# An urn contains 3 white balls and 4 black balls. Second urn contains 6 white balls and 4 black balls.

An urn contains $3$ white balls and $4$ black balls. Second urn contains $6$ white balls and $4$ black balls. From the first urn are draws $2$ balls and they dropped in the second urn. Then from the second urn is drawn one ball which is white.

What is the probability that two balls transferred in second urn were black balls?

My solution for this is problem is $1/4$, but the in math book it is saying that answer is $1/3$

Using Baye's theorem: $$Pr=\frac{\frac{1}{2}\times\frac{6}{21}}{(\frac{1}{2}\times\frac{6}{21})+(\frac{7}{12}\times\frac{12}{21})+(\frac{2}{3}\times\frac{3}{21})}=\frac{3}{12}=\frac{1}{4}$$ in which the first term at denominator is for the case that both ball are black, the second term is for the case that one of them is white, and the other is black, and the last term is for the case that both balls are white.

According to the Baye's theorem, the probability that both ball are black given that the output ball is white is equal to a ratio, that nominator is equal to the probability of the intended event given the two balls are black and the denominator is equal to the summation of the probabilities of the desired event (that output is white) given all the situations, one by one

• Wow, I am so sorry. I thought that the ball drawn from the second urn is black. But if it is white, then you are right. I will edit my answer – CLAUDE Dec 27 '14 at 22:04
• You've identified why the book's answer was different to the question's answer. – Graham Kemp Dec 29 '14 at 3:39

Amir has identified why the book's answer and your's differ, Felix Brauer

The question you asked was:

An urn contains 3 white balls and 4 black balls. Second urn contains 6 white balls and 4 black balls. From the first urn are draws 2 balls and they dropped in the second urn. Then from the second urn is drawn one ball which is white. What is the probability that two balls transferred in second urn were black balls ?

The solution to this is, using Conditional Probability and the Law of Total Probability: \begin{align} \mathsf P(F_{BB}\mid S_W) & = \frac{ \mathsf P(S_W\mid F_{BB})\,\mathsf P(F_{BB}) }{ \mathsf P(S_W\mid F_{BB})\,\mathsf P(F_{BB})+\mathsf P(S_W\mid F_{BW})\,\mathsf P(F_{BW})+\mathsf P(S_W\mid F_{WW})\,\mathsf P(F_{WW}) } \\ & = \frac{\frac 6 {12}{4\choose 2}/{7\choose 2}}{\left(\frac 6 {12}{4\choose 2}+\frac 7{12}{3\choose 1}{4\choose 1}+\frac 8{12}{3\choose 2}\right)/{7\choose 2}} \\ & = \frac{6\cdot 6}{6\cdot 6+7\cdot 12+8\cdot 3} \\ & = 1 / 4 \end{align}

As you, and Amir, obtained.

\begin{align} \mathsf P(F_{BB}\mid S_B) & = \frac{ \mathsf P(S_B\mid F_{BB})\,\mathsf P(F_{BB}) }{ \mathsf P(S_B\mid F_{BB})\,\mathsf P(F_{BB})+\mathsf P(S_B\mid F_{BW})\,\mathsf P(F_{BW})+\mathsf P(S_B\mid F_{WW})\,\mathsf P(F_{WW}) } \\ & = \frac{\frac 6 {12}{4\choose 2}/{7\choose 2}}{\left(\frac 6 {12}{4\choose 2}+\frac 5{12}{3\choose 1}{4\choose 1}+\frac 4{12}{3\choose 2}\right)/{7\choose 2}} \\ & = \frac{6\cdot 6}{6\cdot 6+5\cdot 12+4\cdot 3} \\ & = 1 / 3 \end{align}