philipp1122 wrote:
Ten dice, each fair with six-sides, are rolled simultaneously. What is the probability of getting exactly two fives among them?
The solution using the Binomial formula is:
10C2 * (1/6)^2 * (5/6)^8
However, I was wondering if I could actually use counting here, to find the desired outcomes / total outcomes.
Can someone help me on how I would do that?
Thanks a lot!
Hello,
philipp1122. Sure, there are different ways of arriving at the same solution, but you will notice some overlap. Take matters step by step. First, logically speaking, if you are talking about a desired outcome—two 5s from two dice—then there will be a single way in which that outcome can occur, and you also need to account for the total number of outcomes from ten dice:
\(\frac{1^2}{6^{10}}\)
But now we also have to consider the outcomes of the other eight dice, which
cannot be 5s—i.e. there will be five desirable outcomes for each die:
\(\frac{(1^2)(5^8)}{6^{10}}\)
Finally, you have to account for the number of ways in which these two 5s could be rolled—the first and second dice, the third and tenth, and so on. Yes, you could use 10C2, but you could also rationalize that there would be ten ways to roll the first 5, then nine ways to roll the second. Then, since either die could produce the first or second 5, we have to divide by 2. In other words, we can use
\(\frac{n(n - 1)}{2}\)
\(\frac{(10)((10) - 1)}{2}\)
\(\frac{10(9)}{2}\)
\(45\)
Thus, there are 45 ways in which these two 5s could appear among the ten dice, and multiplying our earlier fraction by 45 yields our answer, even if it would be much quicker to calculate using a calculator or the more formulaic method:
\(\frac{45(1^2)(5^8)}{6^{10}}\)
\(\frac{17578125}{60466176}\)
approx. \(0.2907\)
I am not sure whether the above approach will prove satisfactory to you, but it does answer your question. (Sometimes formulaic approaches prove the most practical.)
Good luck with your studies.
- Andrew