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If X is a random variable with probability mass function
P(x) = kx , x=1,2,3
= 0 , otherwise
then , k=..............
(a) 1/5
(b) 1/4
(c) 1/6
(d) 2/3
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Solution
(c)
x | 1 | 2 | 3 |
P(x) | k | 2k | 3k |
Since, the function is a p.m.f.
∴ ∑P(xi) = 1
∴ k + 2k + 3k = 1
∴ k = 1/6
Concept: Probability Distribution - Probability Mass Function (P.M.F.)
Is there an error in this question or solution?