Class 12: Probability – Exercise 31.5

Question 1:
A bag contains $\displaystyle 6$ black and $\displaystyle 3$ white balls. Another bag contains $\displaystyle 5$ black and $\displaystyle 4$ white balls. If one ball is drawn from each bag, find the probability that these two balls are of the same colour.
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given :}$
$\displaystyle \text{Bag 1} = (3W + 6B)\ \text{balls}$
$\displaystyle \text{Bag 2} = (5B + 4W)\ \text{balls}$
$\displaystyle P(\text{balls of same colour are drawn}) = P(\text{both black}) + P(\text{both white})$
$\displaystyle = \frac{6}{9}\times\frac{5}{9} + \frac{3}{9}\times\frac{4}{9}$
$\displaystyle = \frac{30}{81} + \frac{12}{81}$
$\displaystyle = \frac{42}{81}$
$\displaystyle = \frac{14}{27}$

Question 2:
A bag contains $\displaystyle 3$ red and $\displaystyle 5$ black balls and a second bag contains $\displaystyle 6$ red and $\displaystyle 4$ black balls. A ball is drawn from each bag. Find the probability that one is red and the other is black.
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given :}$
$\displaystyle \text{Bag 1} = (3R + 5B)\ \text{balls}$
$\displaystyle \text{Bag 2} = (6R + 4B)\ \text{balls}$
$\displaystyle P(\text{one is red and one is black}) = P(\text{red from bag 1 and black from bag 2}) + P(\text{red from bag 2 and black from bag 1})$
$\displaystyle = \frac{3}{8}\times\frac{4}{10} + \frac{5}{8}\times\frac{6}{10}$
$\displaystyle = \frac{12}{80} + \frac{30}{80}$
$\displaystyle = \frac{42}{80}$
$\displaystyle = \frac{21}{40}$

Question 3:
Two balls are drawn at random with replacement from a box containing $\displaystyle 10$ black and $\displaystyle 8$ red balls. Find the probability that
(i) both the balls are red.
(ii) first ball is black and second is red.
(iii) one of them is black and other is red.
$\displaystyle \text{Answer:}$

$\displaystyle \text{Total balls}=18$
$\displaystyle P(\text{red})=\frac{8}{18}=\frac{4}{9}$
$\displaystyle P(\text{black})=\frac{10}{18}=\frac{5}{9}$

$\displaystyle (i);P(\text{both red})=\frac{4}{9}\times\frac{4}{9}=\frac{16}{81}$

$\displaystyle (ii);P(\text{first black, second red})=\frac{5}{9}\times\frac{4}{9}=\frac{20}{81}$

$\displaystyle (iii);P(\text{one black and one red})=\frac{5}{9}\times\frac{4}{9}+\frac{4}{9}\times\frac{5}{9}=\frac{40}{81}$

Question 4:
Two cards are drawn successively without replacement from a well-shuffled deck of $\displaystyle 52$ cards. Find the probability of exactly one ace.
$\displaystyle \text{Answer:}$
$\displaystyle P(\text{exactly one ace}) = P(\text{first card is ace and second is not}) + P(\text{first card is not ace and second is ace})$
$\displaystyle = \frac{4}{52}\times\frac{48}{51} + \frac{48}{52}\times\frac{4}{51}$
$\displaystyle = \frac{192 + 192}{52\times51}$
$\displaystyle = \frac{384}{52\times51}$
$\displaystyle = \frac{32}{221}$

Question 5:
A speaks truth in $\displaystyle 75\%$ and B in $\displaystyle 80\%$ of the cases. In what percentage of cases are they likely to contradict each other in narrating the same incident?
$\displaystyle \text{Answer:}$
$\displaystyle P(\text{both narrating different incident}) = P(\text{B lies and A speaks the truth}) + P(\text{A lies and B speaks the truth})$
$\displaystyle = P(A\cap\overline{B}) + P(\overline{A}\cap B)$
$\displaystyle = P(A)P(\overline{B}) + P(\overline{A})P(B)$
$\displaystyle = 0.75(1-0.8) + (1-0.75)0.8$
$\displaystyle = 0.75\times0.2 + 0.25\times0.8$
$\displaystyle = 0.15 + 0.2$
$\displaystyle = 0.35$
$\displaystyle = 35\%$

Question 6:
Kamal and Monica appeared for an interview for two vacancies. The probability of Kamal’s selection is $\displaystyle 1/3$ and that of Monica’s selection is $\displaystyle 1/5$ . Find the probability that
(i) both of them will be selected
(ii) none of them will be selected
(iii) at least one of them will be selected
(iv) only one of them will be selected.
$\displaystyle \text{Answer:}$
$\displaystyle P(\text{Kamal gets selected}) = P(A) = \frac{1}{3}$
$\displaystyle P(\text{Monica gets selected}) = P(B) = \frac{1}{5}$
$\displaystyle (i)\ P(\text{both get selected}) = P(A)\times P(B)$
$\displaystyle = \frac{1}{3}\times\frac{1}{5}$
$\displaystyle = \frac{1}{15}$
$\displaystyle (ii)\ P(\text{none of them get selected}) = P(\overline{A})\times P(\overline{B})$
$\displaystyle = [1-P(A)][1-P(B)]$
$\displaystyle = \left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)$
$\displaystyle = \frac{2}{3}\times\frac{4}{5}$
$\displaystyle = \frac{8}{15}$
$\displaystyle (iii)\ P(\text{at least one of them gets selected}) = P(A\cup B)$
$\displaystyle = P(A)+P(B)-P(A\cap B)$
$\displaystyle = P(A)+P(B)-P(A)\times P(B)$
$\displaystyle = \frac{1}{3}+\frac{1}{5}-\frac{1}{3}\times\frac{1}{5}$
$\displaystyle = \frac{1}{3}+\frac{1}{5}-\frac{1}{15}$
$\displaystyle = \frac{7}{15}$
$\displaystyle (iv)\ P(\text{one of them gets selected}) = P(\overline{A})P(B)+P(\overline{B})P(A)$
$\displaystyle = P(B)[1-P(A)] + P(A)[1-P(B)]$
$\displaystyle = \frac{1}{5}\left(1-\frac{1}{3}\right)+\frac{1}{3}\left(1-\frac{1}{5}\right)$
$\displaystyle = \frac{2}{15}+\frac{4}{15}$
$\displaystyle = \frac{6}{15}$
$\displaystyle = \frac{2}{5}$

Question 7:
A bag contains $\displaystyle 3$ white, $\displaystyle 4$ red and $\displaystyle 5$ black balls. Two balls are drawn one after the other, without replacement. What is the probability that one is white and the other is black?
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given : Box} = (3W + 4R + 5B)\ \text{balls}$
$\displaystyle P(\text{one white and one black}) = P(\text{first white and second black}) + P(\text{first black and second white})$
$\displaystyle = \frac{3}{12}\times\frac{5}{11} + \frac{5}{12}\times\frac{3}{11}$
$\displaystyle = \frac{15}{132} + \frac{15}{132}$
$\displaystyle = \frac{30}{132}$
$\displaystyle = \frac{5}{22}$

Question 8:
A bag contains $\displaystyle 8$ red and $\displaystyle 6$ green balls. Three balls are drawn one after another without replacement. Find the probability that at least two balls drawn are green.
$\displaystyle \text{Answer:}$
$\displaystyle P(\text{at least 2 balls are green}) = 1 - P(\text{at most one ball is green})$
$\displaystyle = 1 - [P(\text{first green}) + P(\text{second green}) + P(\text{third green}) + P(\text{no green})]$
$\displaystyle = 1 - \left[\frac{6}{14}\times\frac{8}{13}\times\frac{7}{12} + \frac{8}{14}\times\frac{6}{13}\times\frac{7}{12} + \frac{8}{14}\times\frac{7}{13}\times\frac{6}{12} + \frac{8}{14}\times\frac{7}{13}\times\frac{6}{12}\right]$
$\displaystyle = 1 - \left[\frac{336}{2184} + \frac{336}{2184} + \frac{336}{2184} + \frac{336}{2184}\right]$
$\displaystyle = 1 - \frac{1344}{2184}$
$\displaystyle = \frac{840}{2184}$
$\displaystyle = \frac{5}{13}$

Question 9:
Arun and Tarun appeared for an interview for two vacancies. The probability of Arun’s selection is $\displaystyle 1/4$ and that of Tarun’s rejection is $\displaystyle 2/3$ . Find the probability that at least one of them will be selected.
$\displaystyle \text{Answer:}$
$\displaystyle P(\text{Arun gets selected}) = P(A) = \frac{1}{4}$
$\displaystyle P(\text{Tarun gets rejected}) = P(\overline{B}) = \frac{2}{3}$
$\displaystyle \Rightarrow P(\text{Tarun gets selected}) = 1-\frac{2}{3} = \frac{1}{3}$
$\displaystyle P(\text{at least one of them is selected}) = P(A\cup B)$
$\displaystyle \Rightarrow P(A\cup B) = P(A)+P(B)-P(A\cap B)$
$\displaystyle = P(A)+P(B)-P(A)\times P(B)$
$\displaystyle = \frac{1}{4}+\frac{1}{3}-\frac{1}{4}\times\frac{1}{3}$
$\displaystyle = \frac{3+4-1}{12}$
$\displaystyle = \frac{1}{2}$

Question 10:
A and B toss a coin alternately till one of them gets a head and wins the game. If A starts the game, find the probability that B will win the game.
$\displaystyle \text{Answer:}$
$\displaystyle P(B\ \text{winning the game}) = P(\text{head at }2^{\text{nd}}\text{ turn}) + P(\text{head at }4^{\text{th}}\text{ turn}) + \cdots$
$\displaystyle = \frac{1}{2}\times\frac{1}{2} + \frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2} + \cdots$
$\displaystyle = \left(\frac{1}{2}\right)^{2} + \left(\frac{1}{2}\right)^{4} + \left(\frac{1}{2}\right)^{6} + \left(\frac{1}{2}\right)^{8} + \cdots$
$\displaystyle = \frac{1}{4}\left[1 + \left(\frac{1}{2}\right)^{2} + \left(\frac{1}{2}\right)^{4} + \left(\frac{1}{2}\right)^{6} + \cdots\right]$
$\displaystyle = \frac{1}{4}\left[\frac{1}{1-\frac{1}{4}}\right]\ \text{[For infinite GP : }1+a+a^{2}+a^{3}+\cdots=\frac{1}{1-a}\text{]}$
$\displaystyle = \frac{1}{4}\times\frac{4}{3}$
$\displaystyle = \frac{1}{3}$

Question 11:
Two cards are drawn from a well shuffled pack of $\displaystyle 52$ cards, one after another without replacement. Find the probability that one of these is red card and the other a black card?
$\displaystyle \text{Answer:}$
$\displaystyle P(\text{one red and one black}) = P(\text{first red and second black}) + P(\text{first black and second red})$
$\displaystyle = \frac{26}{52}\times\frac{26}{51} + \frac{26}{52}\times\frac{26}{51}\ \text{[without replacement]}$
$\displaystyle = \frac{13}{51} + \frac{13}{51}$
$\displaystyle = \frac{26}{51}$

Question 12:
Tickets are numbered from $\displaystyle 1$ to $\displaystyle 10$ . Two tickets are drawn one after the other at random. Find the probability that the number on one of the tickets is a multiple of $\displaystyle 5$ and on the other a multiple of $\displaystyle 4$ .
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that }5\text{ and }10\text{ are multiples of }5,\text{ while }4\text{ and }8\text{ are multiples of }4.$
$\displaystyle P(\text{multiple of }5)=\frac{2}{10}=\frac{1}{5}$
$\displaystyle P(\text{multiple of }4)=\frac{2}{10}=\frac{1}{5}$
$\displaystyle P(\text{multiple of }5\text{ and multiple of }4)=P(\text{multiple of }5\text{ on first card and multiple of }4\text{ on second card})$
$\displaystyle +P(\text{multiple of }4\text{ on first card and multiple of }5\text{ on second card})$
$\displaystyle = \frac{2}{10}\times\frac{2}{9}+\frac{2}{10}\times\frac{2}{9}\ \text{[without replacement]}$
$\displaystyle = \frac{4}{90}+\frac{4}{90}$
$\displaystyle = \frac{8}{90}$
$\displaystyle = \frac{4}{45}$

Question 13:
In a family, the husband tells a lie in $\displaystyle 30\%$ cases and the wife in $\displaystyle 35\%$ cases. Find the probability that both contradict each other on the same fact.
$\displaystyle \text{Answer:}$
$\displaystyle \text{It is given that the husband lies in }30\%\text{ of the cases, while the wife lies in }35\%\text{ cases}$
$\displaystyle P(\text{both will contradict each other on the same fact}) = P(\text{husband lies but wife tells the truth}) + P(\text{wife lies but husband tells the truth})$
$\displaystyle = 0.3(1-0.35) + (1-0.3)0.35$
$\displaystyle = 0.3\times0.65 + 0.7\times0.35$
$\displaystyle = 0.195 + 0.245$
$\displaystyle = 0.44$
$\displaystyle = 44\%$

Question 14:
A husband and wife appear in an interview for two vacancies for the same post. The probability of husband’s selection is $\displaystyle 1/7$ and that of wife’s selection is $\displaystyle 1/5$ . What is the probability that
(i) both of them will be selected?
(ii) only one of them will be selected?
(iii) none of them will be selected?
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }$
$\displaystyle P(\text{husband will be selected}) = P(A) = \frac{1}{7}$
$\displaystyle P(\text{wife will be selected}) = P(B) = \frac{1}{5}$
$\displaystyle P(\text{both will be selected}) = P(A\cap B)$
$\displaystyle = P(A)\times P(B)$
$\displaystyle = \frac{1}{7}\times\frac{1}{5}$
$\displaystyle = \frac{1}{35}$
$\displaystyle \text{(ii) }$
$\displaystyle P(\text{husband will be selected}) = P(A) = \frac{1}{7}$
$\displaystyle P(\text{wife will be selected}) = P(B) = \frac{1}{5}$
$\displaystyle P(\text{only one of them will be selected}) = P(A)P(\overline{B}) + P(\overline{A})P(B)$
$\displaystyle = \frac{1}{7}\left(1-\frac{1}{5}\right)+\frac{1}{5}\left(1-\frac{1}{7}\right)$
$\displaystyle = \frac{4}{35}+\frac{6}{35}$
$\displaystyle = \frac{10}{35}$
$\displaystyle = \frac{2}{7}$
$\displaystyle \text{(iii) }$
$\displaystyle P(\text{husband will be selected}) = P(A) = \frac{1}{7}$
$\displaystyle P(\text{wife will be selected}) = P(B) = \frac{1}{5}$
$\displaystyle P(\text{none of them will be selected}) = P(\overline{A}\cap\overline{B})$
$\displaystyle = P(\overline{A})\times P(\overline{B})$
$\displaystyle = \left(1-\frac{1}{7}\right)\left(1-\frac{1}{5}\right)$
$\displaystyle = \frac{24}{35}$

Question 15:
A bag contains $\displaystyle 7$ white, $\displaystyle 5$ black and $\displaystyle 4$ red balls. Four balls are drawn without replacement. Find the probability that at least three balls are black.
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given : Bag} = (7W + 5B + 4R)\ \text{balls}$
$\displaystyle P(\text{at least 3 balls are black}) = P(\text{exactly 3 black}) + P(\text{all 4 black})$
$\displaystyle = \left(\frac{5}{16}\times\frac{4}{15}\times\frac{3}{14}\times\frac{11}{13}\right) + \left(\frac{5}{16}\times\frac{4}{15}\times\frac{3}{14}\times\frac{2}{13}\right)$
$\displaystyle = \frac{11}{14\times13} + \frac{1}{2\times14\times13}$
$\displaystyle = \frac{22+1}{364}$
$\displaystyle = \frac{23}{364}$

Question 16:
$\displaystyle A,B,$ and $\displaystyle C$ are independent witness of an event which is known to have occurred. A speaks the truth three times out of four, B four times out of five and C five times out of six. What is the probability that the occurrence will be reported truthfully by majority of three witnesses?
$\displaystyle \text{Answer:}$
$\displaystyle P(\text{A speaks truth})=\frac{3}{4}$
$\displaystyle P(\text{B speaks truth})=\frac{4}{5}$
$\displaystyle P(\text{C speaks truth})=\frac{5}{6}$
$\displaystyle P(\text{majority speaks truth})=P(\text{two speak truth})+P(\text{all speak truth})$
$\displaystyle =P(A)P(B)[1-P(C)]+P(A)P(C)[1-P(B)]+P(C)P(B)[1-P(A)]+P(A)P(B)P(C)$
$\displaystyle =\frac{3}{4}\times\frac{4}{5}\left(1-\frac{5}{6}\right)+\frac{3}{4}\times\frac{5}{6}\left(1-\frac{4}{5}\right)+\frac{4}{5}\times\frac{5}{6}\left(1-\frac{3}{4}\right)+\frac{3}{4}\times\frac{4}{5}\times\frac{5}{6}$
$\displaystyle =\frac{12}{120}+\frac{15}{120}+\frac{20}{120}+\frac{60}{120}$
$\displaystyle =\frac{107}{120}$

Question 17:
A bag contains $\displaystyle 4$ white balls and $\displaystyle 2$ black balls. Another bag contains $\displaystyle 3$ white balls and $\displaystyle 5$ black balls. If one ball is drawn from each bag, find the probability that
(i) both are white
(ii) both are black
(iii) one is white and one is black
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given :}$
$\displaystyle \text{Bag 1}=(4W+2B)\ \text{balls}$
$\displaystyle \text{Bag 2}=(3W+5B)\ \text{balls}$
$\displaystyle (i)\ P(\text{both are white})=\frac{4}{6}\times\frac{3}{8}$
$\displaystyle =\frac{1}{4}$
$\displaystyle (ii)\ P(\text{both are black})=\frac{2}{6}\times\frac{5}{8}$
$\displaystyle =\frac{5}{24}$
$\displaystyle (iii)\ P(\text{one is white and one is black})=P(\text{white from bag 1 and black from bag 2})+P(\text{white from bag 2 and black from bag 1})$
$\displaystyle =\frac{4}{6}\times\frac{5}{8}+\frac{3}{8}\times\frac{2}{6}$
$\displaystyle =\frac{20}{48}+\frac{6}{48}$
$\displaystyle =\frac{26}{48}$
$\displaystyle =\frac{13}{24}$

Question 18:
A bag contains $\displaystyle 4$ white, $\displaystyle 7$ black and $\displaystyle 5$ red balls. $\displaystyle 4$ balls are drawn with replacement. What is the probability that at least two are white?
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given : Bag}=(4W+5R+7B)\ \text{balls}$
$\displaystyle P(\text{at least 2 white balls})=1-P(\text{maximum 1 white ball})$
$\displaystyle =1-[P(\text{no white})+P(\text{exactly one white})]$
$\displaystyle =1-\left[\frac{12}{16}\times\frac{12}{16}\times\frac{12}{16}\times\frac{12}{16}+\frac{4}{16}\times\frac{12}{16}\times\frac{12}{16}\times\frac{12}{16}\times4\right]$
$\displaystyle =1-\left[\frac{81}{256}+\frac{108}{256}\right]$
$\displaystyle =1-\frac{189}{256}$
$\displaystyle =\frac{256-189}{256}$
$\displaystyle =\frac{67}{256}$

Question 19:
Three cards are drawn with replacement from a well shuffled pack of $\displaystyle 52$ cards. Find the probability that the cards are a king, a queen and a jack.
$\displaystyle \text{Answer:}$
$\displaystyle P(\text{king})=P(A)=\frac{4}{52}$
$\displaystyle P(\text{queen})=P(B)=\frac{4}{52}$
$\displaystyle P(\text{jack})=P(C)=\frac{4}{52}$
$\displaystyle P(\text{king, queen and jack})=3!\times P(A)\times P(B)\times P(C)$
$\displaystyle =3\times2\times\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}$
$\displaystyle =\frac{6}{2197}$

Question 20:
A bag contains $\displaystyle 4$ red and $\displaystyle 5$ black balls, a second bag contains $\displaystyle 3$ red and $\displaystyle 7$ black balls. One ball is drawn at random from each bag, find the probability that the
(i) balls are of different colours
(ii) balls are of the same colour.
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given :}$
$\displaystyle \text{Bag A}=(4R+5B)\ \text{balls}$
$\displaystyle \text{Bag B}=(3R+7B)\ \text{balls}$
$\displaystyle (i)\ P(\text{balls of different colours})=P(\text{red from bag A and black from bag B})+P(\text{red from bag B and black from bag A})$
$\displaystyle =\frac{4}{9}\times\frac{7}{10}+\frac{3}{10}\times\frac{5}{9}$
$\displaystyle =\frac{28}{90}+\frac{15}{90}$
$\displaystyle =\frac{43}{90}$
$\displaystyle (ii)\ P(\text{balls of same colour})=P(\text{both red})+P(\text{both black})$
$\displaystyle =\frac{4}{9}\times\frac{3}{10}+\frac{5}{9}\times\frac{7}{10}$
$\displaystyle =\frac{12}{90}+\frac{35}{90}$
$\displaystyle =\frac{47}{90}$

Question 21:
A can hit a target $\displaystyle 3$ times in $\displaystyle 6$ shots, B : $\displaystyle 2$ times in $\displaystyle 6$ shots and C : $\displaystyle 4$ times in $\displaystyle 4$ shots. They fix a volley. What is the probability that at least $\displaystyle 2$ shots hit?
$\displaystyle \text{Answer:}$
$\displaystyle P(\text{A hits the target})=\frac{3}{6}$
$\displaystyle P(\text{B hits the target})=\frac{2}{6}$
$\displaystyle P(\text{C hits the target})=\frac{4}{4}=1$
$\displaystyle P(\text{at least 2 shots hit})=P(\text{exactly 2 shots hit})+P(\text{all 3 shots hit})$
$\displaystyle =\frac{3}{6}\left(1-\frac{2}{6}\right)+\frac{2}{6}\left(1-\frac{3}{6}\right)+\frac{3}{6}\times\frac{2}{6}\times1$
$\displaystyle \text{(Here, the probability of C hitting the target is 1, so it will always hit.)}$
$\displaystyle \text{When exactly 2 shots are hit, then either A hits or B hits.}$
$\displaystyle =\frac{3}{6}\times\frac{4}{6}+\frac{2}{6}\times\frac{3}{6}+\frac{6}{36}$
$\displaystyle =\frac{12+6+6}{36}$
$\displaystyle =\frac{24}{36}$
$\displaystyle =\frac{2}{3}$

Question 22:
The probability of student A passing an examination is $\displaystyle 2/9$ and of student B passing is $\displaystyle 5/9$ . Assuming the two events : ‘A passes’, ‘B passes’ as independent, find the probability of :
(i) only A passing the examination
(ii) only one of them passing the examination.
$\displaystyle \text{Answer:}$
$\displaystyle P(\text{A passing examination})=\frac{2}{9}$
$\displaystyle P(\text{B passing examination})=\frac{5}{9}$
$\displaystyle (i)\ P(\text{only A passing examination})=P(\text{A passes})P(\text{B fails})$
$\displaystyle =\frac{2}{9}\left(1-\frac{5}{9}\right)$
$\displaystyle =\frac{2}{9}\times\frac{4}{9}$
$\displaystyle =\frac{8}{81}$
$\displaystyle (ii)\ P(\text{only one of them passing examination})=P(\text{A passes and B fails})+P(\text{B passes and A fails})$
$\displaystyle =\frac{2}{9}\left(1-\frac{5}{9}\right)+\frac{5}{9}\left(1-\frac{2}{9}\right)$
$\displaystyle =\frac{8}{81}+\frac{35}{81}$
$\displaystyle =\frac{43}{81}$

Question 23:
There are three urns $\displaystyle A,B,$ and $\displaystyle C$ . Urn $\displaystyle A$ contains $\displaystyle 4$ red balls and $\displaystyle 3$ black balls. Urn $\displaystyle B$ contains $\displaystyle 5$ red balls and $\displaystyle 4$ black balls. Urn $\displaystyle C$ contains $\displaystyle 4$ red and $\displaystyle 4$ black balls. One ball is drawn from each of these urns. What is the probability that $\displaystyle 3$ balls drawn consist of $\displaystyle 2$ red balls and a black ball?
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given :}$
$\displaystyle \text{Urn A}=(4R+3B)$
$\displaystyle \text{Urn B}=(5R+4B)$
$\displaystyle \text{Urn C}=(4R+4B)$
$\displaystyle P(\text{two red and one black})=P(\text{red from urn A})P(\text{black from urn B})P(\text{red from urn C})+P(\text{red from urn A})P(\text{red from urn B})P(\text{black from urn C})+P(\text{black from urn A})P(\text{red from urn B})P(\text{red from urn C})$
$\displaystyle =\frac{4}{7}\times\frac{4}{9}\times\frac{4}{8}+\frac{4}{7}\times\frac{5}{9}\times\frac{4}{8}+\frac{3}{7}\times\frac{5}{9}\times\frac{4}{8}$
$\displaystyle =\frac{16}{126}+\frac{20}{126}+\frac{15}{126}$
$\displaystyle =\frac{51}{126}$
$\displaystyle =\frac{17}{42}$

Question 24:
X is taking up subjects – Mathematics, Physics and Chemistry in the examination. His probabilities of getting grade A in these subjects are $\displaystyle 0.2, 0.3$ and $\displaystyle 0.5$ respectively. Find the probability that he gets
(i) Grade A in all subjects
(ii) Grade A in no subject
(iii) Grade A in two subjects.
$\displaystyle \text{Answer:}$
$\displaystyle P(\text{A grade in Maths})=P(A)=0.2$
$\displaystyle P(\text{A grade in Physics})=P(B)=0.3$
$\displaystyle P(\text{A grade in Chemistry})=P(C)=0.5$
$\displaystyle (i)\ P(\text{grade A in all subjects})=P(A)\times P(B)\times P(C)$
$\displaystyle =0.2\times0.3\times0.5$
$\displaystyle =0.03$
$\displaystyle (ii)\ P(\text{grade A in no subject})=P(\overline{A})\times P(\overline{B})\times P(\overline{C})$
$\displaystyle =(1-0.2)\times(1-0.3)\times(1-0.5)$
$\displaystyle =0.8\times0.7\times0.5$
$\displaystyle =0.28$
$\displaystyle (iii)\ P(\text{grade A in two subjects})=P(\overline{A})P(B)P(C)+P(A)P(\overline{B})P(C)+P(A)P(B)P(\overline{C})$
$\displaystyle =(1-0.2)\times0.3\times0.5+0.2\times(1-0.3)\times0.5+0.2\times0.3\times(1-0.5)$
$\displaystyle =0.8\times0.3\times0.5+0.2\times0.7\times0.5+0.2\times0.3\times0.5$
$\displaystyle =0.12+0.07+0.03$
$\displaystyle =0.22$

Question 25:
A and B take turns in throwing two dice, the first to throw $\displaystyle 9$ being awarded the prize. Show that their chance of winning are in the ratio $\displaystyle 9:8$ .
$\displaystyle \text{Answer:}$
$\displaystyle \text{Total number of events}=36$
$\displaystyle P(\text{getting }9)=\frac{4}{36}=\frac{1}{9}$
$\displaystyle P(A\ \text{winning})=P(\text{getting }9\ \text{in first throw})+P(\text{getting }9\ \text{in third throw})+\cdots$
$\displaystyle =\frac{1}{9}+\left(1-\frac{1}{9}\right)\left(1-\frac{1}{9}\right)\times\frac{1}{9}+\cdots$
$\displaystyle =\frac{1}{9}\left[1+\frac{64}{81}+\left(\frac{64}{81}\right)^{2}+\cdots\right]$
$\displaystyle =\frac{1}{9}\left[\frac{1}{1-\frac{64}{81}}\right]\ \text{[For infinite GP : }1+a+a^{2}+a^{3}+\cdots=\frac{1}{1-a}\text{]}$
$\displaystyle =\frac{1}{9}\times\frac{81}{17}$
$\displaystyle =\frac{9}{17}$
$\displaystyle P(B\ \text{winning})=P(\text{getting }9\ \text{in second throw})+P(\text{getting }9\ \text{in fourth throw})+\cdots$
$\displaystyle =\left(1-\frac{1}{9}\right)\frac{1}{9}+\left(1-\frac{1}{9}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{9}\right)\times\frac{1}{9}+\cdots$
$\displaystyle =\frac{8}{81}\left[1+\frac{64}{81}+\left(\frac{64}{81}\right)^{2}+\cdots\right]$
$\displaystyle =\frac{8}{81}\left[\frac{1}{1-\frac{64}{81}}\right]$
$\displaystyle =\frac{8}{81}\times\frac{81}{17}$
$\displaystyle =\frac{8}{17}$
$\displaystyle \therefore\ \text{Winning ratio of A to B}=\frac{\frac{9}{17}}{\frac{8}{17}}=\frac{9}{8}$

Question 26:
$\displaystyle A,B$ and $\displaystyle C$ in order toss a coin. The one to throw a head wins. What are their respective chances of winning assuming that the game may continue indefinitely?
$\displaystyle \text{Answer:}$
$\displaystyle P(A\ \text{winning})=P(\text{head in first toss})+P(\text{head in fourth toss})+\cdots$
$\displaystyle =\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}+\cdots$
$\displaystyle =\frac{1}{2}\left[1+\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{2}\right)^{6}+\cdots\right]$
$\displaystyle =\frac{1}{2}\left[\frac{1}{1-\frac{1}{8}}\right]\ \text{[For infinite GP : }1+a+a^{2}+a^{3}+\cdots=\frac{1}{1-a}\text{]}$
$\displaystyle =\frac{1}{2}\times\frac{8}{7}$
$\displaystyle =\frac{4}{7}$
$\displaystyle P(B\ \text{winning})=P(\text{head in second toss})+P(\text{head in fifth toss})+\cdots$
$\displaystyle =\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}+\cdots$
$\displaystyle =\frac{1}{4}\left[1+\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{2}\right)^{6}+\cdots\right]$
$\displaystyle =\frac{1}{4}\left[\frac{1}{1-\frac{1}{8}}\right]\ \text{[For infinite GP : }1+a+a^{2}+a^{3}+\cdots=\frac{1}{1-a}\text{]}$
$\displaystyle =\frac{1}{4}\times\frac{8}{7}$
$\displaystyle =\frac{2}{7}$
$\displaystyle P(C\ \text{winning})=P(\text{head in third toss})+P(\text{head in sixth toss})+\cdots$
$\displaystyle =\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}+\cdots$
$\displaystyle =\frac{1}{8}\left[1+\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{2}\right)^{6}+\cdots\right]$
$\displaystyle =\frac{1}{8}\left[\frac{1}{1-\frac{1}{8}}\right]\ \text{[For infinite GP : }1+a+a^{2}+a^{3}+\cdots=\frac{1}{1-a}\text{]}$
$\displaystyle =\frac{1}{8}\times\frac{8}{7}$
$\displaystyle =\frac{1}{7}$

Question 27:
Three persons $\displaystyle A,B,C$ throw a die in succession till one gets a ‘six’ and wins the game. Find their respective probabilities of winning.
$\displaystyle \text{Answer:}$
$\displaystyle P(\text{six})=\frac{1}{6}$
$\displaystyle P(\text{no six})=\frac{5}{6}$
$\displaystyle P(A\ \text{winning})=P(\text{6 in first throw})+P(\text{6 in fourth throw})+\cdots$
$\displaystyle =\frac{1}{6}+\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}+\cdots$
$\displaystyle =\frac{1}{6}\left[1+\left(\frac{5}{6}\right)^{3}+\left(\frac{5}{6}\right)^{6}+\cdots\right]$
$\displaystyle =\frac{1}{6}\left[\frac{1}{1-\frac{125}{216}}\right]\ \text{[For infinite GP : }1+a+a^{2}+a^{3}+\cdots=\frac{1}{1-a}\text{]}$
$\displaystyle =\frac{1}{6}\times\frac{216}{91}$
$\displaystyle =\frac{36}{91}$
$\displaystyle P(B\ \text{winning})=P(\text{6 in second throw})+P(\text{6 in fifth throw})+\cdots$
$\displaystyle =\frac{5}{6}\times\frac{1}{6}+\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}+\cdots$
$\displaystyle =\frac{5}{36}\left[1+\left(\frac{5}{6}\right)^{3}+\left(\frac{5}{6}\right)^{6}+\cdots\right]$
$\displaystyle =\frac{5}{36}\left[\frac{1}{1-\frac{125}{216}}\right]$
$\displaystyle =\frac{5}{36}\times\frac{216}{91}$
$\displaystyle =\frac{30}{91}$
$\displaystyle P(C\ \text{winning})=P(\text{6 in third throw})+P(\text{6 in sixth throw})+\cdots$
$\displaystyle =\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}+\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}+\cdots$
$\displaystyle =\frac{25}{216}\left[1+\left(\frac{5}{6}\right)^{3}+\left(\frac{5}{6}\right)^{6}+\cdots\right]$
$\displaystyle =\frac{25}{216}\left[\frac{1}{1-\frac{125}{216}}\right]$
$\displaystyle =\frac{25}{216}\times\frac{216}{91}$
$\displaystyle =\frac{25}{91}$

Question 28:
A and B take turns in throwing two dice, the first to throw $\displaystyle 10$ being awarded the prize, show that if A has the first throw, their chance of winning are in the ratio $\displaystyle 12:11$ .
$\displaystyle \text{Answer:}$
$\displaystyle \text{There are only three possible cases, wherein the sum of the numbers obtained after throwing 2 dice is }10,\text{ i.e. }(4,6),(5,5),(6,4).$
$\displaystyle \therefore\ P(\text{sum of the numbers is }10)=\frac{3}{36}=\frac{1}{12}$
$\displaystyle P(\text{sum of the numbers is not }10)=1-\frac{1}{12}=\frac{11}{12}$
$\displaystyle P(\text{any number other than six})=\frac{5}{6}$
$\displaystyle P(A\ \text{winning})=P(10\ \text{in first throw})+P(10\ \text{in third throw})+\cdots$
$\displaystyle =\frac{1}{12}+\frac{11}{12}\times\frac{11}{12}\times\frac{1}{12}+\cdots$
$\displaystyle =\frac{1}{12}\left[1+\left(\frac{11}{12}\right)^{2}+\left(\frac{11}{12}\right)^{4}+\cdots\right]$
$\displaystyle =\frac{1}{12}\left[\frac{1}{1-\frac{121}{144}}\right]\ \text{[For infinite GP : }1+a+a^{2}+a^{3}+\cdots=\frac{1}{1-a}\text{]}$
$\displaystyle =\frac{1}{12}\times\frac{144}{23}$
$\displaystyle =\frac{12}{23}$
$\displaystyle P(B\ \text{winning})=1-P(A\ \text{winning})=\frac{11}{23}$
$\displaystyle \text{Now,}$
$\displaystyle \frac{P(A\ \text{winning})}{P(B\ \text{winning})}=\frac{\frac{12}{23}}{\frac{11}{23}}=\frac{12}{11}$
$\displaystyle \text{Hence proved.}$

Question 29:
There are $\displaystyle 3$ red and $\displaystyle 5$ black balls in bag ‘A’; and $\displaystyle 2$ red and $\displaystyle 3$ black balls in bag ‘B’. One ball is drawn from bag ‘A’ and two from bag ‘B’. Find the probability that out of the $\displaystyle 3$ balls drawn one is red and $\displaystyle 2$ are black.
$\displaystyle \text{Answer:}$
$\displaystyle \text{It is given that bag A contains }3\text{ red and }5\text{ black balls }(3R,5B)\text{ and bag B contains } \\ 2\text{ red and }3\text{ black balls }(2R,3B).$
$\displaystyle \text{Now,}$
$\displaystyle P(\text{one red and two black})=P(\text{one red from bag A and two black from bag B})+P(\text{black ball from bag A and remaining balls from bag B})$
$\displaystyle =\frac{3}{8}\times\frac{3}{5}\times\frac{2}{4}+\frac{5}{8}\times\frac{2}{5}\times\frac{3}{4}\times2$
$\displaystyle =\frac{9}{80}+\frac{30}{80}$
$\displaystyle =\frac{39}{80}$
$\displaystyle \text{Note: }2\text{ is multiplied by second term because there are two ways to select red and black} \\ \text{balls from bag B.}$
$\displaystyle \text{While the first way is to pick black ball first, followed by red, the second way is to pick red} \\ \text{ball first, followed by black.}$

Question 30:
Fatima and John appear in an interview for two vacancies for the same post. The probability of Fatima’s selection is $\displaystyle 1/7$ and that of John’s selection is $\displaystyle 1/5$ . What is the probability that
(i) both of them will be selected?
(ii) only one of them will be selected?
(iii) none of them will be selected?
$\displaystyle \text{Answer:}$
$\displaystyle P(\text{Fatima gets selected})=P(A)=\frac{1}{7}$
$\displaystyle P(\text{John gets selected})=P(B)=\frac{1}{5}$
$\displaystyle (i)\ P(\text{both of them get selected})=P(A\cap B)$
$\displaystyle =P(A)\times P(B)$
$\displaystyle =\frac{1}{7}\times\frac{1}{5}=\frac{1}{35}$
$\displaystyle (ii)\ P(\text{only one of them gets selected})=P(A)P(\overline{B})+P(\overline{A})P(B)$
$\displaystyle =\frac{1}{7}\left(1-\frac{1}{5}\right)+\left(1-\frac{1}{7}\right)\frac{1}{5}$
$\displaystyle =\frac{1}{7}\times\frac{4}{5}+\frac{6}{7}\times\frac{1}{5}$
$\displaystyle =\frac{4}{35}+\frac{6}{35}$
$\displaystyle =\frac{10}{35}=\frac{2}{7}$
$\displaystyle (iii)\ P(\text{none of them get selected})=P(\overline{B})\times P(\overline{A})$
$\displaystyle =\left(1-\frac{1}{5}\right)\times\left(1-\frac{1}{7}\right)$
$\displaystyle =\frac{4}{5}\times\frac{6}{7}$
$\displaystyle =\frac{24}{35}$

Question 31:
A bag contains $\displaystyle 8$ marbles of which $\displaystyle 3$ are blue and $\displaystyle 5$ are red. One marble is drawn at random, its colour is noted and the marble is replaced in the bag. A marble is again drawn from the bag and its colour is noted. Find the probability that the marble will be
(i) blue followed by red.
(ii) blue and red in any order.
(iii) of the same colour.
$\displaystyle \text{Answer:}$
$\displaystyle \text{It is given that the bag contains }3\text{ blue and }5\text{ red marbles.}$
$\displaystyle (i)\ P(\text{blue followed by red})$
$\displaystyle =\frac{3}{8}\times\frac{5}{8}$
$\displaystyle =\frac{15}{64}$
$\displaystyle (ii)\ P(\text{red and blue in any order})=P(\text{blue followed by red})+P(\text{red followed by blue})$
$\displaystyle =\frac{3}{8}\times\frac{5}{8}+\frac{5}{8}\times\frac{3}{8}$
$\displaystyle =\frac{15}{64}+\frac{15}{64}$
$\displaystyle =\frac{30}{64}=\frac{15}{32}$
$\displaystyle (iii)\ P(\text{same colour})=P(\text{both red})+P(\text{both blue})$
$\displaystyle =\frac{5}{8}\times\frac{5}{8}+\frac{3}{8}\times\frac{3}{8}$
$\displaystyle =\frac{25}{64}+\frac{9}{64}$
$\displaystyle =\frac{34}{64}$
$\displaystyle =\frac{17}{32}$

Question 32:
An urn contains $\displaystyle 7$ red and $\displaystyle 4$ blue balls. Two balls are drawn at random with replacement. Find the probability of getting
(i) $\displaystyle 2$ red balls
(ii) $\displaystyle 2$ blue balls
(iii) one red and one blue ball.
$\displaystyle \text{Answer:}$
$\displaystyle \text{It is given that the urn contains }7\text{ red and }4\text{ black balls.}$
$\displaystyle (i)\ P(\text{2 red balls})=\frac{7}{11}\times\frac{7}{11}$
$\displaystyle =\frac{49}{121}$
$\displaystyle (ii)\ P(\text{2 black balls})=\frac{4}{11}\times\frac{4}{11}$
$\displaystyle =\frac{16}{121}$
$\displaystyle (iii)\ P(\text{one red ball and one black ball})=P(\text{black ball followed by red ball})+P(\text{red ball followed by black ball})$
$\displaystyle =\frac{4}{11}\times\frac{7}{11}+\frac{7}{11}\times\frac{4}{11}$
$\displaystyle =\frac{28}{121}+\frac{28}{121}$
$\displaystyle =\frac{56}{121}$

Question 33:
A card is drawn from a well-shuffled deck of $\displaystyle 52$ cards. The outcome is noted, the card is replaced and the deck reshuffled. Another card is then drawn from the deck.
(i) What is the probability that both the cards are of the same suit?
(ii) What is the probability that the first card is an ace and the second card is a red queen?
$\displaystyle \text{Answer:}$

$\displaystyle \text{Since the card is replaced, the two draws are independent.}$

$\displaystyle (i);\text{Probability that second card has the same suit as the first}$
$\displaystyle =\frac{13}{52}=\frac{1}{4}$

$\displaystyle \therefore P(\text{both cards same suit})=\frac{1}{4}$

$\displaystyle (ii);P(\text{first card is an ace})=\frac{4}{52}=\frac{1}{13}$

$\displaystyle P(\text{second card is a red queen})=\frac{2}{52}=\frac{1}{26}$

$\displaystyle \therefore P(\text{first ace and second red queen})=\frac{1}{13}\times\frac{1}{26}=\frac{1}{338}$

Question 34:
Out of $\displaystyle 100$ students, two sections of $\displaystyle 40$ and $\displaystyle 60$ are formed. If you and your friend are among $\displaystyle 100$ students, what is the probability that:
(i) you both enter the same section?
(ii) you both enter the different sections?
$\displaystyle \text{Answer:}$
$\displaystyle (a)\ \text{When both enter the same section.}$
$\displaystyle \text{Here possibilities of two cases.}$
$\displaystyle \text{Case 1: both are in section A.}$
$\displaystyle \text{If both are in section A, }40\text{ students out of }100\text{ can be selected }n(S)={}^{100}C_{40}$
$\displaystyle \text{and }(40-2)=38\text{ students out of }(100-2)=98\text{ can be selected }n(E)={}^{98}C_{38}$
$\displaystyle \text{So, }P(E)=\frac{n(E)}{n(S)}$
$\displaystyle =\frac{{}^{98}C_{38}}{{}^{100}C_{40}}$
$\displaystyle =\frac{98!}{38!\,60!}\div\frac{100!}{40!\,60!}$
$\displaystyle =\frac{98!\times40!}{38!\times100!}$
$\displaystyle =\frac{1}{100\times99}\times40\times39$
$\displaystyle =\frac{26}{165}$
$\displaystyle \text{Case 2: both are in section B.}$
$\displaystyle \text{Here }60\text{ students out of }100\text{ can be selected }n(S)={}^{100}C_{60}$
$\displaystyle \text{and }(60-2)=58\text{ students out of }(100-2)=98\text{ can be selected }n(E)={}^{98}C_{58}$
$\displaystyle \text{So, }P(E)=\frac{n(E)}{n(S)}$
$\displaystyle =\frac{{}^{98}C_{58}}{{}^{100}C_{60}}$
$\displaystyle =\frac{98!}{58!\,40!}\div\frac{100!}{60!\,40!}$
$\displaystyle =\frac{98!\times60!}{58!\times100!}$
$\displaystyle =\frac{1}{100\times99}\times60\times59$
$\displaystyle =\frac{59}{165}$
$\displaystyle \text{Hence, probability that students are either in section A or B}$
$\displaystyle =\frac{26}{165}+\frac{59}{165}$
$\displaystyle =\frac{85}{165}$
$\displaystyle =\frac{17}{33}$
$\displaystyle (b)\ \text{We know, }P(E)=1-P(E')$
$\displaystyle \text{The probability that both enter different sections}=1-\text{Probability that both enter same sections}$
$\displaystyle =1-\frac{17}{33}$
$\displaystyle =\frac{33-17}{33}$
$\displaystyle =\frac{16}{33}$

Question 35:
In a hockey match, both teams A and B scored same number of goals upto the end of the game, so to decide the winner, the referee asked both the captains to throw a die alternately and decide that the team, whose captain gets a first six, will be declared the winner. If the captain of team A was asked to start, find their respective probabilities of winning the match and state whether the decision of the referee was fair or not.
$\displaystyle \text{Answer:}$
$\displaystyle P(\text{a six})=\frac{1}{6}$
$\displaystyle P(\text{not a six})=1-\frac{1}{6}=\frac{5}{6}$
$\displaystyle P(A\ \text{wins})=P(6\ \text{in first throw})+P(6\ \text{in third throw})+\cdots$
$\displaystyle =\frac{1}{6}+\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}+\cdots$
$\displaystyle =\frac{1}{6}\left[1+\left(\frac{5}{6}\right)^{2}+\left(\frac{5}{6}\right)^{4}+\cdots\right]$
$\displaystyle =\frac{1}{6}\left[\frac{1}{1-\frac{25}{36}}\right]\ \text{[For infinite GP : }1+a+a^{2}+a^{3}+\cdots=\frac{1}{1-a}\text{]}$
$\displaystyle =\frac{1}{6}\times\frac{36}{11}$
$\displaystyle =\frac{6}{11}$
$\displaystyle P(B\ \text{wins})=P(6\ \text{in second throw})+P(6\ \text{in fourth throw})+\cdots$
$\displaystyle =\frac{5}{6}\times\frac{1}{6}+\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}+\cdots$
$\displaystyle =\frac{5}{36}\left[1+\left(\frac{5}{6}\right)^{2}+\left(\frac{5}{6}\right)^{4}+\cdots\right]$
$\displaystyle =\frac{5}{36}\left[\frac{1}{1-\frac{25}{36}}\right]\ \text{[For infinite GP : }1+a+a^{2}+a^{3}+\cdots=\frac{1}{1-a}\text{]}$
$\displaystyle =\frac{5}{36}\times\frac{36}{11}$
$\displaystyle =\frac{5}{11}$
$\displaystyle \text{It can be seen that the probability that team A wins is not equal to the probability that team B wins.}$
$\displaystyle \text{Thus, the decision of the referee was not fair.}$

Question 36:
A and B throw a pair of dice alternately. A wins the game if he gets a total of $\displaystyle 7$ and B wins the game if he gets a total of $\displaystyle 10$ . If A starts the game, then find the probability that B wins.
$\displaystyle \text{Answer:}$
$\displaystyle \text{Total of }7\text{ on the dice can be obtained in the following ways: }(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)$
$\displaystyle \text{Probability of getting a total of }7=\frac{6}{36}=\frac{1}{6}$
$\displaystyle \text{Probability of not getting a total of }7=1-\frac{1}{6}=\frac{5}{6}$
$\displaystyle \text{Total of }10\text{ on the dice can be obtained in the following ways: }(4,6),(6,4),(5,5)$
$\displaystyle \text{Probability of getting a total of }10=\frac{3}{36}=\frac{1}{12}$
$\displaystyle \text{Probability of not getting a total of }10=1-\frac{1}{12}=\frac{11}{12}$
$\displaystyle \text{Let }E\text{ and }F\text{ be two events, defined as follows:}$
$\displaystyle E=\text{getting a total of }7\text{ in a single throw of dice}$
$\displaystyle F=\text{getting a total of }10\text{ in a single throw of dice}$
$\displaystyle P(E)=\frac{1}{6},\ P(\overline{E})=\frac{5}{6},\ P(F)=\frac{1}{12},\ P(\overline{F})=\frac{11}{12}$
$\displaystyle \text{A wins if he gets a total of }7\text{ in }1^{\text{st}},3^{\text{rd}}\text{ or }5^{\text{th}}\text{ throws.}$
$\displaystyle \text{Probability of A getting a total of }7\text{ in the }1^{\text{st}}\text{ throw}=\frac{1}{6}$
$\displaystyle \text{Probability of A getting a total of }7\text{ in the }3^{\text{rd}}\text{ throw}=P(\overline{E})P(\overline{F})P(E)$
$\displaystyle =\frac{5}{6}\times\frac{11}{12}\times\frac{1}{6}$
$\displaystyle \text{Probability of A getting a total of }7\text{ in the }5^{\text{th}}\text{ throw}=P(\overline{E})P(\overline{F})P(\overline{E})P(\overline{F})P(E)$
$\displaystyle =\frac{5}{6}\times\frac{11}{12}\times\frac{5}{6}\times\frac{11}{12}\times\frac{1}{6}$
$\displaystyle \text{and so on.}$
$\displaystyle \text{Probability of winning of A}=\frac{1}{6}+\left(\frac{5}{6}\times\frac{11}{12}\times\frac{1}{6}\right)+\left(\frac{5}{6}\times\frac{11}{12}\times\frac{5}{6}\times\frac{11}{12}\times\frac{1}{6}\right)+\cdots$
$\displaystyle =\frac{\frac{1}{6}}{1-\frac{5}{6}\times\frac{11}{12}}$
$\displaystyle =\frac{12}{17}$
$\displaystyle \therefore\ \text{Probability of winning of B}=1-\frac{12}{17}=\frac{5}{17}$