text{ and }P(X=2)=P(X=3)\text{ such that } \\ \sum p_{i}x_{i}^{2}=2\sum p_{i}x_{i},\text{ find the value of }p. \ \ \ \ \ \ \ \ \ \ \ \ \qquad [\text{CBSE 2017}]
\displaystyle \text{Answer:}
\displaystyle  \text{Let }P(X=2)=P(X=3)=\lambda.
\displaystyle \text{It is given that }X\text{ is a random variable taking values }0,1,2\text{ and }3.
\displaystyle \therefore\ P(X=0)+P(X=1)+P(X=2)+P(X=3)=1
\displaystyle \Rightarrow p+p+\lambda+\lambda=1
\displaystyle \Rightarrow \lambda=\frac{1}{2}-p
\displaystyle \text{Thus, the probability distribution of }X\text{ is}
\displaystyle \begin{array}{c|cccc} x_{i}: & 0 & 1 & 2 & 3\\ \hline p_{i}: & p & p & \frac{1}{2}-p & \frac{1}{2}-p \end{array}
\displaystyle \text{It is given that}
\displaystyle \sum p_{i}x_{i}^{2}=2\sum p_{i}x_{i}
\displaystyle \Rightarrow p\times0^{2}+p\times1^{2}+\left(\frac{1}{2}-p\right)\times2^{2}+\left(\frac{1}{2}-p\right)\times3^{2}=2\left\{p\times0+p\times1+\left(\frac{1}{2}-p\right)\times2+\left(\frac{1}{2}-p\right)\times3\right\}
\displaystyle \Rightarrow p+2-4p+\frac{9}{2}-9p=2\left(p+1-2p+\frac{3}{2}-3p\right)
\displaystyle \Rightarrow \left(\frac{13}{2}-12p\right)=2\left(\frac{5}{2}-4p\right)
\displaystyle \Rightarrow 13-24p=10-16p\Rightarrow8p=3\Rightarrow p=\frac{3}{8}

\displaystyle \textbf{Question 7:}\ \ \text{Find the mean, variance and standard deviation of the number of heads} \\ \text{in a simultaneous toss of three coins.}\qquad [\text{CBSE 2007}]
\displaystyle \text{Answer:}
\displaystyle  \text{Let }X\text{ denote the number of heads in a simultaneous toss of three coins. Then, }X\text{ can} \\ \text{take values }0,1,2,3. \text{Now, }
\displaystyle P(X=0)=P(TTT)=\frac{1}{8},\ P(X=1)=P(HTT\text{ or }TTH\text{ or }THT)=\frac{3}{8}
\displaystyle P(X=2)=P(HHT\text{ or }THH\text{ or }HTH)=\frac{3}{8}\ \text{and, }\ P(X=3)=P(HHH)=\frac{1}{8}
\displaystyle \text{Thus, the probability distribution of }X\text{ is given by}
\displaystyle \begin{array}{c|cccc} X: & 0 & 1 & 2 & 3\\ \hline P(X): & \frac{1}{8} & \frac{3}{8} & \frac{3}{8} & \frac{1}{8} \end{array}
\displaystyle \textbf{Computation of mean and variance}
\displaystyle \begin{array}{c|c|c|c} x_{i} & p_{i}=P(X=x_{i}) & p_{i}x_{i} & p_{i}x_{i}^{2}\\ \hline 0 & \frac{1}{8} & 0 & 0\\ 1 & \frac{3}{8} & \frac{3}{8} & \frac{3}{8}\\ 2 & \frac{3}{8} & \frac{6}{8} & \frac{12}{8}\\ 3 & \frac{1}{8} & \frac{3}{8} & \frac{9}{8}\\ \hline & & \sum p_{i}x_{i}=\frac{3}{2} & \sum p_{i}x_{i}^{2}=3 \end{array}
\displaystyle \text{Thus, we have }\sum p_{i}x_{i}=\frac{3}{2}\ \text{and }\sum p_{i}x_{i}^{2}=3
\displaystyle \therefore\ \bar{X}=\text{Mean}=\sum p_{i}x_{i}=\frac{3}{2}\ \text{and }\mathrm{Var}(X)=\sum p_{i}x_{i}^{2}-\left(\sum p_{i}x_{i}\right)^{2}=3-\left(\frac{3}{2}\right)^{2}=\frac{3}{4}
\displaystyle \therefore\ \text{Standard deviation}=\sqrt{\mathrm{Var}(X)}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}=0.87
\displaystyle \text{Hence, Mean}=\frac{3}{2},\ \text{Variance}=\frac{3}{4},\ \text{and, Standard deviation}=0.87

\displaystyle \textbf{Question 8:}\ \ \text{Two numbers are selected at random (without replacement) from the first} \\ \text{six positive integers. Let }X\text{ denote the larger of the two numbers obtained. Find }E(X) \\ \text{ and }\mathrm{Var}(X).\qquad [\text{CBSE 2014}]
\displaystyle \text{Answer:}
\displaystyle  \text{We observe that }X\text{ can take values }2,3,4,5,6\text{ such that}
\displaystyle P(X=2)=\text{Probability that the larger of two numbers is }2
\displaystyle \Rightarrow P(X=2)=\text{Probability of getting 1 in first selection and 2 in second selection} \\ \text{or getting 2 in first selection and 1 in second selection}
\displaystyle \Rightarrow P(X=2)=\frac{1}{6}\times\frac{1}{5}+\frac{1}{6}\times\frac{1}{5}=\frac{2}{30}=\frac{1}{15}
\displaystyle P(X=3)=\text{Probability that the larger of two numbers is }3
\displaystyle \Rightarrow P(X=3)=\text{Probability of getting a number less than 3 in first selection and 3 in second} \\ \text{selection or getting 3 in first selection and a number less than 3 in second selection}
\displaystyle \Rightarrow P(X=3)=\frac{2}{6}\times\frac{1}{5}+\frac{1}{6}\times\frac{2}{5}=\frac{4}{30}=\frac{2}{15}
\displaystyle P(X=4)=\text{(Probability that the larger of two numbers is }4)=\frac{3}{6}\times\frac{1}{5}+\frac{1}{6}\times\frac{3}{5}=\frac{6}{30}=\frac{1}{5}
\displaystyle P(X=5)=\frac{4}{6}\times\frac{1}{5}+\frac{1}{6}\times\frac{4}{5}=\frac{8}{30}=\frac{4}{15}\ \text{and, }\ P(X=6)=\frac{5}{6}\times\frac{1}{5}+\frac{1}{6}\times\frac{5}{5}=\frac{10}{30}=\frac{1}{3}
\displaystyle \text{Thus, the probability distribution of }X\text{ is}
\displaystyle \begin{array}{c|ccccc} X: & 2 & 3 & 4 & 5 & 6\\ \hline P(X): & \frac{1}{15} & \frac{2}{15} & \frac{1}{5} & \frac{4}{15} & \frac{1}{3} \end{array}
\displaystyle \therefore\ E(X)=\frac{1}{15}\times2+\frac{2}{15}\times3+\frac{1}{5}\times4+\frac{4}{15}\times5+\frac{1}{3}\times6=\frac{70}{15}=\frac{14}{3}

\displaystyle \textbf{Question 9:}\ \ \text{Two cards are drawn successively without replacement from a well-} \\ \text{shuffled deck of 52 cards. Compute the variance of the number of aces.}\qquad [\text{CBSE 2010}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }A_{i}\text{ denote the event of getting an ace in }i\text{th draw, where }i=1,2.\text{ Further,} \\ \text{let }X\text{ denote the number of aces in two draws. Then, }X\text{ can take values }0,1,2.
\displaystyle \text{Now, }\ P(X=0)=\text{Probability of getting no ace in two successive draws}
\displaystyle \Rightarrow P(X=0)=P(\overline{A_{1}}\cap\overline{A_{2}})=P(\overline{A_{1}})\,P(\overline{A_{2}}/A_{1})=\frac{48}{52}\times\frac{47}{51}=\frac{564}{663}
\displaystyle P(X=1)=\text{Probability of getting an ace in one of the two draws}
\displaystyle \Rightarrow P(X=1)=P\big((A_{1}\cap\overline{A_{2}})\cup(\overline{A_{1}}\cap A_{2})\big)
\displaystyle \Rightarrow P(X=1)=P(A_{1})P(\overline{A_{2}}/A_{1})+P(\overline{A_{1}})P(A_{2}/\overline{A_{1}})=\frac{4}{52}\times\frac{48}{51}+\frac{48}{52}\times\frac{4}{51}=\frac{96}{663}
\displaystyle P(X=2)=\text{Probability of getting an ace in each draw}
\displaystyle \Rightarrow P(X=2)=P(A_{1}\cap A_{2})=P(A_{1})P(A_{2}/A_{1})=\frac{4}{52}\times\frac{3}{51}=\frac{3}{663}
\displaystyle \text{Thus, the probability distribution of }X\text{ is given by}
\displaystyle \begin{array}{c|ccc} X: & 0 & 1 & 2\\ \hline P(X): & \frac{564}{663} & \frac{96}{663} & \frac{3}{663} \end{array}
\displaystyle \therefore\ \sum p_{i}x_{i}=0\times\frac{564}{663}+1\times\frac{96}{663}+2\times\frac{3}{663}=\frac{102}{663}
\displaystyle \text{and, }\sum p_{i}x_{i}^{2}=\frac{564}{663}\times0+\frac{96}{663}\times1+\frac{3}{663}\times4=\frac{108}{663}
\displaystyle \text{Hence, }\mathrm{Var}(X)=\sum p_{i}x_{i}^{2}-\left(\sum p_{i}x_{i}\right)^{2}=\frac{108}{663}-\left(\frac{102}{663}\right)^{2}=\frac{108\times663-(102)^{2}}{(663)^{2}}=\frac{61200}{663\times663}=\frac{400}{2873}

\displaystyle \textbf{Question 10:}\ \ \text{In a group of 30 scientists working on an experiment, 20 never commit error} \\ \text{in their work and are reporting results elaborately. Two scientists are selected at random} \\ \text{from the group. Find the probability distribution of the number of selected scientists who} \\ \text{never commit error in the work and reporting. Also, find the mean of the distribution.} \\ \text{What values are described in the question?}\qquad [\text{CBSE 2013}]
\displaystyle \text{Answer:}
\displaystyle  \text{Let }X\text{ denote the number of selected scientists who never commit error in the work and} \\ \text{reporting. Clearly, }X\text{ can take values }0,1,2.
\displaystyle \text{Now,}
\displaystyle P(X=0)=\text{Probability that two scientists selected commit error either in the work or in reporting}
\displaystyle =\frac{{}^{10}C_{2}}{{}^{30}C_{2}}=\frac{3}{29}
\displaystyle P(X=1)=\text{Probability that one out of two scientists selected does not commit error in the work and} \\ \text{reporting while the other is not so}
\displaystyle =\frac{{}^{20}C_{1}\times{}^{10}C_{1}}{{}^{30}C_{2}}=\frac{40}{87}
\displaystyle P(X=2)=\text{Probability that two scientists selected do not commit error in the work and reporting}
\displaystyle =\frac{{}^{20}C_{2}}{{}^{30}C_{2}}=\frac{38}{87}
\displaystyle \text{The probability distribution of }X\text{ is as given below:}
\displaystyle \begin{array}{c|ccc} X: & 0 & 1 & 2\\ \hline P(X): & \frac{3}{29} & \frac{40}{87} & \frac{38}{87} \end{array}
\displaystyle \text{Let }\bar{X}\text{ be the mean of the distribution. Then,}
\displaystyle \bar{X}=0\times\frac{3}{29}+1\times\frac{40}{87}+2\times\frac{38}{87}=\frac{116}{87}=1.33
\displaystyle \text{This means that on an average out of two selected scientists one scientists will not commit} \\ \text{error in the work and reporting.}

\displaystyle \textbf{Question 11:}\ \ \text{A pair of dice is thrown 7 times. If getting a total of 7 is considered} \\ \text{a success, what is the probability of}
\displaystyle \text{(i) no success?\qquad (ii) 6 successes?\qquad (iii) at least 6 successes?} \\ \text{(iv) at most 6 successes?}\qquad [\text{CBSE 2004}]
\displaystyle \text{Answer:}
\displaystyle  \text{Let }p\text{ denote the probability of getting a total of 7 in a single throw of a pair of dice.}
\displaystyle \text{Then,}
\displaystyle p=\frac{6}{36}=\frac{1}{6}\qquad \left[\because\ \text{The sum can be 7 in any one of the ways: }(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)\right]
\displaystyle \therefore\ q=1-p=1-\frac{1}{6}=\frac{5}{6}
\displaystyle \text{Let }X\text{ denote the number of successes in 7 throws of a pair of dice. The }X\text{ is a} \\ \text{binomial variate with parameters }n=7\text{ and }p=\frac{1}{6}\text{ such that}
\displaystyle P(X=r)={}^{7}C_{r}\left(\frac{1}{6}\right)^{r}\left(\frac{5}{6}\right)^{7-r},\ r=0,1,2,\ldots,7\qquad \cdots (i)
\displaystyle \text{(i)\quad Probability of no success }=P(X=0)={}^{7}C_{0}\left(\frac{1}{6}\right)^{0}\left(\frac{5}{6}\right)^{7-0}=\left(\frac{5}{6}\right)^{7}\qquad [\text{Using (i)}]
\displaystyle \text{(ii)\quad Probability of 6 successes }=P(X=6)={}^{7}C_{6}\left(\frac{1}{6}\right)^{6}\left(\frac{5}{6}\right)^{7-6}\qquad [\text{Using (i)}]
\displaystyle =35\left(\frac{1}{6}\right)^{7}
\displaystyle \text{(iii)\quad Probability of at least 6 successes }=P(X\ge6)
\displaystyle =P(X=6)+P(X=7)
\displaystyle ={}^{7}C_{6}\left(\frac{1}{6}\right)^{6}\left(\frac{5}{6}\right)^{7-6}+{}^{7}C_{7}\left(\frac{1}{6}\right)^{7}\left(\frac{5}{6}\right)^{0}\qquad [\text{Using (i)}]
\displaystyle =7\left(\frac{1}{6}\right)^{6}\left(\frac{5}{6}\right)+\left(\frac{1}{6}\right)^{7}=\left(\frac{1}{6}\right)^{6}\left(\frac{35}{6}+\frac{1}{6}\right)=\left(\frac{1}{6}\right)^{5}
\displaystyle \text{(iv)\quad Probability of at most 6 successes }=P(X\le6)
\displaystyle =1-P(X>6)
\displaystyle =1-P(X=7)
\displaystyle =1-{}^{7}C_{7}\left(\frac{1}{6}\right)^{7}\left(\frac{5}{6}\right)^{0}\qquad [\text{Using (i)}]
\displaystyle =1-\left(\frac{1}{6}\right)^{7}

\displaystyle \textbf{Question 12:}\ \ \text{A man takes a step forward with probability }0.4\text{ and backwards} \\ \text{with probability }0.6.\text{ Find the probability that at the end of eleven steps he is just} \\ \text{one step away from the starting point.}\qquad [\text{CBSE 2015}]
\displaystyle \text{Answer:}
\displaystyle  \text{Let }p\text{ denote the probability that the man takes a step forward. Then, }p=0.4
\displaystyle \therefore\ q=1-p=1-0.4=0.6
\displaystyle \text{Let }X\text{ denote the number of steps taken in the forward direction. Since the steps are} \\ \text{independent of each other. Therefore, }X\text{ is a binomial variate with parameters } \\ n=11\text{ and }p=0.4\text{ such that}
\displaystyle P(X=r)={}^{11}C_{r}(0.4)^{r}(0.6)^{11-r},\ r=0,1,2,\ldots,11\qquad \cdots (i)
\displaystyle \text{Since the man is one step away from the initial point. Therefore, he is either one step forward} \\ \text{or one step backward from the initial point at the end of eleven steps. If he is one step forward,} \\ \text{then he must have taken six steps forward and five steps backward and if he is} \\ \text{one step backward, then he must have taken five steps forward and six steps backward.} \\ \text{Thus, either }X=6\text{ or }X=5.
\displaystyle \text{Required probability }=P[(X=5)\ \text{or}\ (X=6)]
\displaystyle =P(X=5)+P(X=6)
\displaystyle ={}^{11}C_{5}(0.4)^{5}(0.6)^{11-5}+{}^{11}C_{6}(0.4)^{6}(0.6)^{11-6}\qquad [\text{Using (i)}]
\displaystyle ={}^{11}C_{5}(0.4)^{5}(0.6)^{5}[0.6+0.4]\qquad [\because\ {}^{11}C_{5}={}^{11}C_{6}]
\displaystyle =462(0.4)^{5}(0.6)^{5}=462(0.24)^{5}

\displaystyle \textbf{Question 13:}\ \ \text{An unbiased die is thrown again and again until three sixes are obtained.} \\ \text{Find the probability of obtaining 3rd six in the sixth throw of the die.}\qquad [\text{CBSE 2009}]
\displaystyle \text{Answer:}
\displaystyle  \text{Let }p\text{ be the probability of obtaining a `six' in a single throw of the die. Then,}
\displaystyle p=\frac{1}{6}\ \text{and }\ q=1-\frac{1}{6}=\frac{5}{6}
\displaystyle \text{Obtaining third six in the sixth throw of the die means that in first five throws there are 2} \\ \text{sixes and the third six is obtained in sixth throw. Therefore,}
\displaystyle \text{Required probability }=P(\text{Getting 2 sixes in first 5 throws})\ P(\text{Getting `six' in sixth throw})
\displaystyle =\left({}^{5}C_{2}\,p^{2}\,q^{5-2}\right)(p)={}^{5}C_{2}\left(\frac{1}{6}\right)^{2}\left(\frac{5}{6}\right)^{3}\times\frac{1}{6}=10\times\left(\frac{1}{6}\right)^{3}\left(\frac{5}{6}\right)^{3}=\frac{625}{23328}

\displaystyle \textbf{Question 14:}\ \ \text{For 6 trials of an experiment, let }X\text{ be a binomial variate which} \\ \text{satisfies the relation }9P(X=4)=P(X=2).\text{ Find the probability of success.}\\ \qquad [\text{CBSE 2015}]
\displaystyle \text{Answer:}
\displaystyle  \text{Let }p\text{ be the probability of success and }q\text{ that of failure in a single trial. Then,}
\displaystyle P(X=r)={}^{6}C_{r}\,p^{r}\,q^{6-r}\qquad [\because\ n=6\ (\text{given})]
\displaystyle \text{It is given that}
\displaystyle 9P(X=4)=P(X=2)
\displaystyle \Rightarrow 9\times{}^{6}C_{4}\,p^{4}\,q^{2}={}^{6}C_{2}\,p^{2}\,q^{4}\qquad [\because\ {}^{6}C_{2}={}^{6}C_{4}]
\displaystyle \Rightarrow 9p^{2}=q^{2}
\displaystyle \Rightarrow 3p=q
\displaystyle \Rightarrow 3p=1-p\qquad [\because\ p+q=1]
\displaystyle \Rightarrow p=\frac{1}{4}

\displaystyle \textbf{Question 15:}\ \ \text{Find the probability distribution of the number of doublets in 4 throws} \\ \text{of a pair of dice.}\qquad [\text{CBSE 2008, 2012, 2015}]
\displaystyle \text{Answer:}
\displaystyle  \text{Let }p\text{ denote the probability of getting a doublet in a single throw of a pair of dice.}
\displaystyle \text{Then,}
\displaystyle p=\frac{6}{36}=\frac{1}{6}\ \text{and, }\ q=1-p=1-\frac{1}{6}=\frac{5}{6}
\displaystyle \text{Let }X\text{ denote the number of doublets in 4 throws of a pair of dice. Then, } \\ X\text{ is a binomial variate with parameters }n=4\text{ and }p=\frac{1}{6}\text{ such that}
\displaystyle P(X=r)=\text{Probability of getting }r\text{ doublets}
\displaystyle \text{or, }\ 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 \therefore\ P(X=0)={}^{4}C_{0}\left(\frac{1}{6}\right)^{0}\left(\frac{5}{6}\right)^{4}=\left(\frac{5}{6}\right)^{4},\ P(X=1)={}^{4}C_{1}\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^{3}=\frac{2}{3}\left(\frac{5}{6}\right)^{3}
\displaystyle P(X=2)={}^{4}C_{2}\left(\frac{1}{6}\right)^{2}\left(\frac{5}{6}\right)^{2}=\frac{1}{6}\left(\frac{5}{6}\right)^{2},\ P(X=3)={}^{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{and, }\ P(X=4)={}^{4}C_{4}\left(\frac{1}{6}\right)^{4}\left(\frac{5}{6}\right)^{0}=\left(\frac{1}{6}\right)^{4}
\displaystyle \text{Thus, the probability distribution of }X\text{ is given by}
\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{1}{6}\left(\frac{5}{6}\right)^{2} & \frac{10}{3}\left(\frac{1}{6}\right)^{3} & \left(\frac{1}{6}\right)^{4} \end{array}

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