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Notice that in any urn, we have $r-1+n-r = n-1$ balls. We have to pick 2 balls, so the sample space size is ${n-1 \choose 2 }$. Now if we want the second ball to be mangenta, then we must have either $MM$ or $RM$. So the probability of this is

$$ P = \frac{ MM + RM }{{n-1 \choose 2} } = \frac{ \frac{(n-r)^2}{(n-1)^2} + \frac{r-1}{n-1} \cdot \frac{n-r}{n-1} }{{n-1 \choose 2}} $$

but the answer the lecturer gives is $1/2$, does the above answer seems reasonable?

Now, for second part, we want $P(MM)$ which is just

$$ \frac{ \frac{(n-r)^2}{(n-1)^2} }{{n-1 \choose 2}} $$

is this correct?

  • $\begingroup$ Draws are done without replacement. $\endgroup$ – naive Oct 17 '18 at 16:43

First Magenta Ball

There are an equal number of red and magenta balls, so randomly picking a ball from a random urn has a $\frac12$ chance of being magenta.

Second Magenta Ball

Given that the first ball is magenta, the probability that urn $r$ was chosen is $$ \frac{2(n-r)}{n(n-1)} $$ Given that we've chosen from urn $r$ and we've chosen a magenta ball, the probability that we will choose a second magenta ball is $$ \frac{n-r-1}{n-2} $$ The probability, given that the first ball was magenta, that the second ball is magenta is $$ \begin{align} \sum_{r=1}^n\frac{2(n-r)(n-r-1)}{n(n-1)(n-2)} &=\sum_{r=1}^n\frac{4\binom{n-r}{2}}{6\binom{n}{3}}\\ &=\frac23\frac{\binom{n}{3}}{\binom{n}{3}}\\[6pt] &=\frac23 \end{align} $$


$$\begin{align}\Pr(MM)&=\underset{\text{1st ball M}}{\underbrace{\frac{n-r}{ n-1}}}\quad\underset{\text{2nd ball M}}{\underbrace{\frac{n-r-1}{n-2}}}\\[2ex] \Pr(RM) &= \underset{\text{1st ball R}}{\underbrace{\frac{r-1}{n-1}}}\quad\underset{\text{2nd ball M}}{\underbrace{\frac{n-r}{n-2}}} \end{align}$$

Adding up,

$$\Pr(MM \cup RM)=\Pr(MM)+\Pr(RM)=\frac{n-r}{n-1}$$

And this would be all, IF it wasn't for the fact that at the beginning of the problem, it states that there are $n$ urns, and that they are ordered, so that the number of red and magenta balls in each depends on that order.

Since each urn is chosen at random, the probability is really the sum of

$$\small\begin{align}\Pr(\text {2nd ball M})&=\underset{\text{Pr any urn chosen}}{\underbrace{\quad\frac 1 n\quad}}\left( \underset{\text{Pr(2nd ball M }\mid\text{ 1st urn)}}{\underbrace{\frac{n-1}{n-1}}} +\underset{\text{Pr(2nd ball M }\mid\text{ 2nd urn)}}{\underbrace{\frac{n-2}{n-1}}}+\cdots +\underset{\text{Pr(2nd ball M }\mid\text{ n-th urn)}}{\underbrace{\frac{n-n}{n-1}}}\right)\\[2ex] &=\frac{1}{n}\left(\frac{n^2-\frac{(n+1)n}{2}}{n-1}\right)\\[2ex] &=\frac{n - \frac{n+1}{2}}{n-1}\\[2ex] &=\frac{2n-(n+1)}{2(n-1)}\\[2ex] &=\frac{n-1}{2(n-1)}\\[2ex] &=\frac 1 2 \end{align}$$


In the first urn there are $(r - 1)$ red balls i.e. $0$ red balls and $(n-r)$ i.e. $(n-1)$ magenta balls. There are two ways the second ball drawn will be magenta $RM$ and $MM$. So, the number of ways to select that (the first term in the square brackets shows for the number of favorable ways of $RM$ and the second one reflects for $MM$) is $[0 + \binom{n-1}{1}* \binom{n-2}{1}]$.

Similarly for the second urn $[\binom{1}{1}*\binom{n-2}{1} + \binom{n-2}{1}*\binom{n-3}{1}]$ ways. So, we end up with this sequence:

$[0 + \binom{n-1}{1}* \binom{n-2}{1}]$ - first urn

$+$ $[\binom{1}{1}*\binom{n-2}{1} + \binom{n-2}{1}*\binom{n-3}{1}]$ - second urn

$+$ $[\binom{2}{1}*\binom{n-3}{1} + \binom{n-3}{1}*\binom{n-4}{1}]$ - third urn

$+...+[\binom{n-3}{1}*\binom{2}{1} + \binom{2}{1}*\binom{1}{1}]$ - third from last urn

$+ [\binom{n-2}{1}*\binom{1}{1} + 0 ] $ - second from last

$+ 0$. - last urn

Note that $0$ is for the last urn which does not have any magenta balls.

Adding all of the above terms will give:

$[(n-2)*\frac{n*(n-1)}{2}]$. And the total number of ways the one urn can be selected is $n$. And selecting two balls without replacement will be $(n-1)*(n-2)$ ways. So, the probability will be:

$[(n-2)*\frac{n*(n-1)}{2}]$/$n*(n-1)*(n-2)$ = $0.5$


Each of the $n(n-1)$ balls in all the urns is equally likely to be the second ball from the chosen urn (since all the urns contain the same number of balls). By the symmetry of the red-magenta set-up, half of all the balls are magenta. So the answer to (a) is 1/2.


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