# drawing balls from an urn (conditional probability)

Urn A contains 4 white balls and 2 red balls. Urn B contains 3 red balls and 3 black balls. An urn is randomly selected, and then a ball inside of that urn is removed. We then repeat the process of selecting an urn and drawing out a ball, without returning the first ball. What is the probability that the first ball drawn was red, given that the second ball drawn was black?

I encountered this problem in my AoPS textbook, as a complete newcomer to conditional probability. Is the answer 1/2? I would really appreciate a solution, as I have no way to check my work.

Randomly select an urn, draw a ball without replacement, then again randomly select an urn and draw a second ball. Let $D_n$ be the colour of the $n^{th}$ draw, and $U_n$ be the urn from which it is drawn.

Using k for black, the urns are: $a = \{(w,4), (r,2)\}, b = \{(r,3), (k,3)\}$

Now, if we know a black ball is going to be removed in the second draw, then only one of the other two black balls, or one of the three red balls, could be removed from the second urn during the first draw.

So the probability of drawing a red ball in the first draw give that knowledge is: $$P(D_1=r \mid D_2=k) \\ = P(D_1=r \cap U_1=a \mid D_2=k)+P(D_1=r \cap U_1=b \mid D_2=k) \\ = P(U_1=a)P(D_1=r \mid U_1=a \cap D_2=k)+P(U_1=b)P(D_1=r \mid U_1=b \cap D_2=k) \\ = \frac{1}{2}\frac{2}{6}+\frac{1}{2}\frac{3}{5} \\ = \frac{7}{15}$$

• Thank you so much for responding to an older post of mine! Much appreciated :) Apr 24, 2014 at 2:02
• I am unable to understand this part. P(U1=b)P(D1=r∣U1=b∩D2=k) = 1/2(3/5). Why won't it be 1/2(3/6) ? P(Urn1) = P(Urn2)= 1/2 and the total number of balls = 6 , red balls : 3. ??? Please explain. The required Probabilty must be (7/15) according to my calculation which you can check. Please explain Graham Kemp. Sep 24, 2021 at 10:00
• Under the condition that the second draw will be black and the first draw shall be from the same urn ($b$), then the first ball must be drawn from among the other five balls in that urn (three of which are red). Because we are drawing without replacement. @SoudiptaDutta Sep 24, 2021 at 13:34
• "... if we know a black ball is going to be removed in the second draw, then only one of the other two black balls, or one of the three red balls, could be removed from the second urn during the first draw," means $\mathsf P(D_1{\,=\,}r\mid D_2{\,=\,}k, U_1{\,=\,}a) = 3/5$ Sep 24, 2021 at 13:40

The answer is the Probability that the first ball is red and the second ball is black, divided by the Probability that the second ball is black.

$$\frac{P(\text{First Ball is RED AND Second Ball is BLACK) }}{P(\text{Second Ball is BLACK) }}$$

P(1st ball = Red and 2nd ball = Black )

$$=\frac{1}{2}[ (\frac{2}{6}* \frac{1}{2}*\frac{3}{6}) + ( \frac{3}{6}* \frac{1}{2}*\frac{3}{5} ) ] = \frac{7}{60}$$

P(Second ball is BLACK) = $$\frac{1}{2}[ (Urn A's Redball*\frac{1}{2}*\frac{3}{6}) + ( \frac{3}{6}* \frac{1}{2}*\frac{3}{5} ) + ( \frac{3}{6}* \frac{1}{2}*\frac{2}{5} ) ] = \frac{1}{2}[ (1*\frac{1}{2}*\frac{3}{6}) + ( \frac{3}{6}* \frac{1}{2}*\frac{3}{5} ) + ( \frac{3}{6}* \frac{1}{2}*\frac{2}{5} ) ] = \frac{1}{4}$$

Therefore,

$$\frac{P(\text{First Ball is RED AND Second Ball is BLACK) }}{P(\text{Second Ball is BLACK) }}$$ = $$\frac{\frac{7}{60}}{\frac{1}{4}} = \frac{7}{15}$$