Denote the numbers of sides as m and n. It is given that m, n >= 6, let m <= n.

Because of this, all 6 combinations that give the sum of 7 are possible:

1 + 6, 2 + 5, 3 + 4, 4 + 3, 5 + 2, 6 + 1,

and the probability of sum 7 is `6 / ( m n ) .`

For the sum 10 there are 9 potentially possible combinations:

1 + 9, 2 + 8, 3 + 7, 4 + 6, 5 + 5, 6 + 4, 7 + 3, 8 + 2, 9 + 1.

If some die has less than 9 sides, there will be less combinations, let x of them. Then the probability of sum 10 will be `x / ( m n ) ` and it is given that

`6 / ( m n ) = 3 / 4 * x / ( m n ) , ` so x = 8.

If both m, n <= 8, x will be less than 8 (both 1+9 and 9+1 become impossible), so at least n > 8. If both m, n >= 9, x will be greater than 9, so at least m < 9. Combined, this gives m = 8, n >= 9.

Now use the given fact that the probability of sum 12 is 1/12. There are 11 potentially possible combinations but because m = 8, there are at most 8. Denote the number of such combinations by y, then

`y / ( m n ) = 1 / 12 , ` `12y = 8n , ` `3y = 2n ,`

so y is even and n is divisible by 3. If n = 9 (the least possible value), y must be 6, which is correct (4 + 8, 5 + 7, 6 + 6, 7 + 5, 8 + 4, 9 + 3 are exactly 6 combinations).

This way m is 8 and the smallest n is 9, 17 in total.

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now