P(W)=31P(L)=32x= number of matches that team wins y= number of matches that team loses ∣x−y∣≤2 and x+y=10∣x−y∣=0,1,2x,y∈N Case-I $\begin{aligned}
& -I:|x-y|=0 \Rightarrow x=y \
& \because x+y=10 \Rightarrow x=5=y \
& P(|x-y|=0)={ }^{10} C_5\left(\frac{1}{3}\right)^5\left(\frac{2}{3}\right)^5
\end{aligned}Case−II:|\mathrm{x}-\mathrm{y}|=1 \Rightarrow \mathrm{x}-\mathrm{y}= \pm 1\begin{array}{|c|c|}
\hline \mathrm{x}=\mathrm{y}+1 & \mathrm{x}=\mathrm{y}-1 \
\hline\because \mathrm{x}+\mathrm{y}=10 & \because \mathrm{x}+\mathrm{y}=10 \
\hline 2 \mathrm{y}=9 & 2 \mathrm{y}=11 \
\hline \text{Not possible} & \text{Not possible} \
\hline
\end{array}\begin{aligned} & \text { Case-III : }|\mathrm{x}-\mathrm{y}|=2 \Rightarrow \mathrm{x}-\mathrm{y}= \pm 2 \ & x-y=2 \quad \text { OR } \quad x-y=-2 \ & \because \mathrm{x}+\mathrm{y}=10 \quad \because \mathrm{x}+\mathrm{y}=10 \ & x=6, y=4 \quad x=4, y=6 \ & \mathrm{P}(|\mathrm{x}-\mathrm{y}|=2)={ }^{10} \mathrm{C}_6\left(\frac{1}{3}\right)^6\left(\frac{2}{3}\right)^4+{ }^{10} \mathrm{C}_4\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)^6 \ & \mathrm{p}={ }^{10} \mathrm{C}_5 \frac{2^5}{3^{10}}+{ }^{10} \mathrm{C}_6 \frac{2^4}{3^{10}}+{ }^{10} \mathrm{C}_4 \frac{2^6}{3^{10}} \ & 3^9 \mathrm{p}=\frac{1}{3}\left({ }^{10} \mathrm{C}_5 2^5+{ }^{10} \mathrm{C}_6 2^4+{ }^{10} \mathrm{C}_4 2^6\right) \ & =8288 \ & \end{aligned}$
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…
JEE Main 2024 — Mathematics Probability & Statistics
In a tournament, a team plays 10 matches with probabilities of winning and losing each match as 31 and 32 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∣≤ 2) is p, then 39p equals ______
Held on 4 Apr 2024 · Verified 6 Jul 2026.
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