MCQ
A random variable $X$ has the following probability distribution

$X$ $1$ $2$ $3$ $4$ $5$
$P(X)$ $K^2$ $2K$ $K$ $2K$ $5K^2$

Then $\mathrm{P}(\mathrm{X}> 2)$ is equal to

  • A
    $\frac{7}{12}$
  • $\frac{23}{36}$
  • C
    $\frac{1}{36}$
  • D
    $\frac{1}{6}$

Answer

Correct option: B.
$\frac{23}{36}$
b
$\sum P(X)=1 \Rightarrow K^{2}+2 K+K+2 K+5 K^{2}=1$

$\Rightarrow 6 \mathrm{K}^{2}+5 \mathrm{K}-1=0 \Rightarrow(6 \mathrm{K}-1)(\mathrm{K}+1)=0$

$\Rightarrow \mathrm{K}=-1$ (rejected) $\Rightarrow \mathrm{K}=\frac{1}{6}$

$\mathrm{P}(\mathrm{X}>2)=\mathrm{K}+2 \mathrm{K}+5 \mathrm{K}^{2}=\frac{23}{36}$

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