What is wrong with this combinatorial problem

Question

Suppose $3$ balls are chosen from an urn where $5$ balls are white and $8$ are red. Find the probability that the first ball is red and the second ball is red.

Well, by independence, it is easy to see that the required probability is $\frac{8}{13} \cdot \frac{7}{12} \cdot \frac{11}{11} $. However, here is another argument I tried to use:

We are selecting three balls so size of sample space is ${13 \choose 3}$. Now, the first two balls red is counted in ${8 \choose 2 }$ ways and since the last one can be anything and we only have $11$ left, we count them all. So, we have

$$ Pr = \frac{ {8 \choose 2 } 11 }{{13 \choose 3} } = 3 \cdot \frac{8}{13} \cdot \frac{7}{12} $$

Why do we have the extra three in the above solution? To me it seems pretty solid the counting in both solution.

Answer 1

You correctly found that the probability that the first two of the three balls drawn are red is $$\Pr(\text{first two of three balls selected are red}) = \frac{8}{13} \cdot \frac{7}{12} \cdot \frac{11}{11} = \frac{14}{39}$$

The probability that the first two balls are red is $$\Pr(\text{first two balls are red}) = \frac{\dbinom{8}{2}}{\dbinom{12}{2}} = \frac{14}{39}$$ The fact that you are picking a third ball is irrelevant. If you want to include the third ball in your calculation, observe that you can choose any of the eleven remaining balls after the two red balls have been selected. $$\Pr(\text{first two of three balls selected are red} = \frac{\dbinom{8}{2}}{\dbinom{12}{2}} \cdot \frac{\dbinom{11}{1}}{\dbinom{11}{1}}$$ which simplifies to the first result.

Answer 2

You should note that:

1) The total number of $3$ ordered balls is not $C(13,3)$, but $P(13,3)=\frac{13!}{10!}=13\cdot 12\cdot 11$.

2) There are ${8\choose 1}{7\choose 1}{11\choose 1}$ ways to choose the first $2$ red and the last $1$ any balls. (Note that there is no restriction on the third ball, so it can also be red, therefore it is considered from $11$ remaining balls. If it required red, red and white, then the last factor would be ${5\choose 1}$).

Hence, the required probability follows.