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$X$ is a discrete random variable which takes the values $0, 1, 2 $ and $P(X=0)=\large\frac{144}{169},$$ P(X=1)=\large\frac{1}{169}$ then the value of $P(X=2)$ is

\[\begin{array}{1 1}(1)\frac{145}{169}&(2)\frac{24}{169}\\(3)\frac{2}{169}&(4)\frac{143}{169}\end{array}\]

Can you answer this question?
 
 

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Given $P(X=0)+P(X=1)+P(X=2)=1$
$\large\frac{144}{169} +\frac{1}{169}$$+P(X=2) =1$
$P(X=2)=1- \large\frac{145}{169}$
$\qquad= \large\frac{169-145}{169}$
$P(X=1)=\large\frac{24}{169}$
Hence 2 is the correct answer.
answered May 23, 2014 by meena.p
 

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