question_answer 63)

The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulb (i) one (ii) more than one (iii) one more than one (iv) atleast one will fuse after 150 days of use.  

Answer:

Given that p = 0.05        q = 1 ? 0.05 = 0.95       Let x be the number of fused bulbs, then x has the binomial distribution with n = 5, p = 0.5, q = 0.95        (i) P(none of fused bulb) = P (x = 0)        = 5C­0 (0.95)5-0 (0.05)0 = (0.95)5       (ii) P (not more than one fused bulb)             = 5C0 (0.95)5-0 (0.05)0 + 5C1 (0.05)5?1 (0.05)1                         = (0.95)4 + 5 (0.95)4 (0.05)       = (0.95)4 (0.95 + 0.25)       = 1.2 × (0.95)4       (iii) P (more than one fused bulbs)       = P(x > 1) = 1 [P (x = 0) + P(x = 1)]       = 1 ? (1.2) (0.95)4                 [Using (ii) part]       (iv) P (atleast one fused blub) = P (x  1)       = 1 ? P (x < 1) =1 ? P (x = 0)