Probability
Binomial Distribution — Tournament Matches
nta_pyq_2024_apr
Grade 12
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 team loses. If the probability $P(|x-y|\leq2)$ is $p$, then $3^9p$ equals
Step-by-Step Solution
Key Concept: $x+y=10$, $|x-y|\in\{0,1,2\}$. Case $|x-y|=0$: $x=y=5$. Case $|x-y|=1$: $x-y=\pm1$ with $x+y=10$ gives non-integer $y$, impossible. Case $|x-y|=2$: $(x,y)=(6,4)$ or $(4,6)$.
$3^9p=\frac{1}{3}(^{10}C_5\cdot32+^{10}C_6\cdot16+^{10}C_4\cdot64)=8288$.
Correct Answer: 8288