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