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Urn 1 contains 5 white balls and 8 black balls and Urn 2 contains 7 wh

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Urn 1 contains 5 white balls and 8 black balls and Urn 2 contains 7 wh  [#permalink]

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New post Updated on: 14 Oct 2018, 05:29
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Question Stats:

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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) \(\frac{6}{13}\)

Originally posted by Bismarck on 06 Oct 2018, 03:06.
Last edited by Bismarck on 14 Oct 2018, 05:29, edited 1 time in total.
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Re: Urn 1 contains 5 white balls and 8 black balls and Urn 2 contains 7 wh  [#permalink]

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New post 06 Oct 2018, 19:57
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I am not sure if the picking white ball will have an impact on the probability to pick white ball from Urn 1.

My take is:
* We can select white ball from Urn 1 in 5C1 ways
* Total #ways to pick a ball from Urn 1 is 13C1
* Hence, the probability to pick white ball from Urn 1 = 5C1/13C1 = 5/13.

Option D
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Re: Urn 1 contains 5 white balls and 8 black balls and Urn 2 contains 7 wh  [#permalink]

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New post 14 Oct 2018, 02:44
Bismarck wrote:
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) \(\frac{6}{13}\)


Bismarck.. Is the answer to the question correct.
It does not matter what the probability of ball picked up from Urn 2 is.
chetan2u can you help with this question
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Re: Urn 1 contains 5 white balls and 8 black balls and Urn 2 contains 7 wh  [#permalink]

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New post 14 Oct 2018, 04:17
So the probability that the Ball picked from urn 2 is white, could be
1. either 7/10, if the ball dropped from urn 1 is Black.
2 or, 8/10, if the ball picked from urn 1 is white.

If the ball picked and dropped from urn 1 to urn 2 is white, (as per the ques); then the probability of the ball picked from urn 2 = 8/10 = 4/5

Hence the probability that a ball picked from urn 1 is white AND the ball picked from urn 2 is white = 5/8 * 4/5 = 1/2

Answer is C
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Re: Urn 1 contains 5 white balls and 8 black balls and Urn 2 contains 7 wh  [#permalink]

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New post 14 Oct 2018, 04:41
PALL07 wrote:
So the probability that the Ball picked from urn 2 is white, could be
1. either 7/10, if the ball dropped from urn 1 is Black.
2 or, 8/10, if the ball picked from urn 1 is white.

If the ball picked and dropped from urn 1 to urn 2 is white, (as per the ques); then the probability of the ball picked from urn 2 = 8/10 = 4/5

Hence the probability that a ball picked from urn 1 is white AND the ball picked from urn 2 is white = 5/8 * 4/5 = 1/2

Answer is C

PALL07 I agree with your solution, however, that is the probability of the entire outcome.
Question jut asks for the probability of first urn.
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Re: Urn 1 contains 5 white balls and 8 black balls and Urn 2 contains 7 wh  [#permalink]

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New post 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\(= \frac{WW}{WW+BW} =\frac{\frac{40}{130}}{\frac{40}{130}+\frac{56}{130}}=\frac{40}{96}=\frac{5}{12}\)
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Re: Urn 1 contains 5 white balls and 8 black balls and Urn 2 contains 7 wh  [#permalink]

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New post 04 Oct 2019, 04:11
Bismarck wrote:
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\(= \frac{WW}{WW+BW} =\frac{\frac{40}{130}}{\frac{40}{130}+\frac{56}{130}}=\frac{40}{96}=\frac{5}{12}\)


Why did you divide WW?
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Re: Urn 1 contains 5 white balls and 8 black balls and Urn 2 contains 7 wh  [#permalink]

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New post 04 Oct 2019, 05:17
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Bismarck wrote:
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) \(\frac{6}{13}\)


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.veritasprep.com/blog/2012/0 ... onditions/
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Re: Urn 1 contains 5 white balls and 8 black balls and Urn 2 contains 7 wh   [#permalink] 04 Oct 2019, 05:17
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