SOLUTION: An urn contains 4 white balls and 6 red balls. A second urn contains 6 white balls and 4 red balls. An urn is selected, and a ball is randomly drawn from the selected urn. The prob

Question 1203813: An urn contains 4 white balls and 6 red balls. A second urn contains 6 white balls and 4 red balls. An urn is selected, and a ball is randomly drawn from the selected urn. The probability of selecting the first urn is 0.8. If the ball is white, find the probability that the second urn was selected.
Found 2 solutions by greenestamps, math_tutor2020:
Answer by greenestamps(13198) (Show Source): You can put this solution on YOUR website!

P(first urn selected and ball is white) = (0.8)(4/10) = (0.8)(0.4) = 0.32

P(second urn selected and ball is white) - (0.2)(6/10) =(0.2)(0.6) = 0.12

P(ball is white) = 0.32+0.12 = 0.44

P(ball is white, given that it came from second urn) = 0.12/0.44 = 12/44 = 3/11

ANSWER: 3/11

Convert to decimal or percentage if required....

Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!

A = 1st urn
B = 2nd urn
W = ball is white
R = ball is red

P(A) = probability of selecting 1st urn
P(A) = 0.8 = 4/5
P(B) = 0.2 = 1/5

P(W given A) = probability the ball is white given the 1st urn is selected
P(W given A) = 4/(4+6) = 4/10 = 2/5
P(W given B) = 6/(6+4) = 6/10 = 3/5

P(A and W) = probability of selecting 1st urn and white ball
P(A and W) = P(A)*P(W given A)
P(A and W) = (4/5)*(2/5)
P(A and W) = 8/25

P(B and W) = probability of selecting 2nd urn and white ball
P(B and W) = P(B)*P(W given B)
P(B and W) = (1/5)*(3/5)
P(B and W) = 3/25

P(W) = probability the ball is white
P(W) = P(W and A) + P(W and B) ....... law of total probability
P(W) = P(A and W) + P(B and W)
P(W) = 8/25 + 3/25
P(W) = (8+3)/25
P(W) = 11/25

P(B given W) = probability the 2nd urn was chosen given we know a white ball was picked
P(B given W) = P(B and W)/P(W)
P(B given W) = (3/25) divide (11/25)
P(B given W) = (3/25) * (25/11)
P(B given W) = 3/11 is the final answer.

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Another approach

Imagine that we had 5 urns.
4 of those urns are labeled "A"
The last urn is labeled "B"
Note how P(A) = 4/5 = 0.8

Since we know the selected ball is white, we can ignore the red balls.
The first four urns labeled "A" chip in 4*4 = 16 white balls.

The last urn labeled "B" chips in 6 white balls to get 16+6 = 22 white balls total.

Of those 22 white balls, 6 came from that last urn (labeled "B")
6/22 = 3/11 is the probability of selecting the 2nd urn given that we know the selected ball is white.

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Answer: 3/11

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