Question 1172381: A bag contains 4 red chips and 4 blue chips. Another bag contains 2 red chips and 6 blue chips. A chip is randomly selected from one of the bags, and found to be blue. What is the probability that the chip is from the first bag?
Answer by math_tutor2020(3817)   (Show Source):
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Quick Method:
Ignore the red chips since we know the chip is blue.
We have 4 blue chips in bag A and 6 blue chips in bag B. There are 4+6 = 10 blue chips total.
The probability the blue chip is from bag A is 4/10 = 2/5, which is the answer.
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A more in-depth look:
Use the law of total probability to say that
P(blue) = P(blue and bag A) + P(blue and bag B)
P(blue) = P(blue, given bag A)*P(bag A) + P(blue, given bag B)*P(bag B)
P(blue) = (4/8)*(1/2) + (6/8)*(1/2)
P(blue) = (4/8+6/8)*(1/2)
P(blue) = (10/8)*(1/2)
P(blue) = 10/16
P(blue) = 5/8
Now use conditional probability
P(blue, given bag A) = P(bag A and blue)/P(bag A)
P(blue, given bag A)*P(bag A) = P(bag A and blue)
P(bag A and blue) = P(blue, given bag A)*P(bag A)
P(bag A, given blue) = P(bag A and blue)/P(blue)
P(bag A, given blue) = P(blue, given bag A)*P(bag A)/P(blue)
P(bag A, given blue) = (1/2)*(1/2)/(5/8)
P(bag A, given blue) = (1/4)/(5/8)
P(bag A, given blue) = (1/4)*(8/5)
P(bag A, given blue) = 8/20
P(bag A, given blue) = 2/5
If we know the chip is blue, the probability of it being from bag A is 2/5.
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Answer: 2/5
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