While reviewing in preparation for an exam, three students: Amy, Bob, and Chris
ID: 3311897 • Letter: W
Question
While reviewing in preparation for an exam, three students: Amy, Bob, and Chris are sitting together. The probabilities that Amy, bob and Chris can Solve a certain problem independently
Question 2.2 (OPEN BOOK EXAM) While reviewing, in preparation for an exam, three students: Amy, Bob, and Chris are sitting together. The probabilities that Amy, Bob, and Chris can solve a certain problem independently are 0.2+ respectively 0.4, and 0.3 O. 5584 Fig. 2.2. Preparing for final exam ( a) Find the probability that the problem will be olved by a least one of them if they try independently (b) In part (a), if the probabilities that Amy and Bob together can solve the problem is 0.45; that Amy the problem is 0.65; that Bob and Chris can solve the problem is 0.35: and that all together can solve the problem is 0.26. What is the probability that the problem will be solved by at least one of them?Explanation / Answer
Let A shows the event that Amy will solve the problem, B shows the event that Bob will solve the problem and C shows the event that Chris will solve the problem.
So we have
P(A) = 0.2+0.5584 = 0.7584
P(B) = 0.4
P(C) = 0.3
(a)
Since they solve the problems independently so
P(A and B) = P(A)P(B) = 0.7584 * 0.4 = 0.30336
P(B and C) = P(B)P(C) = 0.4 *0.3 = 0.12
P(A and C) = P(A)P(C) = 0.7584 *0.3 = 0.22752
and
P(A and B and C) = P(A)P(B)P(C) = 0.7584 *0.4 *0.3 = 0.091008
Therefore, the probability that probalem will be solved at least one of them is
P(A or B or C) = P(A) + P(B) +P(C) -P(A and B)-P(A and C)-P(B and C)+-P(A and B and C) = 0.7584 +0.4 +0.3 - 0.30336- 0.12 - 0.22752 + 0.091008 = 0.898528
(b)
Now we have
P(A and B) = 0.45
P(B and C) =0.35
P(A and C) = 0.65
and
P(A and B and C) = 0.26
Therefore, the probability that probalem will be solved at least one of them is
P(A or B or C) = P(A) + P(B) +P(C) -P(A and B)-P(A and C)-P(B and C)+-P(A and B and C) = 0.7584 +0.4 +0.3 - 0.45- 0.35 - 0.65 + 0.26 = 0.2684
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