D6C y12 Question 2.2 (OPEN BOOK EXAM) While reviewing, in preparation for an exa
ID: 3312503 • Letter: D
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D6C y12 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 DGC 0.4, and 0.3 problem indepenDGC+200 respectively Fig. 2.2. Preparing for final exam ( a ) Find the probability that the problem will be solved by at 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 and Chris can solve 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
Here, we are given the 3 probabilities as:
P(A) = 0.2 + ( 412 ) / ( 412 + 200) = 0.8732
P(B) = 0.4
P(C) = 0.3
a) Probability that the problem will be solved by at least one of them
= 1 - Probability that the problem wont be solved by anyone
= 1 - ( 1- 0.8732)*(1 - 0.4)*(1 - 0.3)
= 0.9467
Therefore 0.9467 is the required probability here.
b) Here we are given that:
P(A and B) = 0.45,
P(A and C) = 0.65,
P(B and C) = 0.35,
P( A and B and C) = 0.26
Using the above given probabilities we compute:
P( A and B only ) = P(A and B) - P( A and B and C) = 0.45 - 0.26 = 0.19
P( A and C only ) = P(A and C) - P( A and B and C) = 0.65 - 0.26 = 0.39
P( C and B only ) = P(C and B) - P( A and B and C) = 0.35 - 0.26 = 0.09
Now, computing P(A only ) = P(A) - P( A and B only ) - P( A and C only ) - P( A and B and C)
Therefore, P( solved by exactly 2) = 0.19 + 0.39 + 0.09 = 0.67 and P( all 3 ) = 0.26
Therefore, P( exactly 2 or all 3 ) = 0.67 + 0.26 = 0.93
Therefore the probability that it is solved by at least one of them is computed as:
= 1 - 0.93
= 0.07
Therefore 0.07 is the required probability here.
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