Let U be a geometric random variable with p=.3, V be a geometric random variable with p=.2, X be a geometric random variable with p=.5, Y be a geometric random variable with p=.4. Let U, V, X, Y be independent and W=U+V+Y+X. Find P(W=k) for k=6, 7, 8. Solution W is also geometric with parameter p=[1 - (1 - .3)(1 - .2 )(1 - .5)(1 - .6)] = [1 - (0.7)(0.8)(0.5)(0.4)]= 0.832 Then P(W=k) is the probability the first successful outcome occurs on the kth trial. P(W=6) = 5 failures followed by a success on the 6th trial = (1 - 0.832)5•(0.832) = 0.0001113 P(W=7) = 6 failures followed by a success on the 7th trial = (1 - 0.832)6•(0.832) = 0.0000187 P(W=8) = 7 failures followed by a success on the 8th trial = (1 - 0.832)7•(0.832) = 0.00000314 *Note: There are other ways to correctly define a geometric distribution. This is the method I see most often..