There are two boxes. The first box contains 4 red chips and 3 blue chips. The se
ID: 3203569 • Letter: T
Question
There are two boxes. The first box contains 4 red chips and 3 blue chips. The second box contains 1 red chips and 5 blue chips. First, a chip is selected from the first box and put into the second box; then a chip is selected from the second box. Let B_1 be the event that the chip selected from the first box is red. Let B_2 be the event that the chip selected from the first box is blue. Let A be the event that the chip selected from the second box is red. Find P(B_1) P(B_2), P(A), P(B_1|A), and P(B_2|A).Explanation / Answer
Total number of balls in the first box is : 4 +3 = 7
P(B1) = 4/7
After placing red chip in the second box number of chips in the second box is 7 out of which 2 are red and 5 are blue. So
P(A|B1) = 2/7
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Now
P(B2) = 3/7
After placing blue chip in the second box number of chips in the second box is 7 out of which 1 is red and 6 are blue. So
P(A|B2) = 1/7
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By the law of total probability, the probability of event A is
P(A) = P(A|B1)P(B1)+P(A|B2)P(B2) = (2/7)(4/7) + (1/7)(3/7) = 11/49
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Now
P(B1|A) = [P(A|B1)P(B1)]/P(A) = 8 / 11
P(B2|A) = [P(A|B2)P(B2)]/P(A) = 3 / 11