# Is this a trick question? Balls into urns probability

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!

• 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.
– lulu
Jan 13 '17 at 22:21
• i just found the solution to a similar problem here; math.stackexchange.com/questions/770773/…
– MugB
Jan 13 '17 at 22:44
• 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. Jan 13 '17 at 23:11
• 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. Jan 13 '17 at 23:13
• @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
– MugB
Jan 13 '17 at 23:31

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 $\frac 12$.

• Hi, thanks for the explanation but can I ask why in the second event, drawing a red ball becomes 3/4?
– MugB
Jan 13 '17 at 22:35
• 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! Jan 13 '17 at 23:35
• @MugB If we have transferred a red ball to urn $2$, that would mean that the urn contained three red balls and one white...
– lulu
Jan 13 '17 at 23:52

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}

$\blacksquare$

• thank you, you have all been very kind to take the time to explain. I learned something new.
– MugB
Jan 14 '17 at 9:45

A Pretty Straightforward question. Please go through the diagram. Another way to think is if 1W transferred to U2= (1W + 2R) makes U2 = (2R + 2W). Now, # White Balls = # Red Balls . Then P(W)= P(R) = 1/2 for Urn2.