JEE Main 2024 (Online) 4th April Evening ShiftQuestion 1
In a tournament, a team plays $10$ matches with probabilities of winning and losing each match as $\dfrac{1}{3}$ and $\dfrac{2}{3}$ respectively. Let $x$ be the number of matches that the team wins, and $y$ be the number of matches that team loses. If the probability $P(|x-y|\le2)$ is $p$, then $3^{9}p$ equals $\_\_\_\_\_\_\_\_$ .
Probability
Answer$8288$
Solution
$|x-y|\le2$ with $x+y=10$ means $x\in\{4,5,6\}$: $p=\frac{\binom{10}4 2^6+\binom{10}5 2^5+\binom{10}6 2^4}{3^{10}}=\frac{24864}{3^{10}}$, so $3^9p=8288$.