Question

Let X ~ B(10, 0.2). Find P(X ≥ 1).

Answer 1

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) = 1 - P(X < 1)

= 1 - P(X = 0)

= 1 - `"^10C_0 (0.2)^0 (0.8)^(10 - 0)`

`= 1 - 1 xx 1 xx (0.8)^10`

`= 1 - (0.8)^10`