What is the probability of exactly 2 balls of the same colour are picked from 4 picks.

Question

I've been trying to solve a question about probability and I can't seem to wrap my head around how to do it.

The question is: Imagine an urn with 4 red balls and 4 blue balls. You pick out two balls and leave them to the side, making a pair. You continue doing this until there are no balls left in the urn; what is the probability that exactly one pair of balls are red?

Thanks in advance!

Answer 1

I would say that the probability is $24/35$. For this, number the balls $b_1,\cdots,b_8$. The ball $b_1$ has 7 possible peers. Once the peer is chosen, the next free ball in this order has 5 possible peers, again once it is chosen, the next free ball has 3 possible peers. Hence there are $7\times5\times3$ ways to group the balls in pairs. The order of appearance of the pairs can be ignored.

Now suppose that the red balls are $b_1,\cdots,b_4$. There are $\binom{4}{2} = 6$ ways to choose the pair of red balls in the result. Once this pair is chosen, the first free red ball has 4 possible blue peers, and the next free red ball has 3 possible blue peers. Hence $6\times 4 \times 3$ configurations give a unique pair of red balls.

Hence the probability is \begin{equation} p = \frac{6\times 4 \times 3}{7\times5\times3} = \frac{24}{35} \end{equation}

Answer 2

Here is a solution using a variant of the Principle of Inclusion / Exclusion.

Let's say a sequence of draws has "Property $i$" if the $i$th pair is red, for $i=1,2,3,4$. Define $S_j$ to be the total of the probabilities of the sequences with $j$ of the properties (with over-counting) for $j=1,2$.

For $S_1$, there are $4$ ways to pick the red pair. A pair can be chosen in $\binom{8}{2}$ ways, of which $\binom{4}{2}$ are all red. So $$S_1 = 4 \binom{4}{2} / \binom{8}{2} = \frac{24}{28}$$ For $S_2$, there are $\binom{4}{2}$ ways to pick the red pair. Two pairs can be chosen in $\binom{8}{4}$ ways, and only one of those is all red. So $$S_2 = \binom{4}{2}/\binom{8}{4} = \frac{6}{70}$$ The variant of Inclusion/Exclusion we want to use is that if there are $N$ properties then the probability of exactly $m$ occurring is $$P_{[m]} = S_m - \binom{m+1}{m} S_{m+1} + \binom{m+2}{m} S_{m+2} -\dots \pm \binom{N}{m} S_N$$ (Reference: A Introduction to Probability Theory and Its Applications, Volume I, Third Edition by William Feller, Section IV.3) Ours is the case $m=1$, $N=2$, so $$P_{[1]} = S_1 - \binom{2}{1} S_2 = \frac{24}{35}$$