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The probability of a bulb produced by a factory will fuse after 150 days of use is 0.05. Find 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.

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Best answer

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.

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