MCQ
A random variable $X$ has the following probability distribution:
$X = x$01234
P$(X = x)$k3k5k2kk
Then the value of $\mathrm{P}(\mathrm{X} \geq 2)$ is
  • A
    $\frac{1}{3}$
  • $\frac{2}{3}$
  • C
    $\frac{3}{4}$
  • D
    $\frac{1}{4}$

Answer

Correct option: B.
$\frac{2}{3}$
(B)
Since, $\sum_{x=0}^4 \mathrm{P}(\mathrm{X}=x)=1$,
$\begin{aligned}& k+3 k+5 k+2 k+k=1 \\& \Rightarrow 12 k=1 \quad \Rightarrow k=\frac{1}{12}\end{aligned}$
Now, $P(X \geq 2)=P(X=2)+P(X=3)+P(X=4)$
$\begin{aligned} & =5 \mathrm{k}+2 \mathrm{k}+\mathrm{k} \\ & =8 \mathrm{k}=8\left(\frac{1}{12}\right)=\frac{2}{3}\end{aligned}$

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