Question

Two players A and B play a game which consist of a series of matches independent of each other. Whoever first wins two matches, not necessarily consecutive, will win the game. The probability of A's winning, drawing or losing a match against B are $\frac{1}{2},\frac{1}{3},\frac{1}{6}$ respectively. Which of the following is/are true?

• A. Probability of A winning the game at the end of ${11}^{\text{th}}$ match is ${}^{10}{C}_1{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{3}\right)}^9+{}^{10}{C}_2{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{3}\right)}^9$
• B. Probability of A winning the game at the end of ${11}^{\text{th}}$ match is ${}^{10}{C}_1{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{3}\right)}^9+{}^{10}{C}_2\left(\frac{1}{2}\right)\left(\frac{1}{6}\right){\left(\frac{1}{3}\right)}^9$
• C. If it is known that A wins the game at the end of ${11}^{\text{th}}$ match, then the probability that B will win only one match is $\frac{9}{11}$
• D. If it is known that A win the game at the end of ${11}^{th}$ match then the probability that B will win only one match is $\frac{9}{13}$

Answer 1

Answer: (A), (C)

The correct options are
A Probability of A winning the game at the end of ${11}^{\text{th}}$ match is ${}^{10}{C}_1{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{3}\right)}^9+{}^{10}{C}_2{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{3}\right)}^9$
C If it is known that A wins the game at the end of ${11}^{\text{th}}$ match, then the probability that B will win only one match is $\frac{9}{11}$

$P\left(W\right)=\frac{1}{2},P\left(L\right)=\frac{1}{6},P\left(D\right)=\frac{1}{3}$

$P$(A wins) $=$ (Till ${10}^{\text{th}}$ match, $1$ win & $9$ draw)$({11}^{\text{th}}$ match win)
+ (Till ${10}^{\text{th}}$ match 1 win, 8 draw, 1 lost) $({11}^{\text{th}}$ match win)

$=\frac{10!}{9!1!}{\left(\frac{1}{2}\right)}^1{\left(\frac{1}{3}\right)}^9.\frac{1}{2}+\frac{10!}{8!}\left(\frac{1}{2}\right){\left(\frac{1}{3}\right)}^8\left(\frac{1}{6}\right)\left(\frac{1}{2}\right)={}^{10}{C}_1{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{3}\right)}^9{+}^{10}{C}_2{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{3}\right)}^{9}P\left(\text{B will win only one match given that}\text{A will win the game after 11th match}\right)=\frac{{}^{10}{C}_2}{{}^{10}{C}_2+{}^{10}{C}_1}=\frac{9}{11}$