Question

The probability that a bulb produced by a factory will fuse after 200 days of use is 0.2. Let X denote the number of bulbs (out of 5) that fuse after 200 days of use. Find the probability of X = 0

Answer 1

Let X denote the number of bulbs that will fuse after 200 days.

P(bulb will fuse after 200 days) = p = 0.2

∴ q = 1 – p = 1 – 0.2 = 0.8

Given, n = 5

∴ X ∼ B(5, 0.2)

The p.m.f. of X is given by

P(X = x) = `""^5"C"_x (0.2)^x (0.8)^(5 - x), x` = 0, 1, ..., 5

P(X = 0) = `""^5"C"_0 (0.2)^0 (0.8)^5` = (0.8)⁵