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Two urns contain 5 white and 7 black balls and 3 white and 9 black balls respectively. One ball is transferred to the second urn and then one ball is drawn from the second urn. Find the probability that the first ball transferred is black, given that the ball drawn is black?









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Option(C) is correct

$P(A) =$ probability that the ball transferred from urn first to second is black:


$P(B) =$ probability that the ball drawn from second urn is black.

Case I: If white ball goes to urn II

\(P(C)=\dfrac{5}{12}\times \dfrac{9}{13}\)

Case II: If black ball goes to urn II

\(P(D)=\dfrac{7}{12}\times \dfrac{10}{13}\)





\(P(A/B)=\) Probability of event $A$ when $B$ has occurred

= Probability that ball drawn from II urn is black

\(\Rightarrow \dfrac{\text{A intersects B}}{B}\)

\(=\dfrac{\dfrac{7}{12}\times \dfrac{10}{13}}{115/156}\)


(2) Comment(s)


While calculating the probability of drawing a black ball from second urn we are making cases and in those cases like in case 1 you are multiplying 9/13 with 5/12 which is not needed and is incorrect since we just have to calculate the probability of drawing black ball . adding one more white ball to the urn will just increase the total number of balls.

I think it should be just 9/13 and similarly with another case


Drawing a ball from urn 1 will sure affect the probability of drawing the black ball from urn 2. That multiplication is very much needed.

Otherwise, it would be no different than adding a KNOWN COLOR new ball to urn 2 without have to pick it from urn 2.

I believe you have to take care the color of the ball which is taken from urn 1 as this will affect the probability of drawing black ball from urn 2.