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Probability Distributions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsProbability Distributions2 Marks
MCQ
If a random variable X has the probability distribution given by
$\mathrm{P}(\mathrm{X}=0)=3 \mathrm{C}^3, \mathrm{P}(\mathrm{X}=2)=5 \mathrm{C}-10 \mathrm{C}^2$ and $P(X=4)=4 C-1$, then the variance of that distribution is
  • A
    $\frac{68}{9}$
  • B
    $\frac{22}{9}$
  • C
    $\frac{612}{81}$
  • ✓
    $\frac{128}{81}$
✓

Answer

Correct option: D.
$\frac{128}{81}$
(D)
Given probability distribution of a r.v.X.
$\begin{array}{ll}\therefore & P(X=0)+P(X=2)+P(X=4)=1 \\ \therefore & 3 C^3+5 C-10 C^2+4 C-1=1 \\ \therefore & 3 C^3-10 C^2+9 C-2=0\end{array}$
$\Rightarrow \mathrm{C}=2$ or $\frac{1}{3}$ or 1
C cannot be 2 or 1 as $0 \leq$ probability $\leq 1$.
$\therefore \quad C=\frac{1}{3}$
$\Rightarrow \mathrm{P}(\mathrm{X}=0)=\frac{1}{9}, \mathrm{P}(\mathrm{X}=2)=\frac{5}{9}, \mathrm{P}(\mathrm{X}=4)=\frac{1}{3}$
$\therefore \quad E(X)=0 \times \frac{1}{9}+2 \times \frac{5}{9}+4 \times \frac{1}{3}=\frac{22}{9}$
$E\left(X^2\right)=0 \times \frac{1}{9}+4 \times \frac{5}{9}+16 \times \frac{1}{3}=\frac{68}{9}$
$\therefore \quad \operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2=\frac{68}{9}-\left(\frac{22}{9}\right)^2=\frac{128}{81}$

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