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 (i) X = 0, (ii) X ≤ 1, (iii) X > 1, (iv) X ≥ 1.

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) = ${\displaystyle {}^5{\text{C}}_x{\left(0.2\right)}^x{\left(0.8\right)}^{5-x},x}$ = 0, 1, ... ,5

(i) P(X = 0) = ${\displaystyle {}^5{\text{C}}_0{\left(0.2\right)}^0{\left(0.8\right)}^5}$ = (0.8)⁵

(ii) P(X ≤ 1) = P(X = 0 or X = 1)

= P(X = 0) + P(X = 1)

= ${\displaystyle {\left(0.8\right)}^5+{}^5{\text{C}}_1\left(0.2\right){\left(0.8\right)}^4}$      ...[From (i)]

= (0.8)⁴ [0.8 + 5 x 0.2]

= (1.8) (0.8)⁴ 

(iii) P(X > 1) = 1 – p(X ≤ 1)

= 1 – (1.8) (0.8)⁴     ...[From (ii)]

(iv) P(X ≥ 1) = 1 – P(X < 1)

= 1 – P(X = 0)

= 1 – (0.8)⁵     ......[From (ii)]