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    				| 1. | If the probability of success is 0.09, how many trials are needed to have a probability of at least one success as 1/3 or more? | 
| Answer» Given p = 0.09 (success) q = 0.91 (failure) We have to find number of trials ‘n.’ According to the problem, P(X ≥ 1 ) > 1/3 (We must have atleast one success) 1 – P(X < 1) > 1/3 1 – P(X = 0) > 1/3 (or) P(X = 0) < 2/3 Using p.m.f, we have, nC0(0.09) 0 (0.91) n < 2/3 (0.91)n < 2/3 we can use log tables to calculate (or) by trial method try for n = 1, 2,…… using calculator. We observe that (0.91)5 < 2/3 but (0.91)4 = 0.6857 > 2/3. Thus we need a minimum of 5 trials or more. | |