## Thinking Mathematically (6th Edition)

If A and B are independent events, then P(E) = P(A)*P(B) A: frequent hangovers B: frequent hangovers C: frequent hangovers P(A) = 0.10 P(B) =0.10 P(C)= 0.10 P(E) = P(A)$\times$P(B)$\times$P(C) P(E) =0.10 x 0.10 x 0.10 =0.001 Thus, the probability that three randomly selected adults will all suffer from frequent hangovers is .001.