Question

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 bulbs. (i) none (ii) not more than one (iii) more than one (iv) at least one will fuse after 150 days of use Select the correct answer from above options

Answer 1

Probability that a bulb gets fuse after 150 days of its use = 0.05 Probability that the bulb will not fuse after 150 days of its use = 1 – 0.05 = 0.95 (i) Probability that no bulb will fuse after 150 days of its use = P(none) = (0.95)5 = 0.7738 = 0.77(approx) (ii) P(not more than one ) = O(0) + P(1) = (0.95)5 + 5C1 x (0.95)4 x (0.05) = (0.95)5[0.95 + 5 x 0.05] = (0.95)4 x 1.2 (iii) P(more than one) = P(2) + P(3) + P(4) + P(5) = [P(0) + P(1) + P(2) + P(3) + P(4) + P(5)] - [P(0) + P(1)] = 1 - [P(0) + P(1)] = 1 - (0.95)4 x 1.2