Answer to Question #5218 in Statistics and Probability for cash
A group of chimpanzees are given two tasks to do, the second of which is harder than the first. A proportion 3/5 get the first one right and 7/15 get the second one right. If a chimpanzee gets the first one right, it has a conditional probability of 5/8 of also performing the second task right. One chimpanzee is chosen at random from the group.
(i) What is the probability that it gets both tasks right?
(ii) What is the probability that it gets at least one right?
(iii) If the chimpanzee got the first task wrong, what is the conditional probability that it will also get the second one wrong?
(iv) If the chimpanzee got the second task wrong, what is the probability that its first task was correctly carried out?
Let B be the event of getting the second task right.
We know that P(A) = 3/5, P(B) = 7/15, P(B given A) = 5/8.
(i) P(both tasks right) = P(A and B) = P(A)P(B given A) = (3/5)(5/8) = 3/8
(ii) P(at least one right) = P(A or B) = P(A) + P(B) - P(A and B) = 3/5 + 7/15 - 3/8, since P(A and B) = 3/8 from part (i) = 83/120
(iii) P(got second one wrong, given first one wrong) = P(not B given not A) = P(not A and not B) / P(not A) = (1 - P(A or B)) / (1 - P(A)) = (1 - (83/120)) / (1 - (3/5)), since P(A or B) = 83/120 from part (ii) = (37/120) / (2/5) = 37/48
(iv) P(got first task right, given second task wrong) = P(A given not B) = P(A and not B) / P(not B) = (P(A) - P(A and B)) / (1 - P(B)) = (3/5 - 3/8) / (1 - (7/15)), since P(A and B) = 3/8 from part (i) = (9/40) / (8/15) = 27/64