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A random variable $X$ has the following probability distribution
Then $P\left(X^3 2\right)=$
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Then $P\left(X^3 2\right)=$
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Verified Answer
The correct answer is:
$\frac{45}{49}$
$\begin{aligned} & \sum P(x)=1 \\ & \Rightarrow 49 k=1\end{aligned}$
Now, $P\left(x^3 2\right)=1-P(X<2)$
$\begin{aligned} & =1-\{P(X=0)+P(x=1)\} \\ & =1-\left\{\frac{1}{49}+\frac{3}{49}\right\} \\ & =\frac{45}{49}\end{aligned}$
Now, $P\left(x^3 2\right)=1-P(X<2)$
$\begin{aligned} & =1-\{P(X=0)+P(x=1)\} \\ & =1-\left\{\frac{1}{49}+\frac{3}{49}\right\} \\ & =\frac{45}{49}\end{aligned}$
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