JEE MAIN - Mathematics (2024 - 4th April Evening Shift - No. 25)

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 _________.
Answer
8288

Explanation

$$\begin{aligned} & x+y=10 \\ & A=x-y \\ & P(|A|<2) \text { is } P \\ & \Rightarrow|A|=2,1,0 \Rightarrow A=0,1,-1,2,-2 \\ & \Rightarrow x=\frac{10+A}{2} \Rightarrow A \in \text { even as } x \in \text { integer } \\ & \Rightarrow A=0,-2,2 \\ & \Rightarrow P(|A| \leq 2)=P(A=0)+P(A=-2)+P(A=2) \end{aligned}$$

(1) $$A=0 \Rightarrow x=5=y$$

$$P(A=0)={ }^{10} C_5\left(\frac{1}{3}\right)^5\left(\frac{2}{3}\right)^5$$

(2) $$A=-2$$

$$\Rightarrow x=4$$ and $$y=6$$

$$P(A=-2)={ }^{10} C_4 \cdot\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)^6$$ and

$$\begin{aligned} & \text { Similarly, } P(A=2)={ }^{10} C_6\left(\frac{1}{3}\right)^6\left(\frac{2}{3}\right)^4 \\ & \Rightarrow P(|A| \leq 2) 3^9=3\left({ }^{10} C_5 \cdot 2^5+{ }^{10} C_4 \cdot 2^6+{ }^{10} C_6 \cdot 2^4\right) \\ & =8288 \end{aligned}$$

Comments (0)

Advertisement