Answer
0.001
Work Step by Step
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.