# Answer to Question #5218 in Statistics and Probability for cash

Question #5218
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? [12 marks]
1
2012-07-20T08:43:00-0400
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.

= 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

= 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

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!