Sum
Let X ~ B(10, 0.2). Find P(X = 1).
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Solution
X ~ B(10, 0.2)
∴ n = 10, p = 0.2
∴ q = 1 - p = 1 - 0.2 = 0.8
The p,m.f. of X is given by
P(X = x) = `"^nC_x p^x q^(n - x)`
∴ P(X = x) = `"^10C_x (0.2)^x (0.8)^(10 - x)`, x = 0, 1, 2, 3,....,10
P(X = 1) = `"^10C_1 (0.2)^1 (0.8)^(10 - 1)`
`= 10 (0.2)^1 (0.8)^9 = 2 (0.8)^9`
Concept: Binomial Distribution
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