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Maria Alcala · Jul 9, 2011, 04:34 PM

What is the probability of obtaining exactly two 5's in six rolls of a fair die?

jcaron2 · Jul 10, 2011, 06:52 AM

First you have to find the probability of getting exactly two fives in some particular order. For example, what are the chances of rolling a five on the first two rolls, but not on the remaining four rolls?

That probability would be $\frac16 \cdot \frac16 \cdot \frac56 \cdot \frac56 \cdot \frac56 \cdot \frac56 = $\frac16$^2 \cdot $\frac56$^4 = \frac{625}{46656} \approx 1.34%$

Now you need to multiply that answer by the number of possible ways you could choose the order to be in. In other words, if 'x' represents a non-five, we just found the chances of rolling 55xxxx. Now we need to account for the probabilities of rolling 5x5xxx, x55xxx, xx5xx5, etc. So how many ways can you choose two dice out of six?

The answer is $_6C_2=\frac{6!}{2!(6-2)!} =15$ different ways.

Since each of those different unique ways of rolling exactly two fives has the same probability of occurring, the total probability is

$P=\frac{625}{46656}\cdot 15 = \frac{3125}{15552}\approx 20.1%$