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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.
Solution
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) = nCx px qn−x
i.e., p(x) = 5Cx (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.
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