If $\mathrm{x}$ is a random variable with p.m.f. as follows. $P(X=x)=\frac{5}{16}, x=0,1 =\frac{k x}{48},…
If $\mathrm{x}$ is a random variable with p.m.f. as follows.
$P(X=x)=\frac{5}{16}, x=0,1 =\frac{k x}{48}, x=2,$ then $E(x)=\frac{1}{4}, x=3$
- 1.1875
- 1.3125
- 1.5625
- 0.5625
Solution
From given data, we write
When $\mathrm{x}=0, \mathrm{P}=\frac{5}{16}=\frac{15}{48}$
$\begin{aligned}
& x=1, P=\frac{5}{16}=\frac{15}{48} \\
& x=2, P=\frac{2 k}{48} \\
& x=3, P=\frac{1}{4}=\frac{12}{48}
\end{aligned}$
Here $\sum \mathrm{Pi}=1$
$\begin{aligned}
& \therefore \frac{15}{48}+\frac{15}{48}+\frac{2 \mathrm{k}}{48}+\frac{12}{48}=1 \Rightarrow \mathrm{k}=3 \\
& \therefore \text { When } \mathrm{x}=2, \mathrm{P}=\frac{6}{48}=\frac{1}{8}
\end{aligned}$
$\begin{aligned}
& \text { Now } \mathrm{E}=\sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}} \\
& =\left[\left(\frac{5}{16}\right)(0)\right]+\left[\left(\frac{5}{16}\right)(1)\right]+\left[\left(\frac{1}{8}\right)(2)\right]+\left[\left(\frac{1}{4}\right)(3)\right]=0+\frac{5}{16}+\frac{1}{4}+\frac{3}{4}=\frac{21}{16}=1.3125
\end{aligned}$
Asked in: MHT CET 2021 (21 Sep Shift 1)
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