A random variable $X$ has the following probability distribution Then $P\left(X^3 2\right)=$

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Aicharya · Accepted Answer

$\begin{aligned} & \sum P(x)=1 \ & \Rightarrow 49 k=1\end{aligned}$ Now, $P\left(x^3 2ight)=1-P(X<2)$ $\begin{aligned} & =1-\{P(X=0)+P(x=1)\} \ & =1-\left\{\frac{1}{49}+\frac{3}{49}ight\} \ & =\frac{45}{49}\end{aligned}$

A random variable $X$ has the following probability distribution Then $P\left(X^3 2\right)=$

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