# Urn A has 4 red, 3 yellow, and 7 blue balls. Urn B has 3 red, 4 yellow, and 12 blue balls.

Urn A has 4 red, 3 yellow, and 7 blue balls. Urn B has 3 red, 4 yellow, and 12 blue balls. A ball is first selected at random from urn A and put into urn B. Then a ball is selected at random from urn B. If the ball selected from urn B is blue, what is the probability that the ball transfered from urn A to urn B is red?

The answer is $$48/175$$ but I have no idea how that answer was derived. Could someone please explain?

• Have you heard of conditional probabilities before? Do you know how to rewrite $P(X\mid Y)$ in terms of $P(X),P(Y),P(X\cap Y)$? Oct 7, 2019 at 19:38
• Do you know and understand Bayes' theorem? Oct 7, 2019 at 19:42
• I am familiar with Bayes theorem, I just don't know how to apply it to this question.
– Mark
Oct 7, 2019 at 20:06
• $X$ is an event, not a number. You mean to say $Pr(X)=\frac{4}{14}$ instead. Now, $Pr(Y)=Pr(Y\cap X)+Pr(Y\cap X^c) = Pr(X)Pr(Y\mid X)+Pr(X^c)Pr(Y\mid X^c)$. You should be able to find each of the above numbers and you should be able to see why the above expansion was valid to do. Oct 7, 2019 at 20:20
• Define the following events: $$\begin{array}{cc}RR & \text{xfer red, choose red} \\ RY & \text{xfer red, choose yellow} \\ \vdots & \vdots \\ BY & \text{xfer blue, choose yellow} \\ BB & \text{xfer blue, choose blue}\end{array}$$ Then $$P(X) = \dfrac{4}{14}, P(Y) = P(RB)+P(YB)+P(BB)$$ Oct 7, 2019 at 20:21

Urn A has 4 red, 3 yellow, and 7 blue balls. Urn B has 3 red, 4 yellow, and 12 blue balls. A ball is first selected at random from urn A and put into urn B. Then a ball is selected at random from urn B. If the ball selected from urn B is blue, what is the probability that the ball transfered from urn A to urn B is red?

If the second ball is blue, it is either originally from urn B, or originally from urn A.

The probability for drawing a blue ball that is from the 12 originally among urn B, when drawing from the $$19$$ balls in urn B after the transfer is: ___

The probability for first transfering one from the 7 blue ball from urn A (among the 14 balls there) and then drawing it again is: ___

So the probability for drawing a blue balls is: ___. Call this $$\mathsf P(Y)$$

The probability for transferring one from the 4 red balls in urn A and then drawing one from the 12 blue balls in urn B is: ___. Call this $$\mathsf P(X\cap Y)$$.

You seek $$\mathsf P(X\mid Y)$$, the probability for transferring a red ball when given that a blue ball will be drawn from the second urn. Use the definition of conditional probability.