Which statement correctly describes a binomial distribution?

• A. X ~ Bin(n,p); mean np; variance np(1-p)
• B. X ~ Bin(n,p); mean p; variance p(1-p)
• C. X ~ Geom(p); mean 1/p; variance (1-p)/p^2
• D. X ~ N(mu, sigma^2); mean mu; variance sigma^2

Answer: (A)

The main idea is that a binomial distribution counts how many successes occur in a fixed number of independent Bernoulli trials, each with probability p of success. If you perform n such trials, the expected number of successes is n times p, and the variability around that expectation is n p (1 − p). That makes X ~ Bin(n, p) have mean np and variance np(1 − p). So the statement with X ~ Bin(n,p); mean np; variance np(1-p) matches these properties precisely.

For context, think of flipping a coin 10 times. If p = 0.5, you’d expect about 5 heads (mean 10 × 0.5 = 5) and the spread around that expectation would be 10 × 0.5 × (1 − 0.5) = 2.5. The other options describe different things: a single Bernoulli trial has mean p and variance p(1−p); the geometric distribution refers to the number of trials until the first success; and the normal distribution is a continuous distribution with its own parameters mu and sigma^2.