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Probability

MHT CET 2025 20th April Evening Shift

MCQ+2 / -02025

A random variable $X$ takes the values $0,1,2,3$, $\qquad$ with probability


$\mathrm{P}(\mathrm{X}=x)=\mathrm{k}(x+1)\left(\frac{1}{5}\right)^x$, where k is a constant.


Then $\mathrm{P}(\mathrm{X}=0)$ is

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