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    				| 1. | The probability that a bulb produced in a factory will fuse after 10 days is 0.05. Find the probability that out of 5 such bulbs, not more than 1 will fuse after 400 days of use. | 
| Answer» Given that X ~ B (5, 0.05) (i.e.) n = 5, p = 0.05, q = 0.95 P(X ≤ 1) = P(X = 0) + P(X = 1) = 5C0(0.05)0(0.95)5 + 5C1(0.05)1(0.95)4 = (0.95)5 + 5 (0.05) (0.95)4 = (0.95)4 + [0.95 + 0.25] = 0.9774 | |