I am working through Grimmett and Strizaker, Probability and Random Processes. Problem 1.4.7 states:

There are n urns of which the rth contains r-1 red balls and n-r magenta balls. You pick an urn at random and remove two balls without replacement. Find the probability that:

a) the second ball is magenta

For completeness, part b asks for the probability that given the first is red, is not relevant for my question.

The solution manual for the book, entitled 1000 Exercises in probability (by the same authors), states the following:

Since each urn contains the same number of n-1 balls, the second ball picked is equally likely to be any of the n(n-1) balls available. One half of these balls are magenta thus P(second ball is magenta) = 1/2

The section is for Conditional Probability, so I had attempted my solution with it.

My Attempt:

Let $B_i$ be the $i^{th}$ draw of the ball. Let U be the selected Urn. Let $B_i$ = R represent the event that the $i^{th}$ draw is a red ball and $B_i=M$ be likewise for magenta. Then we can calculate the probability of the second ball selected being magenta with marginal probability:

$P(B_2=M) = \sum_{r=1}^n$ $P(B_2=M, U=r) = \sum_{r=1}^n P(B_2=M | U=r)P(U=r)$

The urns are chosen at random so $P(U=r)$ = $\frac{1}{n}$

Applying marginal probability on $B_1$:

$P(B_2=M) = \frac{1}{n}\sum_{r=1}^nP(B_2=M|U=r, B_1=R)P(B_1=R) + P(B_2=M|U=r, B_1=R)P(B_1=M)$

Reasoning that the draws from the urn are uniform, we have:

$P(B_1=R) = \frac{r-1}{n-1}$

$P(B_1=M) = \frac{n-r}{n-1}$

Now, the number of balls to pick from for the second ball is $n-2$ (after the first is picked). Thus we calculate the corresponding probabilities for the second ball:

$P(B_2=M|U=r, B_1=R) = \frac{n-r}{n-2}$

$P(B_2=M|U=r, B_1=M) = \frac{n-r-1}{n-2}$

Substituting these probabilities yields:

$P(B_2=M) = \frac{1}{n}\sum_{r=1}^n\frac{n-r}{n-2}\frac{r-1}{n-1} + \frac{n-r-1}{n-2}\frac{n-r}{n-1}$

Reducing gives us:

$P(B_2=M) = \frac{n+1}{2(n-1)}$

Which approaches 1/2 as $n\rightarrow\infty$.

In writing a python script to simulate this situation I got numbers close to 0.5 as one would expect. However, the simulation does not help me for this case.

My solution breaks from requested answer because of my interpretation of removing the balls "without replacement". So I take the total number of balls for my second draw to be n-2. If I change my second ball draw total to n-1 and $P(B_2=M|U=r, B_1=M) = \frac{n-r}{n-2}$, I get $P(B_2=M) = \frac{1}{2}$. However, this seems to be drawing the balls with replacement.

Where am I going wrong?

  • $\begingroup$ Another solution: symmetry. For the red balls the situation is exactly the same as for the magenta balls. So $P(B_2=R)=P(B_2=M)$ wich can be combined with $P(B_2=R)+P(B_2=M)=1$. $\endgroup$ – drhab Feb 21 '17 at 8:36

Seems to be an algebraic erro, because: $$\begin{align}P(B_2=M) ~&=~ \frac 1n \sum^n_{ r=1 }\left( \frac{n−r}{n−2}\frac{r−1}{n−1} +\frac{n−r−1}{n−2}\frac{ n−r}{n−1}\right) \\[1ex] &=~\frac{1}{n(n-1)}\sum_{r=1}^n(n-r)\cdot\frac{r-1+n-r-1}{n-2} \\[1ex] &=~\frac{n^2-\sum_{r=1}^n r}{n^2-n} \\[1ex] &= \frac{n^2 - \frac 12 n(n+1)}{n^2-n} \\[1ex] &= \frac 12 \end{align}$$

  • $\begingroup$ Thank you, I need to be more thorough in reviewing my own work. $\endgroup$ – ArtisanV Feb 21 '17 at 4:06

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