SOLUTION: One urn contains 3 read and 7 white marbles. A second urn contains 2 red and 8 white marbles. One urn is selected and a ball drawn. The probability of selecting the first urn is
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Question 92278: One urn contains 3 read and 7 white marbles. A second urn contains 2 red and 8 white marbles. One urn is selected and a ball drawn. The probability of selecting the first urn is 0.4 and the probability of selecting the second is 0.6. If a white ball is drawn, find the probability it came from the second urn. Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! One urn contains 3 read and 7 white marbles. A second urn contains 2 red and 8 white marbles. One urn is selected and a ball drawn. The probability of selecting the first urn is 0.4 and the probability of selecting the second is 0.6. If a white ball is drawn, find the probability it came from the second urn.
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P(2nd urn|white) = P(2nd and white)/P(w)
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But P(2nd and white) = 8/20 = 2/5 = 0.4
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And P(w) = P(w |1st urn)+P(w |2nd urn)
=P(w AND 1st)/P(1st)+P(w AND 2nd)/P(2nd)
= (7/20)/(8/20 + (8/20)/(12/20)
= (7/8) + (2/3)
= 37/24
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Final: P(2nd urn |white) = 0.4/(37/24) = 0.259...
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If you have another answer I would like to see it.
Why? Probability problems are always tricky.
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Cheers,
Stan H
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