A geometric distribution models the number of trials until the first success. What is its mean?

A geometric distribution models the number of trials until the first success. What is its mean?

• A. 1/p
• B. p
• C. 1-p
• D. 1/p^2

Answer: (C)

The key idea is the expected number of trials needed to get the first success when each trial has probability p of success. If K is the trial on which the first success occurs, then K takes values 1, 2, 3, … and has probability P(K = k) = p(1−p)^(k−1). The mean is the sum E[K] = sum_{k≥1} k p (1−p)^(k−1). Using the standard series sum for k r^(k−1) = 1/(1−r)^2 with r = 1−p, we get E[K] = p * 1/(1−(1−p))^2 = p * 1/p^2 = 1/p. So the mean number of trials until the first success is 1/p. (If you instead count the number of failures before the first success, the mean would be (1−p)/p; the question specifies counting trials, so the correct mean is 1/p.)

The key idea is the expected number of trials needed to get the first success when each trial has probability p of success. If K is the trial on which the first success occurs, then K takes values 1, 2, 3, … and has probability P(K = k) = p(1−p)^(k−1). The mean is the sum E[K] = sum_{k≥1} k p (1−p)^(k−1). Using the standard series sum for k r^(k−1) = 1/(1−r)^2 with r = 1−p, we get E[K] = p * 1/(1−(1−p))^2 = p * 1/p^2 = 1/p.

So the mean number of trials until the first success is 1/p. (If you instead count the number of failures before the first success, the mean would be (1−p)/p; the question specifies counting trials, so the correct mean is 1/p.)