In a tournament, a team plays 10 matches with probabilities of winning and losing each match as (1)/(3) and (2)/(3) respectively. Let x be the number of matches that the team wins, and y be the number of matches that

Correct answer: 0 aligned & P(W)=(1)/(3) P(L)=(2)/(3) & x= number of matches that team wins & y= number of matches that team loses & |x-y| ≤ 2 and x+y=10 & |x-y|=0,1,2 x, y ∈ N aligned Case-I aligned & -I:|x-y|=0 x=y & x+y=10 x=5=y & P(|x-y|=0)= ^ 10 C_5((1)/(3))^5((2)/(3))^5 aligned Case-II : |x-y|=1 x-y= ± 1 x=y+1 x=y-1 x+y=10 x+y=10 2 y=9 2 y=11 Not possible Not possible aligned & Case-III : |x-y|=2 x-y= ± 2 & x-y=2 OR x-y=-2 & x+y=10 x+y=10 & x=6, y=4 x=4, y=6 & P(|x-y|=2)= ^ 10 C_6((1)/(3))^6((2)/(3))^4+ ^ 10 C_4((1)/(3))^4((2)/(3))^6 & p= ^ 10 C_5 2^5 3^ 10 + ^ 10 C_6 2^4 3^ 10 + ^ 10 C_4 2^6 3^ 10 & 3^9 p=(1)/(3)( ^ 10 C_5 2^5+ ^ 10 C_6 2^4+ ^ 10 C_4 2^6) & =8288 & aligned

JEE Main 20244 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 4 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Probability

JEE Main 2024 · 4 April · Evening Shift · Mathematics · Probability

In a tournament, a team plays 10 matches with probabilities of...

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

\mathrm{P}(|x-y| \leq

2) is

p

, then

3^9 p

equals ______

Correct answer

Step-by-step solution

\begin{aligned} & P(W)=\frac{1}{3} \quad P(L)=\frac{2}{3} \\ & x=\text{ number\text{ of } matches that team wins } \\ & y=\text{ number of matches that team loses } \\ & |x-y| \leq 2 \text{ and } x+y=10 \\ & |x-y|=0,1,2 \quad x, y \in N\end{aligned}

Case-I

\begin{aligned} & -I:|x-y|=0 \Rightarrow x=y \\ & \because x+y=10 \Rightarrow x=5=y \\ & P(|x-y|=0)={ }^{10} C_5\left(\frac{1}{3}\right)^5\left(\frac{2}{3}\right)^5 \end{aligned}

Case-II :

|\mathrm{x}-\mathrm{y}|=1 \Rightarrow \mathrm{x}-\mathrm{y}= \pm 1 <table class="qx-match-list qx-match-table" border="1" cellpadding="6" cellspacing="0"><tr><td>x=y+1</td><td>x=y-1</td></tr><tr><td>\because x+y=10</td><td>\because x+y=10</td></tr><tr><td>2 y=9</td><td>2 y=11</td></tr><tr><td>Not possible</td><td>Not possible</td></tr></table>

\begin{aligned} & \text{ Case-III : }|\mathrm{x}-\mathrm{y}|=2 \Rightarrow \mathrm{x}-\mathrm{y}= \pm 2 \\ & x-y=2 \quad \text{\text{ OR } } \quad x-y=-2 \\ & \because \mathrm{x}+\mathrm{y}=10 \quad \because \mathrm{x}+\mathrm{y}=10 \\ & x=6, y=4 \quad x=4, y=6 \\ & \mathrm{P}(|\mathrm{x}-\mathrm{y}|=2)={ }^{10} \mathrm{C}_6\left(\frac{1}{3}\right)^6\left(\frac{2}{3}\right)^4+{ }^{10} \mathrm{C}_4\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)^6 \\ & \mathrm{p}={ }^{10} \mathrm{C}_5 \frac{2^5}{3^{10}}+{ }^{10} \mathrm{C}_6 \frac{2^4}{3^{10}}+{ }^{10} \mathrm{C}_4 \frac{2^6}{3^{10}} \\ & 3^9 \mathrm{p}=\frac{1}{3}\left({ }^{10} \mathrm{C}_5 2^5+{ }^{10} \mathrm{C}_6 2^4+{ }^{10} \mathrm{C}_4 2^6\right) \\ & =8288 \\ & \end{aligned}

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