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

Correct Answer: 8288

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

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: \[ 3^9 p = 8288. \]