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Urn 1 contains 5 white balls and 8 black balls and Urn 2 contains 7 white balls and 2 black balls. One ball is picked from urn 1 and dropped in urn 2. Then one ball is picked from urn 2. If this ball turns out to be white, find the probability that the ball taken from urn 1 and dropped in urn 2 is also white.
(A) \(\frac{5}{12}\)
(B) \(\frac{7}{12}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{5}{13}\)
(E) \(\frac{6}{13}\)
(A) \(\frac{5}{12}\)
(B) \(\frac{7}{12}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{5}{13}\)
(E) \(\frac{6}{13}\)
04 Oct 2019, 05:17
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Bismarck
This is a question on conditional probability.
P (X given Y) = P("first ball is white" given "second ball is white")
P(first ball is white and second is white ) = (5/13) * (8/10) = 40/130
P (second ball is white) = (5/13)*(8/10) + (8/13)*(7/10) = (40 + 56)/130 = 96/130
P(X given Y) = P(first ball is white and second is white )/P (second ball is white) = (40/130) / (96/130) = 5/12
Check this post on conditional probability: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... onditions/
14 Oct 2018, 05:29
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OA: A
Following combinations are possible : WW,WB,BW,BB
Probability of each case
WW ( 1st Urn : White ;2nd Urn : White) \(: \frac{5}{13}*\frac{8}{10}=\frac{40}{130}\)
WB ( 1st Urn : White ; 2nd Urn : Black) \(: \frac{5}{13}*\frac{2}{10}=\frac{10}{130}\)
BW ( 1st Urn : Black ; 2nd Urn : White) \(: \frac{8}{13}*\frac{7}{10}=\frac{56}{130}\)
BB ( 1st Urn : Black ; 2nd Urn : Black) \(: \frac{8}{13}*\frac{3}{10}=\frac{24}{130}\)
Probability of 1 st ball being white , given 2nd ball is white\(= \frac{WW}{WW+BW} =\frac{\frac{40}{130}}{\frac{40}{130}+\frac{56}{130}}=\frac{40}{96}=\frac{5}{12}\)
Following combinations are possible : WW,WB,BW,BB
Probability of each case
WW ( 1st Urn : White ;2nd Urn : White) \(: \frac{5}{13}*\frac{8}{10}=\frac{40}{130}\)
WB ( 1st Urn : White ; 2nd Urn : Black) \(: \frac{5}{13}*\frac{2}{10}=\frac{10}{130}\)
BW ( 1st Urn : Black ; 2nd Urn : White) \(: \frac{8}{13}*\frac{7}{10}=\frac{56}{130}\)
BB ( 1st Urn : Black ; 2nd Urn : Black) \(: \frac{8}{13}*\frac{3}{10}=\frac{24}{130}\)
Probability of 1 st ball being white , given 2nd ball is white\(= \frac{WW}{WW+BW} =\frac{\frac{40}{130}}{\frac{40}{130}+\frac{56}{130}}=\frac{40}{96}=\frac{5}{12}\)
