JEE Main 2024 — Probability Question with Solution

$\begin{array}{rl}& P(W)=\frac{1}{3}P(L)=\frac{2}{3}\\ & x=\text{ number of matches that team wins }\\ & y=\text{ number of matches that team loses }\\ & |x-y|\le 2\text{ and }x+y=10\\ & |x-y|=0,1,2x,y\in N\end{array}$ Case-I $\begin{aligned} & -I:|x-y|=0 \Rightarrow x=y \\ & \because x+y=10 \Rightarrow x=5=y \\ & P(|x-y|=0)={ }^{10} C_5\left(\frac{1}{3}\right)^5\left(\frac{2}{3}\right)^5 \end{aligned}$ Case-II : $|\mathrm{x}-\mathrm{y}|=1\Rightarrow \mathrm{x}-\mathrm{y}=\pm 1$ $\begin{array}{|c|c|} \hline \mathrm{x}=\mathrm{y}+1 & \mathrm{x}=\mathrm{y}-1 \\ \hline\because \mathrm{x}+\mathrm{y}=10 & \because \mathrm{x}+\mathrm{y}=10 \\ \hline 2 \mathrm{y}=9 & 2 \mathrm{y}=11 \\ \hline \text{Not possible} & \text{Not possible} \\ \hline \end{array}$ $\begin{array}{rl}& \text{ Case-III : }|\mathrm{x}-\mathrm{y}|=2\Rightarrow \mathrm{x}-\mathrm{y}=\pm 2\\ & x-y=2\text{ OR }x-y=-2\\ & \because \mathrm{x}+\mathrm{y}=10\because \mathrm{x}+\mathrm{y}=10\\ & x=6,y=4x=4,y=6\\ & \mathrm{P}(|\mathrm{x}-\mathrm{y}|=2)={}^{10}{\mathrm{C}}_6{\left(\frac{1}{3}\right)}^6{\left(\frac{2}{3}\right)}^4+{}^{10}{\mathrm{C}}_4{\left(\frac{1}{3}\right)}^4{\left(\frac{2}{3}\right)}^6\\ & \mathrm{p}={}^{10}{\mathrm{C}}_5\frac{2^5}{3^{10}}+{}^{10}{\mathrm{C}}_6\frac{2^4}{3^{10}}+{}^{10}{\mathrm{C}}_4\frac{2^6}{3^{10}}\\ & 3^9\mathrm{p}=\frac{1}{3}\left({}^{10}{\mathrm{C}}_5{2}^5+{}^{10}{\mathrm{C}}_6{2}^4+{}^{10}{\mathrm{C}}_4{2}^6\right)\\ & =8288\\ & \end{array}$