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Probability — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSProbability5 Marks
Question
Two probability distributions of the discrete random variable $X$ and $Y$ are given below.
$\text{X}$ $0$ $1$ $2$ $3$
$\text{P}(\text{X})$ $\frac{1}{5}$ $\frac{2}{5}$ $\frac{1}{5}$ $\frac{1}{5}$
 
$\text{Y}$ $0$ $1$ $2$ $3$
$\text{P}(\text{Y})$ $\frac{1}{5}$ $\frac{3}{10}$ $\frac{2}{5}$ $\frac{1}{10}$
Prove that $E(Y^2) = 2E(X)$.
✓

Answer

Since, we have to prove that $, E(Y^2) = 2E(X) .......(i)$
Taking $\text{L.H.S}$. of equation $(i),$ we have
$\text{E}(\text{Y}^2)=\sum\text{Y}^2\text{P}(\text{Y})$
$=0\cdot\frac{1}{5}+1\cdot\frac{3}{10}+4\cdot\frac{2}{5}+9\cdot\frac{1}{10}$
$=\frac{3}{10}+\frac{8}{5}+\frac{9}{10}$
$=\frac{28}{10}=\frac{14}{5}$
$\Rightarrow\text{E}(\text{Y})^2=\frac{14}{5}\ ......(\text{ii})$
Now taking $\text{R.H.S}$. of equation $(i),$ we get
$\text{E}(\text{X})=\sum\text{XP}(\text{X})$
$=0.\frac{1}{5}+1\cdot\frac{2}{5}+2\cdot\frac{1}{5}+3\cdot\frac{1}{5}=\frac{7}{5}$
$2\text{E}(\text{X})=\frac{14}{5}\ ......(\text{iii})$
Thus, from equations $(ii)$ and $(iii),$ we get
$E(Y^2) = 2E(X)$
Hence proved.

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