plans to have children and will continue until the first girl. What is the probability that there are zero boys before the first girl, one boy before the first girl, two boys before the first girl, and so on? The probability of having a girl (success) is p= 0.5 and the probability of having a boy (failure) is q = 1 − p = 0.5. The probability of no boys before the first girl is The probability of one boy before the first girl is The probability of two boys before the first girl is and so on. Properties Moments and cumulants The expected value for the number of independent trials to get the first success, and the variance of a geometrically distributed random variable X is: Similarly, the expected value and variance of the geometrically distributed random variable Y = X - 1 (See definition of distribution ) is: Proof That the expected value is (1 − p)/p can be shown in the following way. Let Y be as above. Then The interchange of summation and differentiation is justified by the fact that convergent power series converge uniformly on compact subsets of the set of points where they converge. Let μ = (1 − p)/p be the expected value of Y. Then the cumulants of the probability distribution of Y satisfy the recursion Expected value examples E3) A patient is waiting for a suitable matching kidney donor for a transplant. If the probability that a randomly selected donor is a suitable match is p = 0.1, what is the expected number of donors who will be tested before a matching donor is found? With p = 0.1, the mean number of failures before the first success is E(Y) = (1 − p)/p =(1 − 0.1)/0.1 = 9. For the alternative formulation, where X is the number of trials up to and including the first success, the expected value is E(X) = 1/p = 1/0.1 = 10. For example 1 above, with p = 0.6, the mean number of failures before the first success is E(Y) = (1 − p)/p = (1 − 0.6)/0.6 = 0.67. Higher-order moments The moments for the number of failures before the first success are given by where is the polylogarithm function. General properties The probability-generating functions of X and Y are, respectively, Like its continuous analogue (the exponential distribution), the geometric distribution is memoryless.