Binomial Distribution

For a binomial distribution with parameters n and p, the mean is np and the variance is np(1 − p), so the ratio of variance to mean reveals 1 − p directly. This mean–variance back-solving is a standard JEE Advanced technique. Given np = 4 and np(1 − p) = 2, dividing the variance by the mean gives np(1 − p)/(np) = 1 − p = 2/4 = 1/2, hence p = 1/2. Substituting into np = 4 yields n × (1/2) = 4, so n = 8. Option 6 results from misusing variance = np instead of np(1 − p). Option 4 confuses n with the mean value itself. Option 16 comes from solving n × (1/4) = 4 with an incorrect p. The solution relies on the binomial moment formulas mean = np and variance = npq with q = 1 − p, and recognising their ratio isolates q. Plausibility check: with n = 8 and p = 1/2, mean = 8 × 0.5 = 4 and variance = 8 × 0.5 × 0.5 = 2, both matching the given data exactly, so n = 8 is verified.