- #1
dfraser
- 4
- 0
I'm struggling with this problem:
Suppose we throw 10 standard dice. Find the probability that faces 2 and 4 occur 3 times each.
I think the solution should look something like this:
$$4*{10\choose3,3,4}(1/6)^3(1/6)^3(1/6)^4 + 4*{10\choose3,3,3,1}(1/6)^3(1/6)^3(1/6)^3(1/6)^1$$
$$+ 6*{10\choose3,3,2,2}(1/6)^3(1/6)^3(1/6)^2(1/6)^2 + {10\choose3,3,1,1,1,1}(1/6)^3(1/6)^3(1/6)^1(1/6)^1(1/6)^1(1/6)^1$$
In particular, I reason that there are four ways that we can use up the remaining 4 faces: (1) they can all be the same (2) three can be the same and one different (3) we can have two pairs, and (4) all the faces can be different.
(1) can happen in 4 ways, (2) can happen in 4 ways, (3) can happen in 6 ways, and (4) can happen in 1 way. Hence my solution.
However, I'm told that the correct answer corresponds to .0178, which differs from my solution. I think I might be getting confused about the number of combinations for each option; can anyone give me some pointers?
Suppose we throw 10 standard dice. Find the probability that faces 2 and 4 occur 3 times each.
I think the solution should look something like this:
$$4*{10\choose3,3,4}(1/6)^3(1/6)^3(1/6)^4 + 4*{10\choose3,3,3,1}(1/6)^3(1/6)^3(1/6)^3(1/6)^1$$
$$+ 6*{10\choose3,3,2,2}(1/6)^3(1/6)^3(1/6)^2(1/6)^2 + {10\choose3,3,1,1,1,1}(1/6)^3(1/6)^3(1/6)^1(1/6)^1(1/6)^1(1/6)^1$$
In particular, I reason that there are four ways that we can use up the remaining 4 faces: (1) they can all be the same (2) three can be the same and one different (3) we can have two pairs, and (4) all the faces can be different.
(1) can happen in 4 ways, (2) can happen in 4 ways, (3) can happen in 6 ways, and (4) can happen in 1 way. Hence my solution.
However, I'm told that the correct answer corresponds to .0178, which differs from my solution. I think I might be getting confused about the number of combinations for each option; can anyone give me some pointers?