\displaystyle \textbf{Question 1. }\text{If }A\text{ and }B\text{ are events such that }P\left(\frac{A}{B}\right)=P\left(\frac{B}{A}\right),\text{ then}
\displaystyle \text{(a) }A\subset B\text{ but }A\neq B
\displaystyle \text{(b) }A=B
\displaystyle \text{(c) }A\cap B=\phi
\displaystyle \text{(d) }P(A)=P(B)
\displaystyle \text{Answer:}
\displaystyle P\left(\frac{A}{B}\right)=P\left(\frac{B}{A}\right)
\displaystyle \therefore \frac{P(A\cap B)}{P(B)}=\frac{P(A\cap B)}{P(A)}
\displaystyle \therefore P(A\cap B)\left[\frac{1}{P(B)}-\frac{1}{P(A)}\right]=0
\displaystyle \therefore P(A\cap B)=0\text{ or }P(A)=P(B)
\displaystyle \text{Since the expected option is based on }P(A\cap B)\neq0,
\displaystyle \therefore P(A)=P(B)
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 2. }\text{If }A\text{ and }B\text{ are two events such that }P(B)=\frac{3}{5}, \\ P\left(\frac{A}{B}\right)=\frac{1}{2}\text{ and }P(A\cup B)=\frac{4}{5},\text{ then }P(A)\text{ equals to}
\displaystyle \text{(a) }\frac{3}{10}\qquad\text{(b) }\frac{1}{5}\qquad\text{(c) }\frac{1}{2}\qquad\text{(d) }\frac{3}{5}
\displaystyle \text{Answer:}
\displaystyle P(B)=\frac{3}{5},\qquad P\left(\frac{A}{B}\right)=\frac{1}{2}
\displaystyle \therefore P(A\cap B)=P\left(\frac{A}{B}\right)\cdot P(B)  =\frac{1}{2}\times\frac{3}{5}=\frac{3}{10}
\displaystyle \text{Now, }P(A\cup B)=P(A)+P(B)-P(A\cap B)  \frac{4}{5}=P(A)+\frac{3}{5}-\frac{3}{10}
\displaystyle \therefore P(A)=\frac{4}{5}-\frac{3}{5}+\frac{3}{10}  =\frac{1}{5}+\frac{3}{10}  =\frac{5}{10}=\frac{1}{2}
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 3. }\text{If }A\text{ and }B\text{ are two events such that }A\subset B\text{ and }P(B)\neq0, \\ \text{then which of the following is correct?}
\displaystyle \text{(a) }P\left(\frac{B}{A}\right)=\frac{P(A)}{P(B)} \qquad \text{(b) }P\left(\frac{A}{B}\right)<P(A)
\displaystyle \text{(c) }P\left(\frac{A}{B}\right)\geq P(A) \qquad \text{(d) }\text{None of these}
\displaystyle \text{Answer:}
\displaystyle A\subset B
\displaystyle \therefore A\cap B=A
\displaystyle \therefore P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P(B)}  =\frac{P(A)}{P(B)}
\displaystyle \text{Since }P(B)\leq1,
\displaystyle \therefore \frac{P(A)}{P(B)}\geq P(A)
\displaystyle \therefore P\left(\frac{A}{B}\right)\geq P(A)
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 4. }\text{A flashlight has }8\text{ batteries out of which }3\text{ are dead. If two batteries are}
\displaystyle \text{selected without replacement and tested, then probability that both are dead is}
\displaystyle \text{(a) }\frac{33}{56}\qquad\text{(b) }\frac{9}{64}\qquad\text{(c) }\frac{1}{14}\qquad\text{(d) }\frac{3}{28}
\displaystyle \text{Answer:}
\displaystyle \text{Number of ways of selecting 2 batteries from 8 batteries}={}^{8}C_{2}=28
\displaystyle \text{No. of ways of selecting 2 dead batteries from 3 dead batteries}={}^{3}C_{2}=3
\displaystyle \therefore P(\text{both batteries are dead})=\frac{{}^{3}C_{2}}{{}^{8}C_{2}} =\frac{3}{28}
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 5. }\text{The probability of obtaining an even prime number on each die} \\ \text{when a pair of dice is rolled is}
\displaystyle \text{(a) zero}\qquad\text{(b) }\frac{1}{3}\qquad\text{(c) }\frac{1}{12}\qquad\text{(d) }\frac{1}{36}
\displaystyle \text{Answer:}
\displaystyle \text{The only even prime number is }2.
\displaystyle \text{Probability of getting }2\text{ on one die}=\frac{1}{6}
\displaystyle \text{Since both dice must show }2,
\displaystyle P(\text{2 on both dice})=\frac{1}{6}\times\frac{1}{6}  =\frac{1}{36}
\displaystyle \therefore \text{Correct option is (d).}
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\displaystyle \textbf{Question 6. }\text{Out of }50\text{ tickets numbered }00,01,02,\ldots,49,\text{ one ticket is}
\displaystyle  \text{drawn randomly, the probability } \text{of the ticket having the product of its digits }7,
\displaystyle \text{ given that the sum of the digits is }8,\text{ is}
\displaystyle \text{(a) }\frac{1}{14}\qquad\text{(b) }\frac{3}{14}\qquad\text{(c) }\frac{1}{5}\qquad\text{(d) }\text{None of these}
\displaystyle \text{Answer:}
\displaystyle \text{Tickets from }00\text{ to }49\text{ whose sum of digits is }8\text{ are } 08,\ 17,\ 26,\ 35,\ 44
\displaystyle \text{Out of these, product of digits is }7\text{ only for }17.
\displaystyle \therefore P(\text{product of digits is }7\mid\text{sum of digits is }8)=\frac{1}{5}
\displaystyle \therefore \text{Correct option is (c).}
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\displaystyle \textbf{Question 7. }\text{If }P(A\cup B)=0.83,\ P(A)=0.3\text{ and }P(B)=0.6,\text{ then the events will be}
\displaystyle \text{(a) dependent}\qquad\text{(b) independent}
\displaystyle \text{(c) cannot say anything}\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle 0.83=0.3+0.6-P(A\cap B)
\displaystyle \therefore P(A\cap B)=0.9-0.83=0.07
\displaystyle \text{Now, }P(A)P(B)=0.3\times0.6=0.18
\displaystyle \therefore P(A\cap B)\neq P(A)P(B)
\displaystyle \therefore A\text{ and }B\text{ are dependent events.}
\displaystyle \therefore \text{Correct option is (a).}
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\displaystyle \textbf{Question 8. }\text{If }P(A)=0.5,\ P(B)=0.4\text{ and }P(A\cap B)=0.3,\text{ then }P\left(\frac{A'}{B'}\right)\text{ is equal to}
\displaystyle \text{(a) }\frac{1}{3}\qquad\text{(b) }\frac{1}{2}\qquad\text{(c) }\frac{2}{3}\qquad\text{(d) }\frac{3}{4}
\displaystyle \text{Answer:}
\displaystyle P(B')=1-P(B)=1-0.4=0.6
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)  =0.5+0.4-0.3=0.6
\displaystyle \therefore P(A'\cap B')=1-P(A\cup B) =1-0.6=0.4
\displaystyle P\left(\frac{A'}{B'}\right)=\frac{P(A'\cap B')}{P(B')}  =\frac{0.4}{0.6} =\frac{2}{3}
\displaystyle \therefore \text{Correct option is (c).}
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\displaystyle \textbf{Question 9. }\text{It is given that the events }A\text{ and }B\text{ are such that }P(A)=\frac{1}{4},
\displaystyle P\left(\frac{A}{B}\right)=\frac{1}{2}  \text{and }P\left(\frac{B}{A}\right)=\frac{2}{3}.\text{ Then, }P(B)\text{ is equal to}
\displaystyle \text{(a) }\frac{1}{2}\qquad\text{(b) }\frac{1}{6}\qquad\text{(c) }\frac{1}{3}\qquad\text{(d) }\frac{2}{3}
\displaystyle \text{Answer:}
\displaystyle P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P(A)}=\frac{2}{3}
\displaystyle \therefore P(A\cap B)=\frac{2}{3}\times\frac{1}{4}=\frac{1}{6}
\displaystyle P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P(B)}=\frac{1}{2}
\displaystyle \therefore \frac{\frac{1}{6}}{P(B)}=\frac{1}{2}
\displaystyle \therefore P(B)=\frac{1}{3}
\displaystyle \therefore \text{Correct option is (c).}
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\displaystyle \textbf{Question 10. }\text{A and B throw a pair of dice alternately. A wins the game, if he gets a}
\displaystyle \text{total of }7\text{ and B wins the game,}   \text{if he gets a total of }10.\text{ If A starts the game, then find}
\displaystyle \text{the probability that B wins.}
\displaystyle \text{Answer:}
\displaystyle P(\text{A gets }7)=\frac{6}{36}=\frac{1}{6}
\displaystyle P(\text{B gets }10)=\frac{3}{36}=\frac{1}{12}
\displaystyle P(\text{A does not win})=1-\frac{1}{6}=\frac{5}{6}
\displaystyle P(\text{B does not win})=1-\frac{1}{12}=\frac{11}{12}
\displaystyle P(\text{B wins in first round})=\frac{5}{6}\times\frac{1}{12}=\frac{5}{72}
\displaystyle P(\text{No one wins in one complete round})=\frac{5}{6}\times\frac{11}{12}=\frac{55}{72}
\displaystyle \therefore P(\text{B wins})=\frac{5}{72}\left[1+\frac{55}{72}+\left(\frac{55}{72}\right)^2+\cdots\right]
\displaystyle =\frac{5}{72}\cdot\frac{1}{1-\frac{55}{72}}
\displaystyle =\frac{5}{72}\cdot\frac{72}{17}
\displaystyle =\frac{5}{17}
\displaystyle \therefore \text{Probability that B wins }=\frac{5}{17}.
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\displaystyle \textbf{Question 11. }\text{Probabilities of solving a specific problem independently by} \\ \text{A and B are }\frac{1}{2}\text{ and }\frac{1}{3}\text{ respectively.}  \text{If both try to solve the problem independently, } \\ \text{then find the probability that}
\displaystyle \text{(i) problem is solved.}
\displaystyle \text{(ii) exactly one of them solves the problem.}
\displaystyle \text{Answer:}
\displaystyle P(A)=\frac{1}{2},\qquad P(B)=\frac{1}{3}
\displaystyle \text{Since the attempts are independent,}
\displaystyle P(A\cap B)=\frac{1}{2}\times\frac{1}{3}=\frac{1}{6}
\displaystyle \text{(i) Probability that the problem is solved}
\displaystyle =P(A\cup B)
\displaystyle =P(A)+P(B)-P(A\cap B)
\displaystyle =\frac{1}{2}+\frac{1}{3}-\frac{1}{6}
\displaystyle =\frac{2}{3}
\displaystyle \therefore P(\text{problem is solved})=\frac{2}{3}
\displaystyle \text{(ii) Probability that exactly one solves the problem}
\displaystyle =P(A\cap B')+P(A'\cap B)
\displaystyle =\frac{1}{2}\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{2}\right)\frac{1}{3}
\displaystyle =\frac{1}{3}+\frac{1}{6}
\displaystyle =\frac{1}{2}
\displaystyle \therefore P(\text{exactly one solves the problem})=\frac{1}{2}
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\displaystyle \textbf{Question 12. }\text{A couple has }2\text{ children. Find the probability that both} \\ \text{are boys, if it is known that}
\displaystyle \text{(i) one of them is a boy.}
\displaystyle \text{(ii) the older child is a boy.}
\displaystyle \text{Answer:}
\displaystyle \text{The sample space for two children is }S=\{BB,BG,GB,GG\}
\displaystyle \text{(i) Given that one of them is a boy.}
\displaystyle \therefore \text{Possible outcomes are }BB,BG,GB
\displaystyle \therefore P(\text{both are boys}\mid\text{one of them is a boy})=\frac{1}{3}
\displaystyle \text{(ii) Given that the older child is a boy.}
\displaystyle \therefore \text{Possible outcomes are }BB,BG
\displaystyle \therefore P(\text{both are boys}\mid\text{older child is a boy})=\frac{1}{2}
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\displaystyle \textbf{Question 13. }P\text{ speaks truth in }70\%\text{ of the cases and }Q\text{ in }80\%\text{ of the cases. In what }
\displaystyle \text{per cent of cases are they likely to agree in stating the same fact?}
\displaystyle \text{Answer:}
\displaystyle P(\text{P speaks truth})=\frac{70}{100}=\frac{7}{10}
\displaystyle P(\text{Q speaks truth})=\frac{80}{100}=\frac{4}{5}
\displaystyle P(\text{P speaks false})=1-\frac{7}{10}=\frac{3}{10}
\displaystyle P(\text{Q speaks false})=1-\frac{4}{5}=\frac{1}{5}
\displaystyle \text{They agree when both speak truth or both speak false.}
\displaystyle \therefore P(\text{they agree})=\frac{7}{10}\times\frac{4}{5}+\frac{3}{10}\times\frac{1}{5}
\displaystyle =\frac{28}{50}+\frac{3}{50}
\displaystyle =\frac{31}{50}
\displaystyle =62\%
\displaystyle \therefore \text{They are likely to agree in }62\%\text{ of the cases.}
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\displaystyle \textbf{Question 14. }\text{If }A\text{ and }B\text{ are two independent events such that}
\displaystyle P(\overline{A}\cap B)=\frac{2}{15}\text{ and }P(A\cap\overline{B})=\frac{1}{6},\text{ then find }P(A)\text{ and }P(B).
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(A)=p\text{ and }P(B)=q
\displaystyle \text{Since }A\text{ and }B\text{ are independent,}
\displaystyle P(\overline{A}\cap B)=P(\overline{A})P(B)
\displaystyle \therefore (1-p)q=\frac{2}{15}\qquad (1)
\displaystyle P(A\cap\overline{B})=P(A)P(\overline{B})
\displaystyle \therefore p(1-q)=\frac{1}{6}\qquad (2)
\displaystyle \text{From (1), }q-pq=\frac{2}{15}
\displaystyle \text{From (2), }p-pq=\frac{1}{6}
\displaystyle \therefore q-p=\frac{2}{15}-\frac{1}{6}=-\frac{1}{30}
\displaystyle \therefore q=p-\frac{1}{30}
\displaystyle \text{Substituting in (2),}
\displaystyle p\left(1-p+\frac{1}{30}\right)=\frac{1}{6}
\displaystyle p\left(\frac{31}{30}-p\right)=\frac{1}{6}
\displaystyle 30p^2-31p+5=0
\displaystyle (5p-1)(6p-5)=0
\displaystyle \therefore p=\frac{1}{5}\text{ or }\frac{5}{6}
\displaystyle \text{Taking the required solution, }P(A)=\frac{1}{5}
\displaystyle \therefore P(B)=p-\frac{1}{30}=\frac{1}{5}-\frac{1}{30}=\frac{1}{6}
\displaystyle \therefore P(A)=\frac{1}{5},\qquad P(B)=\frac{1}{6}
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\displaystyle \textbf{Question 15. }\text{There are 4 cards numbered }1,3,5\text{ and }7,\text{ one number on one}
\displaystyle \text{card. Two cards are drawn at random without replacement. Let }X\text{ denote}
\displaystyle \text{the sum of the numbers on the two drawn cards. Find the mean.}
\displaystyle \text{Answer:}
\displaystyle \text{Possible pairs are }(1,3),(1,5),(1,7),(3,5),(3,7),(5,7)
\displaystyle \text{Corresponding values of }X\text{ are }4,6,8,8,10,12
\displaystyle \therefore E(X)=\frac{4+6+8+8+10+12}{6}
\displaystyle =\frac{48}{6}
\displaystyle =8
\displaystyle \therefore \text{Mean of }X=8
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\displaystyle \textbf{Question 16. }\text{The random variable }X\text{ can take only the values }0,1,2,3.\text{ Given}
\displaystyle \text{that } P(X=0)=P(X=1)=p\text{ and }P(X=2)=P(X=3)\text{ such that }
\displaystyle \sum p_i x_i^2=2\sum p_i x_i,\text{ find}  \text{ the value of }p.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(X=2)=P(X=3)=q
\displaystyle \therefore p+p+q+q=1
\displaystyle \therefore p+q=\frac{1}{2} \qquad (1)
\displaystyle \sum p_i x_i=0\cdot p+1\cdot p+2q+3q=p+5q
\displaystyle \sum p_i x_i^2=0^2\cdot p+1^2\cdot p+2^2q+3^2q=p+13q
\displaystyle \text{Given, }\sum p_i x_i^2=2\sum p_i x_i
\displaystyle \therefore p+13q=2(p+5q)
\displaystyle \therefore p+13q=2p+10q
\displaystyle \therefore p=3q \qquad (2)
\displaystyle \text{From (1) and (2), }3q+q=\frac{1}{2}
\displaystyle \therefore q=\frac{1}{8}
\displaystyle \therefore p=\frac{3}{8}
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\displaystyle \textbf{Question 17. }\text{In a game, a man wins Rs }5\text{ for getting a number greater}
\displaystyle \text{than }4\text{ and loses Rs }1\text{ otherwise,}   \text{when a fair die is thrown. The man decided to} 
\displaystyle \text{throw a die thrice but to quit as and when he gets}   \text{a number greater than }4. 
\displaystyle \text{Find the } \text{expected value of what he wins/loses.}
\displaystyle \text{Answer:}
\displaystyle P(\text{number greater than }4)=\frac{2}{6}=\frac{1}{3}
\displaystyle P(\text{number not greater than }4)=\frac{4}{6}=\frac{2}{3}
\displaystyle \text{If he wins on first throw, gain }=5,\text{ probability }=\frac{1}{3}
\displaystyle \text{If he wins on second throw, gain }=-1+5=4,\text{ probability }=\frac{2}{3}\cdot\frac{1}{3}=\frac{2}{9}
\displaystyle \text{If he wins on third throw, gain }=-1-1+5=3,\text{ probability }=\frac{2}{3}\cdot\frac{2}{3}\cdot\frac{1}{3}=\frac{4}{27}
\displaystyle \text{If he loses on all three throws, gain }=-3,\text{ probability }=\left(\frac{2}{3}\right)^3=\frac{8}{27}
\displaystyle \therefore E(X)=5\cdot\frac{1}{3}+4\cdot\frac{2}{9}+3\cdot\frac{4}{27}-3\cdot\frac{8}{27}
\displaystyle =\frac{5}{3}+\frac{8}{9}+\frac{12}{27}-\frac{24}{27}
\displaystyle =\frac{19}{9}
\displaystyle \therefore \text{Expected gain }=\text{Rs }\frac{19}{9}
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\displaystyle \textbf{Question 18. }\text{If }P(A)=\frac{6}{11},\ P(B)=\frac{5}{11}\text{ and }P(A\cup B)=\frac{7}{11},\text{ then find}
\displaystyle \text{(i) }P(A\cap B)\text{ (ii) }P\left(\frac{A}{B}\right).
\displaystyle \text{Answer:}
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle \frac{7}{11}=\frac{6}{11}+\frac{5}{11}-P(A\cap B)
\displaystyle \therefore P(A\cap B)=\frac{11}{11}-\frac{7}{11}=\frac{4}{11}
\displaystyle \text{(i) }P(A\cap B)=\frac{4}{11}
\displaystyle \text{(ii) }P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P(B)}  =\frac{\frac{4}{11}}{\frac{5}{11}} =\frac{4}{5}
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\displaystyle \textbf{Question 19. }\text{There are }25\text{ tickets bearing numbers from }1\text{ to }25.\text{ One ticket is drawn }
\displaystyle \text{at random. Find the probability that the number on it is a multiple of }5\text{ or }6.
\displaystyle \text{Answer:}
\displaystyle \text{Multiples of }5\text{ from }1\text{ to }25\text{ are }5,10,15,20,25
\displaystyle \text{Multiples of }6\text{ from }1\text{ to }25\text{ are }6,12,18,24
\displaystyle \text{Thus, favourable numbers are }5,6,10,12,15,18,20,24,25
\displaystyle \therefore \text{Number of favourable outcomes}=9
\displaystyle \text{Total number of outcomes}=25
\displaystyle \therefore P(\text{multiple of }5\text{ or }6)=\frac{9}{25}
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\displaystyle \textbf{Question 20. }\text{A die has two faces each with number }1,\text{ three faces each with number }2 
\displaystyle \text{ and one face with number 3. If die is rolled once, then determine probability of not getting }3.
\displaystyle \text{Answer:}
\displaystyle \text{Total number of faces}=6
\displaystyle \text{Number of faces marked }3=1
\displaystyle \therefore P(\text{getting }3)=\frac{1}{6}
\displaystyle \therefore P(\text{not getting }3)=1-\frac{1}{6}  =\frac{5}{6}
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\displaystyle \textbf{Question 21. }\text{Evaluate }P(A\cup B),\text{ if }2P(A)=P(B)=\frac{5}{13}\text{ and }P\left(\frac{A}{B}\right)=\frac{2}{5}.
\displaystyle \text{Answer:}
\displaystyle 2P(A)=P(B)=\frac{5}{13}
\displaystyle \therefore P(A)=\frac{5}{26},\qquad P(B)=\frac{5}{13}
\displaystyle P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P(B)}=\frac{2}{5}
\displaystyle \therefore P(A\cap B)=\frac{2}{5}\times\frac{5}{13}=\frac{2}{13}
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle =\frac{5}{26}+\frac{5}{13}-\frac{2}{13}
\displaystyle =\frac{5}{26}+\frac{10}{26}-\frac{4}{26}
\displaystyle =\frac{11}{26}
\displaystyle \therefore P(A\cup B)=\frac{11}{26}
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\displaystyle \textbf{Question 22. }\text{A can hit a target }4\text{ times out of }5\text{ times, }B\text{ can hit the target }
\displaystyle 3\text{ times out of }4\text{ times} \text{ and }C\text{ can hit the target }2\text{ times out of }3\text{ times. They fire}
\displaystyle \text{simultaneously. Find the probability that any two out of }A,B\text{ and }C\text{ will hit the target.}
\displaystyle \text{Answer:}
\displaystyle P(A)=\frac{4}{5},\qquad P(B)=\frac{3}{4},\qquad P(C)=\frac{2}{3}
\displaystyle P(A')=\frac{1}{5},\qquad P(B')=\frac{1}{4},\qquad P(C')=\frac{1}{3}
\displaystyle \text{Probability that exactly two hit the target}
\displaystyle =P(A\cap B\cap C')+P(A\cap B'\cap C)+P(A'\cap B\cap C)
\displaystyle =\frac{4}{5}\cdot\frac{3}{4}\cdot\frac{1}{3}+\frac{4}{5}\cdot\frac{1}{4}\cdot\frac{2}{3}+\frac{1}{5}\cdot\frac{3}{4}\cdot\frac{2}{3}
\displaystyle =\frac{1}{5}+\frac{2}{15}+\frac{1}{10}
\displaystyle =\frac{6+4+3}{30}
\displaystyle =\frac{13}{30}
\displaystyle \therefore \text{Required probability }=\frac{13}{30}
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\displaystyle \textbf{Question 23. }\text{Given two independent events }A\text{ and }B\text{ such that }P(A)=0.3\text{ and } \\ P(B)=0.6.\text{ Find}
\displaystyle \text{(i) }P(A\text{ and not }B)\qquad\text{(ii) }P(A\text{ or }B)
\displaystyle \text{Answer:}
\displaystyle \text{Since }A\text{ and }B\text{ are independent,}
\displaystyle P(A\cap B)=P(A)P(B)=0.3\times0.6=0.18
\displaystyle \text{(i) }P(A\cap B')=P(A)P(B')
\displaystyle =0.3\times(1-0.6)
\displaystyle =0.3\times0.4
\displaystyle =0.12
\displaystyle \text{(ii) }P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle =0.3+0.6-0.18
\displaystyle =0.72
\displaystyle \therefore P(A\text{ and not }B)=0.12,\qquad P(A\text{ or }B)=0.72
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\displaystyle \textbf{Question 24. }\text{Given that }E\text{ and }F\text{ are events such that }P(E)=\frac{6}{11},
\displaystyle P(F)=\frac{5}{11} \text{and }P(E\cap F)=\frac{4}{11}.\text{ Find }P\left(\frac{E}{F}\right)\text{ and }P\left(\frac{F}{E}\right).
\displaystyle \text{Answer:}
\displaystyle P\left(\frac{E}{F}\right)=\frac{P(E\cap F)}{P(F)}
\displaystyle =\frac{\frac{4}{11}}{\frac{5}{11}}
\displaystyle =\frac{4}{5}
\displaystyle P\left(\frac{F}{E}\right)=\frac{P(E\cap F)}{P(E)}
\displaystyle =\frac{\frac{4}{11}}{\frac{6}{11}}
\displaystyle =\frac{2}{3}
\displaystyle \therefore P\left(\frac{E}{F}\right)=\frac{4}{5},\qquad P\left(\frac{F}{E}\right)=\frac{2}{3}
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\displaystyle \textbf{Question 25. }\text{In answering a question on a multiple choice test, a student either}
\displaystyle \text{knows the answer or guesses. Let }\frac{3}{5}\text{ be the probability that he knows the answer and }
\displaystyle \frac{2}{5}\text{ be the} \text{probability that he guesses. Assuming that a student who guesses at the}
\displaystyle \text{answer will be correct with probability }\frac{1}{3}.\text{ What is the probability that the student}
\displaystyle \text{knows the answer given that he answered it correctly?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }K=\text{event that the student knows the answer}
\displaystyle \text{and }C=\text{event that the answer is correct}
\displaystyle P(K)=\frac{3}{5},\qquad P(K')=\frac{2}{5}
\displaystyle P(C|K)=1
\displaystyle P(C|K')=\frac{1}{3}
\displaystyle \text{By Bayes' theorem,}
\displaystyle P(K|C)=\frac{P(K)\,P(C|K)}{P(K)\,P(C|K)+P(K')\,P(C|K')}
\displaystyle =\frac{\frac{3}{5}\times1}{\frac{3}{5}\times1+\frac{2}{5}\times\frac{1}{3}}
\displaystyle =\frac{\frac{3}{5}}{\frac{3}{5}+\frac{2}{15}}
\displaystyle =\frac{\frac{3}{5}}{\frac{11}{15}}
\displaystyle =\frac{9}{11}
\displaystyle \therefore \text{The required probability is }\frac{9}{11}.
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\displaystyle \textbf{Question 26. }\text{Assume that the chances of a patient having a heart attack is } 40\%. 
\displaystyle \text{Assuming that a}   \text{meditation and yoga course reduces the risk of heart attack by } 
\displaystyle 30\%\text{ and prescription of certain drug}  \text{reduces its chance by }25\%. \text{ At a time,} 
\displaystyle \text{a patient can choose anyone of the two options with equal probabilities.}    
\displaystyle \text{It is given that after going through one of the two options, the patient selected} 
\displaystyle \text{at random suffers a heart attack.}   \text{Find the probability that the patient followed} 
\displaystyle \text{a course of meditation and yoga.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }M=\text{patient followed meditation and yoga course}
\displaystyle \text{and }D=\text{patient followed drug prescription}
\displaystyle P(M)=P(D)=\frac{1}{2}
\displaystyle \text{Initial probability of heart attack }=\frac{40}{100}=\frac{2}{5}
\displaystyle P(H|M)=\frac{2}{5}\left(1-\frac{30}{100}\right)=\frac{2}{5}\cdot\frac{7}{10}=\frac{7}{25}
\displaystyle P(H|D)=\frac{2}{5}\left(1-\frac{25}{100}\right)=\frac{2}{5}\cdot\frac{3}{4}=\frac{3}{10}
\displaystyle \text{By Bayes' theorem,}
\displaystyle P(M|H)=\frac{P(M)P(H|M)}{P(M)P(H|M)+P(D)P(H|D)}
\displaystyle =\frac{\frac{1}{2}\cdot\frac{7}{25}}{\frac{1}{2}\cdot\frac{7}{25}+\frac{1}{2}\cdot\frac{3}{10}}
\displaystyle =\frac{\frac{7}{25}}{\frac{7}{25}+\frac{3}{10}}
\displaystyle =\frac{\frac{7}{25}}{\frac{29}{50}}
\displaystyle =\frac{14}{29}
\displaystyle \therefore \text{Required probability }=\frac{14}{29}
\\

\displaystyle \textbf{Question 27. }\text{A girl throws a die. If she gets a }5\text{ or }6,\text{ she tosses a}
\displaystyle \text{coin three times and notes the number of heads.}  \text{If she gets }1,2,3\text{ or }4,\text{ she tosses} 
\displaystyle \text{a coin two times and notes the number of heads obtained.}   \text{If she obtained exactly} 
\displaystyle \text{two heads, what is the probability that she threw }1,2,3\text{ or }4\text{ with the die?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=\text{getting }1,2,3\text{ or }4\text{ on the die}
\displaystyle \text{and }B=\text{getting }5\text{ or }6\text{ on the die}
\displaystyle P(A)=\frac{4}{6}=\frac{2}{3},\qquad P(B)=\frac{2}{6}=\frac{1}{3}
\displaystyle \text{Let }E=\text{event of getting exactly two heads}
\displaystyle P(E|A)=\frac{{}^{2}C_{2}}{2^{2}}=\frac{1}{4}
\displaystyle P(E|B)=\frac{{}^{3}C_{2}}{2^{3}}=\frac{3}{8}
\displaystyle \text{By Bayes' theorem,}
\displaystyle P(A|E)=\frac{P(A)P(E|A)}{P(A)P(E|A)+P(B)P(E|B)}
\displaystyle =\frac{\frac{2}{3}\cdot\frac{1}{4}}{\frac{2}{3}\cdot\frac{1}{4}+\frac{1}{3}\cdot\frac{3}{8}}
\displaystyle =\frac{\frac{1}{6}}{\frac{1}{6}+\frac{1}{8}}
\displaystyle =\frac{\frac{1}{6}}{\frac{7}{24}}
\displaystyle =\frac{4}{7}
\displaystyle \therefore \text{Required probability }=\frac{4}{7}
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\displaystyle \textbf{Question 28. }\text{A card from a pack of }52\text{ cards is lost. From the remaining cards}
\displaystyle \text{of the pack, two cards are drawn at random}   \text{and are found to both of clubs. Find the}
\displaystyle \text{probability of the lost card being of clubs.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=\text{lost card is a club}
\displaystyle \text{and }E=\text{two drawn cards are clubs}
\displaystyle P(A)=\frac{13}{52}=\frac{1}{4},\qquad P(A')=\frac{39}{52}=\frac{3}{4}
\displaystyle P(E|A)=\frac{{}^{12}C_{2}}{{}^{51}C_{2}}
\displaystyle P(E|A')=\frac{{}^{13}C_{2}}{{}^{51}C_{2}}
\displaystyle \text{By Bayes' theorem,}
\displaystyle P(A|E)=\frac{P(A)P(E|A)}{P(A)P(E|A)+P(A')P(E|A')}
\displaystyle =\frac{\frac{1}{4}\cdot\frac{{}^{12}C_{2}}{{}^{51}C_{2}}}{\frac{1}{4}\cdot\frac{{}^{12}C_{2}}{{}^{51}C_{2}}+\frac{3}{4}\cdot\frac{{}^{13}C_{2}}{{}^{51}C_{2}}}
\displaystyle =\frac{{}^{12}C_{2}}{{}^{12}C_{2}+3\cdot{}^{13}C_{2}}
\displaystyle =\frac{66}{66+3\cdot78}
\displaystyle =\frac{66}{300}
\displaystyle =\frac{11}{50}
\displaystyle \therefore \text{Required probability }=\frac{11}{50}
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\displaystyle \textbf{Question 29. }\text{A man is known to speak truth }3\text{ out of }5\text{ times. He throws a}
\displaystyle\text{die and reports that it is a number}   \text{greater than }4.\text{ Find the probability that it is}
\displaystyle \text{actually a number greater than }4.
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=\text{number greater than }4
\displaystyle \text{and }E=\text{man reports that the number is greater than }4
\displaystyle P(A)=\frac{2}{6}=\frac{1}{3},\qquad P(A')=\frac{2}{3}
\displaystyle P(E|A)=\frac{3}{5}
\displaystyle P(E|A')=\frac{2}{5}
\displaystyle \text{By Bayes' theorem,}
\displaystyle P(A|E)=\frac{P(A)P(E|A)}{P(A)P(E|A)+P(A')P(E|A')}
\displaystyle =\frac{\frac{1}{3}\cdot\frac{3}{5}}{\frac{1}{3}\cdot\frac{3}{5}+\frac{2}{3}\cdot\frac{2}{5}}
\displaystyle =\frac{\frac{1}{5}}{\frac{1}{5}+\frac{4}{15}}
\displaystyle =\frac{\frac{1}{5}}{\frac{7}{15}}
\displaystyle =\frac{3}{7}
\displaystyle \therefore \text{Required probability }=\frac{3}{7}
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\displaystyle \textbf{Question 30. }\text{A box contains }4\text{ orange and }4\text{ green balls, another box contains }3\text{ orange}
\displaystyle \text{and 5 green balls. One of the two boxes is selected at random and a ball is drawn from the box,}
\displaystyle \text{which is found to be orange. Find the probability that the ball is drawn from the first box.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }B_{1}=\text{selection of first box},\qquad B_{2}=\text{selection of second box}
\displaystyle \text{and }O=\text{event of drawing an orange ball}
\displaystyle P(B_{1})=P(B_{2})=\frac{1}{2}
\displaystyle P(O|B_{1})=\frac{4}{8}=\frac{1}{2}
\displaystyle P(O|B_{2})=\frac{3}{8}
\displaystyle \text{By Bayes' theorem,}
\displaystyle P(B_{1}|O)=\frac{P(B_{1})P(O|B_{1})}{P(B_{1})P(O|B_{1})+P(B_{2})P(O|B_{2})}
\displaystyle =\frac{\frac{1}{2}\cdot\frac{1}{2}}{\frac{1}{2}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{3}{8}}
\displaystyle =\frac{\frac{1}{4}}{\frac{1}{4}+\frac{3}{16}}
\displaystyle =\frac{\frac{1}{4}}{\frac{7}{16}}
\displaystyle =\frac{4}{7}
\displaystyle \therefore \text{Required probability }=\frac{4}{7}
\\

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