Question

In a tournament, a team plays 10 matches with probabilities of winning and losing each match as $\frac{1}{3}$ and $\frac{2}{3}$, respectively. Let $x$ be the number of matches that the team wins, and $y$ be the number of matches that the team loses. If the probability $P(|x - y| \leq 2)$ is $p$, then $3^9 p$ equals .

Answer 1

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

Let $x =$ number of matches the team wins, $y =$ number of matches the team loses.
The conditions are: $|x - y| \leq 2 \quad \text{and} \quad x + y = 10.$ This implies: $|x - y| = 0, 1, 2, \quad x, y \in \mathbb{N}.$

Case-I: $|x - y| = 0 \implies x = y$
From $x + y = 10$, we have: $x = y = 5.$ The probability for this case is: $P(|x - y| = 0) = \binom{10}{5} \left(\frac{1}{3}\right)^5 \left(\frac{2}{3}\right)^5.$

Case-II: $|x - y| = 1 \implies x - y = \pm 1$
For $x = y + 1$: $x + y = 10 \implies 2y = 9, \; \text{Not possible.}$

Case-III: $|x - y| = 2 \implies x - y = \pm 2$
For $x - y = 2$: $x + y = 10 \implies x = 6, \; y = 4.$ For $x - y = -2$: $x + y = 10 \implies x = 4, \; y = 6.$ The probability for this case is: $P(|x - y| = 2) = \binom{10}{6} \left(\frac{1}{3}\right)^6 \left(\frac{2}{3}\right)^4 + \binom{10}{4} \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^6.$

Total Probability: $p = \binom{10}{5} \left(\frac{1}{3}\right)^5 \left(\frac{2}{3}\right)^5 + \binom{10}{6} \left(\frac{1}{3}\right)^6 \left(\frac{2}{3}\right)^4 + \binom{10}{4} \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^6.$ Simplify: $3^9 p = \frac{1}{3} \left[\binom{10}{5} \left(\frac{1}{3}\right)^5 \left(\frac{2}{3}\right)^5 + \binom{10}{6} \left(\frac{1}{3}\right)^6 \left(\frac{2}{3}\right)^4 + \binom{10}{4} \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^6\right].$

Final Result: $3^9 p = 8288.$