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There are two urns. Urn 1 contains 3 white and 2 red ball, urn 2 one white and two red. First, a ball from urn 1 is randomly chosen and placed into urn 2. Finally, a ball from urn 2 is picked. This ball be red: What is the probability that the ball transferred from urn 1 to urn 2 was white?

My answer is 3/5 - 3W over 5 balls in urn 1 as the second event does not tell me anything to influence which ball was transferred since urn 2 already has a red ball anyways. Hope to know if my reasoning is valid!

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    Well....observing a red ball from urn $2$ makes it somewhat more likely that a red one was moved. Write out the tree...analyze the events that lead to observing a red ball and compute how many of those events start with a red ball being transferred. – 2017-01-13
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    i just found the solution to a similar problem here; http://math.stackexchange.com/questions/770773/what-is-the-probability-that-the-first-ball-drawn-from-urn-a-was-also-red – 2017-01-13
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    It is absolutely a trick question. You are told that a red ball was picked from urn 2. That has absolutely no bearing on the probability that the ball moved from urn 1 to urn 2 was white. Instead: since there were 3 white and 2 red balls in urn 1, if each ball was equally likely to be moved from urn 1 to urn 2 (a reasonable assumption which was NOT expressed explicitly), the probability the transferred ball was white is 3/5 or 0.6. – 2017-01-13
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    It's different from "the probability that the ball picked from urn 2 was red was 0.75." Something like that WOULD tell you something about which ball was moved. But in that case, the "probability" they ask for would be either 0 or 1 - you would know with certainty whether the ball moved between urns was white or red. – 2017-01-13
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    @mathguy that's what i thought at first but then given the Bayes formula, this question can be transformed into what is the probability that the ball that was transferred from Urn 1 to Urn 2 was white given that the ball drawn from urn 2 is found to be red. How about that? :D – 2017-01-13
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    Absolutely not. The Bayes formula relates different probabilities, it has nothing to do with "a red ball has been drawn from urn 2". There is nothing in the problem about what the probability of that event was, before the event occurred. – 2017-01-13
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    @mathguy Not following you at all. Suppose that urn $2$ had initially contained only white balls. Then drawing the red would be absolute proof that the ball transferred from $1$ was red, yes? Therefore you can't claim that the draw from urn two "had no bearing" on the probability. – 2017-01-13
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    @mathguy Try it this way: Suppose we did this game $100$ times. Then suppose that $60$ of those times involve a white transfer and $40$ would be red. Of course this matches the desired probability. Now in the $60$ white transfer cases, $30$ should have a red observation. In the $40$ red transfer cases, $30$ have a red observation. Thus there are $60$ instances wherein a red ball is observed...half come from the white transfer and half from the red. – 2017-01-13
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    @mathguy As a third approach (more or less parallel to the one just sketched, but with a different feel) try sampling the problem and see for yourself. – 2017-01-13
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    @MugB the other question you found is indeed parallel. You can recast my solution below as a straight application of Bayes' Theorem if you like (the steps are identical). I assumed from your question that you weren't familiar with the statement of that theorem so I phrased my argument without simply quoting the formal result. – 2017-01-14
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    I'll give one more way to picture the problem and then I'll stop: a priori, four events are possible (according to which color is transferred and which is drawn). It is easy to compute the probabilities: $(W,R), (R,R),(W,W)$ each have probability $\frac 3{10}$. $(R,W)$ has probability $\frac 1{10}$. Easy to see now that, conditioned on the event that the draw was $R$, the probability that the transfer was $W$ is $\frac 12$. – 2017-01-14

2 Answers 2

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Confirming lulu's answer.

You begin with ($3$white and $2$red) balls in urn#1, then select $1$ at random(without bias) and move it into urn#2.   Urn#2 then contains either ($2$white and $2$red) or ($1$white and $3$red) depending on which colour of ball was moved into it (white or red respectively).   Then a ball is extracted(without bias) from urn#2 and it turns out to be red.

Let $W_1$ be the event that a white ball was moved into urn#2, and $R_2$ the event that a red ball was extracted from urn#2.   We wish to determine $\mathsf P(W_1\mid R_2)$, the conditional probability that a white ball had been transferred into urn#2 when given that a red ball was subsequently extracted.

We know $~\mathsf P(W_1)=3/5~,~ \mathsf P(R_2\mid W_1)=2/4~,~ \\\mathsf P(W_1^\complement)=2/5~,~ \mathsf P(R_2\mid W_1^\complement)=3/4~$.

Then we apply Bayes' Rule:$$\begin{align}\mathsf P(W_1\mid R_2) ~&=~\dfrac{\mathsf P(W_1)\,\mathsf P(R_2\mid W_1)}{~\mathsf P(W_1)\,\mathsf P(R_2\mid W_1)+\mathsf P(W_1^\complement)\,\mathsf P(R_2\mid W_1^\complement)~}\\[1ex]&=~\dfrac{3\cdot 2}{~3\cdot 2+2\cdot 3~}\\[1ex]&=~\dfrac{~1~}{2}\end{align}$$

$\blacksquare$

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    thank you, you have all been very kind to take the time to explain. I learned something new. – 2017-01-14
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Method I: (Bayes) There are two scenarios in which a red ball is observed:

I: A white ball is transferred (probability $\frac 35$). In this case, a red ball is observed with probability $\frac 12$. Thus this scenario has probability $$\frac 35 \times \frac 12 = \frac 3{10}$$

II: A red ball is transferred (probability $\frac 25$). Now the probability of drawing a red ball is $\frac 34$ Thus this scenario has probability $$\frac 25\times \frac 34 =\frac 3{10}$$

We see that the two scenarios contribute equally, thus the probability that it was a white ball that was transferred initially is $\boxed {\frac 12}$

Note: as our prior was that the probability the transferred was white was $\frac 35$ we see that the observation of the red ball has caused us to lower our estimate for the probability

Method II (Conditional Probability). A priori, the universe here consists of four possible events: $(W,W),(W,R),(R,R),(R,W)$ according to the color of the transferred ball and the color of the drawn ball. A routine calculation shows that $(W,W),(W,R),(R,R)$ each have probability $\frac 3{10}$ and $(R,W)$ has probability $\frac 1{10}$. We are asked for the probability that the transferred ball is $W$ conditioned on the fact that the drawn ball is $R$ and inspection now shows the answer to be $\frac 12$.

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    Hi, thanks for the explanation but can I ask why in the second event, drawing a red ball becomes 3/4? – 2017-01-13
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    That makes absolutely no sense. Why, from the fact that the two scenarios contribute equally (to a computation that has nothing to do with the given fact that "a red ball was drawn from urn 2" anyway), does it follow that the probability that it was a white ball that was transferred initially is 1/2? Utter nonsense! – 2017-01-13
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    @MugB If we have transferred a red ball to urn $2$, that would mean that the urn contained three red balls and one white... – 2017-01-13