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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?

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    $\begingroup$ Have you heard of conditional probabilities before? Do you know how to rewrite $P(X\mid Y)$ in terms of $P(X),P(Y),P(X\cap Y)$? $\endgroup$ – JMoravitz Oct 7 '19 at 19:38
  • $\begingroup$ Do you know and understand Bayes' theorem? $\endgroup$ – B.Swan Oct 7 '19 at 19:42
  • $\begingroup$ I am familiar with Bayes theorem, I just don't know how to apply it to this question. $\endgroup$ – Mark Oct 7 '19 at 20:06
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    $\begingroup$ $X$ 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. $\endgroup$ – JMoravitz Oct 7 '19 at 20:20
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    $\begingroup$ 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 $$P(X) = \dfrac{4}{14}, P(Y) = P(RB)+P(YB)+P(BB)$$ $\endgroup$ – InterstellarProbe Oct 7 '19 at 20:21
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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.

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