Question 1203813
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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) = <font color=red>3/11</font> 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 = <font color=red>3/11</font> is the probability of selecting the 2nd urn given that we know the selected ball is white.


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Answer:  <font color=red size=4>3/11</font>
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