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
Given: n = 5
∴ X ~ B(5, 0.05)
The p.m.f. of X is given by
(i) P(none of a bulb produced by a factory will fuse after 150 days of use) = P[X = 0]
Hence, the probability that none of the bulbs will fuse after 150 days = (0.95)5.
(ii) P(not more than one bulb will fuse after 150 days of j use) = P[X ≤ 1]
= p(0) + p(1)
Hence, the probability that not more than one bulb will fuse after 150 days = (1.2)(0.95)4.
(iii) P(more than one bulb fuse after 150 days) = 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 bulb fuse after 150 days)
Hence, the probability that at least one bulb fuses after 150 days = 1 – (0.95)5.