Sum of independent Binomial variables with the same success probability q is distributed as which?

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Multiple Choice

Sum of independent Binomial variables with the same success probability q is distributed as which?

When you sum independent binomial counts that share the same success probability, you can view all trials together as one big binomial experiment. If each X_i ~ Binomial(m_i, q) and the trials are independent, then the total number of successes X = X_1 + X_2 + ... + X_k behaves like a Binomial with the total number of trials M = sum m_i and the same success probability q. In other words, X ~ Binomial(M, q). This works because each of the M individual trials has probability q of success, and the total number of successes is just counting successes across all these trials. The probability mass function is P(X = k) = C(M, k) q^k (1−q)^{M−k}, with mean E[X] = M q and variance Var[X] = M q (1−q). If the probabilities q differed across X_i, the sum would not be binomial; it would follow a more general Poisson-binomial distribution. The given setup requires the same q, which is why the binomial form with the total number of trials is the correct description.

When you sum independent binomial counts that share the same success probability, you can view all trials together as one big binomial experiment. If each X_i ~ Binomial(m_i, q) and the trials are independent, then the total number of successes X = X_1 + X_2 + ... + X_k behaves like a Binomial with the total number of trials M = sum m_i and the same success probability q. In other words, X ~ Binomial(M, q).

This works because each of the M individual trials has probability q of success, and the total number of successes is just counting successes across all these trials. The probability mass function is P(X = k) = C(M, k) q^k (1−q)^{M−k}, with mean E[X] = M q and variance Var[X] = M q (1−q).

If the probabilities q differed across X_i, the sum would not be binomial; it would follow a more general Poisson-binomial distribution. The given setup requires the same q, which is why the binomial form with the total number of trials is the correct description.

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