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Probability Distribution

A random vari...

Question

A random variable X has the following probability distribution.
X 0 1 2 3 4 5 6
P(X) K 3K 5K 7K 9K 11K 13K
Find K, $P (X \geq 2)$

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Solution

Sum of probabilities is $1$
$⟹ k + 3 k + 5 k + 7 k + 9 k + 11 k + 13 k = 49 k = 1 k = \frac{1}{49} p (X \geq 2) = 5 k + 7 k + 9 k + 11 k + 13 k = 45 k = \frac{45}{49}$

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