Let's see, Solveit.....
P(A and B) = P(A) * P(B) [ if A, B are independent events ]
"At least one" means that we have three possibilities:
Let A = probability that Tom passes= .3 and B = probability that John passes = .4
Both pass = P (A and B) = .3 * . 4 = .12
Tom passes, and John doesn't = P (A and not B) =.3 * .6 = .18
John passes and Tom doesn't = P(not A and B) = .7 * .4 = .28
Summing these, we have .12 + .18 + .28 = .58 (D)
