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Sum
The probability distribution of a discrete random variable X is:
X=x | 1 | 2 | 3 | 4 | 5 |
P(X=x) | k | 2k | 3k | 4k | 5k |
find P(X≤4)
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Solution
Given X is discreate r . v.
∴ P(x) is p.m.f.
p.m.f = ∑ pi = 1
i.e. k + 2k + 3k + 4k + 5k = 1
i.e. 15 k = 1
i.e. k = `1/15`
P(X≤ 4) = P(X =1) + P(X = 2) + P(X = 3) + P(X = 4)
= k + 2k + 3k + 4k
= 10k
=`10 xx 1/15`
= `2/3`
Concept: Probability Distribution of a Discrete Random Variable
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