Search any question & find its solution
Question:
Answered & Verified by Expert
If $X$ is a poisson variate such that $2 \mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=2)=2 \mathrm{P}(\mathrm{X}=1)$ then $\mathrm{E}(\mathrm{X})$ is equal to
Options:
Solution:
1721 Upvotes
Verified Answer
The correct answer is:
2
If $2 \mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=2)=2 \mathrm{P}(\mathrm{X}=1)$
Let probability distribution of $X$ be given by
$\begin{array}{l}
\mathrm{P}(\mathrm{X}=\mathrm{r})=\frac{\mathrm{m}^{\mathrm{r}} \cdot \mathrm{e}^{-\mathrm{m}}}{\mathrm{r} !} \text { where } \mathrm{r}=0,1,2, \ldots \ldots \\
\therefore 2\left[\frac{\mathrm{m}^{0} \mathrm{e}^{-\mathrm{m}}}{0 !}\right]+\left[\frac{\mathrm{m}^{2} \cdot \mathrm{e}^{-\mathrm{m}}}{2 !}\right]=2\left[\frac{\mathrm{m} \cdot \mathrm{e}^{-\mathrm{m}}}{1 !}\right] \\
\Rightarrow 2+\frac{\mathrm{m}^{2}}{2}=2 \mathrm{~m} \Rightarrow \mathrm{m}^{2}-4 \mathrm{~m}+4=0 \\
\Rightarrow(\mathrm{m}-2)^{2}=0 \Rightarrow \mathrm{m}=2
\end{array}$
Let probability distribution of $X$ be given by
$\begin{array}{l}
\mathrm{P}(\mathrm{X}=\mathrm{r})=\frac{\mathrm{m}^{\mathrm{r}} \cdot \mathrm{e}^{-\mathrm{m}}}{\mathrm{r} !} \text { where } \mathrm{r}=0,1,2, \ldots \ldots \\
\therefore 2\left[\frac{\mathrm{m}^{0} \mathrm{e}^{-\mathrm{m}}}{0 !}\right]+\left[\frac{\mathrm{m}^{2} \cdot \mathrm{e}^{-\mathrm{m}}}{2 !}\right]=2\left[\frac{\mathrm{m} \cdot \mathrm{e}^{-\mathrm{m}}}{1 !}\right] \\
\Rightarrow 2+\frac{\mathrm{m}^{2}}{2}=2 \mathrm{~m} \Rightarrow \mathrm{m}^{2}-4 \mathrm{~m}+4=0 \\
\Rightarrow(\mathrm{m}-2)^{2}=0 \Rightarrow \mathrm{m}=2
\end{array}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.