In a tournament, a team plays 10 matches with probabilities of winning and losing each match as and respectively. Let be the number of matches that the team wins, and be the number of matches that the team loses. If the probability is , then equals

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 _______ . [2024]

Smart Achievers · Accepted Answer

(8288)

$| x - y | \leq 10$ and $0 \leq x \leq 10$ and $0 \leq y \leq 10$

$∴$ $p = P (| x - y | \leq 2)$

$= P (x = 4, y = 6) + P (x = 5, y = 5) + P (x = 6, y = 4)$

$= {}^{10}{C_{4}} {(\frac{1}{3})}^{4} {(\frac{2}{3})}^{6} + {}^{10}{C_{5}} {(\frac{1}{3})}^{5} {(\frac{2}{3})}^{5} + {}^{10}{C_{6}} {(\frac{1}{3})}^{6} {(\frac{2}{3})}^{4}$

$= 210 (\frac{2^{6}}{3^{10}}) + 252 (\frac{2^{5}}{3^{10}}) + 210 (\frac{2^{4}}{3^{10}})$

$= \frac{2^{5}}{3^{10}} [420 + 252 + 105] \Rightarrow p = \frac{2^{5}}{3^{10}} \times 777$

$Hence, 3^{9} p = 3^{9} \times \frac{2^{5}}{3^{10}} \times 777 = \frac{2^{5}}{3} \times 777 = 8288$

Solution

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Solution

Ans. (8288) |x-y|≤10 and 0≤x≤10 and 0≤y≤10 ∴ p=P(|x-y|≤2) =P(x=4,y=6)+P(x=5,y=5)+P(x=6,y=4) =C410(13)4(23)6+C510(13)5(23)5+C610(13)6(23)4 =210(26310)+252(25310)+210(24310) =25310[420+252+105]⇒p=25310×777 Hence, 39p=39×25310×777=253×777=8288

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