Q63 / 200 · TS EAMCET 2002 · Mathematics
A random variable $\mathrm{X}$ takes the values 0,1 and 2. If $P(X=1)=P(X=2)$ and $P(X=0)=0.4$, then the mean of the random variable $\mathrm{X}$ is
- A$0.2$
- B$0.7$
- C$0.5$
- Correct answerD$0.9$
Solution
We have, $P(X=1)=P(X=2)$
$$
\frac{\lambda^1}{1 !} e^\lambda=\frac{\lambda^2}{2 !} e^\lambda \Rightarrow \lambda=2
$$
Also, $P(X=0)+P(X=1)+P(X=2)=1$
$$
\begin{aligned}
& \Rightarrow \quad 0.4+P(X=1)+P(X=2)=1 \\
& \Rightarrow \quad P(X=1)+P(X=2)=0.6=\frac{6}{10}=\frac{3}{5}
\end{aligned}
$$
Also, $P(X=1)+P(X=1)=\frac{3}{5}$
[from (i)]
$\begin{aligned} & \Rightarrow \quad P(X=1)=\frac{3}{10} \\ & \Rightarrow \quad P(X=1)=P(X=2)=\frac{3}{10} \\ & \text { Mean } X_0 P(X=0)+X_1 P(X=1)+X_2 P(X=2) \\ & =0+1 \cdot \frac{3}{10}+2 \frac{3}{10}=\frac{9}{10}=0.9\end{aligned}$
Hence, the correct answer is option: 'D'.