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If the sum and product of the mean and variance of a binomial distribution are 18 and 72 respectively, then the probability of obtaining atmost one success is

Solution

✅ Correct Option: 1

Mean and variance are roots of t2−18t+72=0t^2 - 18t + 72 = 0, giving t=12t = 12 or 66. Since variance << mean for binomial: variance =6= 6, mean =12= 12. So npq=6npq = 6, np=12np = 12, giving q=1/2q = 1/2, p=1/2p = 1/2, n=24n = 24. P(X≤1)=(1/2)24+24(1/2)24=25(1/2)24P(X \leq 1) = (1/2)^{24} + 24(1/2)^{24} = 25(1/2)^{24}.

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