max96hsv wrote:
Hi, I was wandering if the approach of "1 - counterprobability ( =no rotten apple)" is valid? I.e. 1 - 4/5*3/4= 1 - 12/20= 8/20 = 2/5
Yes! Great and totally valid solution. While this method is most commonly used with more complex "at least/more than" problems, any time we can break the scenario down into complementary events (two events where one and only one event must occur), we can solve for one event by finding 1 - the probability of its complement.
In this case, the complement of picking the bad apple is NOT picking the bad apple, so the probability of picking the bad apple = 1 - the probability of NOT picking the bad apple.
On the first pick, we have a 4/5 chance of NOT picking the bad apple (out of the 5 total apples in the bag, 4 are NOT bad). On the second pick, we have a 3/4 chance of NOT picking the bad apple (of the 4 apples left in the bag after the first pick, 3 are NOT bad). So we have a 4/5*3/4 = 12/20 = 3/5 chance of NOT picking the bad apple in either of the two picks. So the chance of picking the bad apple in either of the two picks is 1 - 3/5 = 2/5.