is the geometric distribution an appropriate model? The geometric distribution is an appropriate model if the following assumptions are true. The phenomenon being modelled is a sequence of independent trials. There are only two possible outcomes for each trial, often designated success or failure. The probability of success, p, is the same for every trial. If these conditions are true, then the geometric random variable Y is the count of the number of failures before the first success. The possible number of failures before the first success is 0, 1, 2, 3, and so on. In the graphs above, this formulation is shown on the right. An alternative formulation is that the geometric random variable X is the total number of trials up to and including the first success, and the number of failures is X − 1. In the graphs above, this formulation is shown on the left. Probability outcomes examples The general formula to calculate the probability of k failures before the first success, where the probability of success is p and the probability of failure is q = 1 − p, is for k = 0, 1, 2, 3, ... E1) A doctor is seeking an antidepressant for a newly diagnosed patient. Suppose that, of the available anti-depressant drugs, the probability that any particular drug will be effective for a particular patient is p = 0.6. What is the probability that the first drug found to be effective for this patient is the first drug tried, the second drug tried, and so on? What is the expected number of drugs that will be tried to find one that is effective? The probability that the first drug works. There are zero failures before the first success. Y = 0 failures. The probability Pr(zero failures before first success) is simply the probability that the first drug works. The probability that the first drug fails, but the second drug works. There is one failure before the first success. Y = 1 failure. The probability for this sequence of events is Pr(first drug fails) p(second drug succeeds), which is given by The probability that the first drug fails, the second drug fails, but the third drug works. There are two failures before the first success. Y = 2 failures. The probability for this sequence of events is Pr(first drug fails) p(second drug fails) Pr(third drug is success) E2) A newlywed couple