Question

The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. What is the probability that out of 5 such bulbs
(i) none
(ii) not more than one
(iii) more than one
(iv) at least one, will fuse after 150 days of use.

Answer 1

Let X = number of fuse bulbs.

p = probability of a bulb produced by a factory will fuse after 150 days of use.

∴ p = 0.05 and q = 1 – p = 1 – 0.05 = 0.95

X has a binomial distribution with n = 5 and p = 0.05

∴ X ~ B(5, 0.05)

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

P(X = x) = ⁿC_x p^x q^n−x

i.e., p(x) = ⁵C_x (0.05)^x (0.95)^5−x, x = 0, 1, 2, 3, 4, 5

(i) P(none) = P[X = 0]

= p(0)

= ⁵C₀ (0.05)⁰ (0.95)^{5 - 0}

= 1 × 1 × (0.95)⁵

= (0.95)⁵ 

Hence, the probability that none of the bulbs will fuse after 150 days = (0.95)⁵.

(ii) P(not more than one) = P(X ≤ 1)

= p(0) + p(1)

= ⁵C₀ (0.05)⁰ (0.95)^{5 - 0} + ⁵C₁ (0.05)¹ (0.95)^{5 - 1}

= ⁵C₀ (0.05)⁰ (0.95)⁵ + ⁵C₁ (0.05)¹ (0.95)⁴

= 1 × 1 × (0.95)⁵ + 5 × (0.05) × (0.95)⁴

= (0.95)⁴ [0.95 + 5(0.05)]

= (0.95)⁴ [0.95 + 0.25]

= (0.95)⁴ (1.20)

= (1.2)(0.95)⁴

Hence, the probability that not more than one bulb will fuse after 150 days = (1.2)(0.95)⁴

(iii) P (more than 1) = P(X > 1)

= 1 - P[X ≤ 1]

= 1 - (1.2)(0.95)⁴ 

Hence, the probability that more than one bulb fuse after 150 days = 1 – (1.2)(0.95)⁴.

(iv) P (at least one) = P(X ≥ 1)

= 1 - P[X = 0]

= 1 - p(0)

= 1 - ⁵C₀ (0.05)⁰ (0.95)^{5 - 0}

= 1 - 1 × 1 × (0.95)⁵

= 1 - (0.95)⁵ 

Hence, the probability that at least one bulb fuses after 150 days = 1 - (0.95)⁵.