JEE Main 2024 — 4 April, Shift 2 — Question 27

$P(W)=\frac{1}{3} ,\quad P(L)=\frac{2}{3}$

$\mathrm{x}=$ number of matches that team wins $y=$ number of matches that team loses $|x-y| \leq 2$ and $x+y=10$

$|x-y|=0,1,2 \quad x, y \in N$

Case-I : $|x-y|=0 \Rightarrow x=y$

$\because x+y=10 \Rightarrow x=5=y$

$\mathrm{P}(|\mathrm{x}-\mathrm{y}|=0)={ }^{10} \mathrm{C}_{5}\left(\frac{1}{3}\right)^{5}\left(\frac{2}{3}\right)^{5}$

Case-II : $|x-y|=1 \Rightarrow x-y= \pm 1$

Case-III : $|x-y|=2 \Rightarrow x-y= \pm 2$

$\begin{array}{ccc} \mathrm{x}-\mathrm{y}=2 \end{array}$ OR

$\mathrm{x}-\mathrm{y}=-2$

$\because \mathrm{x}+\mathrm{y}=10$

$\because \mathrm{x}+\mathrm{y}=10$

$\mathrm{x}=6, \mathrm{y}=4$

$\mathrm{x}=4, \mathrm{y}=6$

$\mathrm{p}(|\mathrm{x}-\mathrm{y}|=2)={ }^{10} \mathrm{C}_{6}\left(\frac{1}{3}\right)^{6}\left(\frac{2}{3}\right)^{4}+{ }^{10} \mathrm{C}_{4}\left(\frac{1}{3}\right)^{4}\left(\frac{2}{3}\right)^{6}$

$\mathrm{p}={ }^{10} \mathrm{C}_{5} \frac{2^{5}}{3^{10}}+{ }^{10} \mathrm{C}_{6} \frac{2^{4}}{3^{10}}+{ }^{10} \mathrm{C}_{4} \frac{2^{6}}{3^{10}}$

$3^{9} \mathrm{p}=\frac{1}{3}\left({ }^{10} \mathrm{C}_{5} 2^{5}+{ }^{10} \mathrm{C}_{6} 2^{4}+{ }^{10} \mathrm{C}_{4} 2^{6}\right)$

$=8288$