Probability Question

[shu132]

Two Dice are thrown. What is the probability that a number between ten and twenty inclusive is obtained when the numbers in the two dice are multiplied?

My answer (14/36=7/18) differs from that provdied (13/36). I assume this is because the solution may have forgotten to include (4,4) twice? I'm open to correction if anyone can answer this..

Thanks

[tcd252]

1 x 1 = 1
1 x 2 = 2
1 x 3 = 3
1 x 4 = 4
1 x 5 = 5
1 x 6 = 6
2 x 2 = 4
2 x 3 = 6
2 x 4 = 8
2 x 5 = 10 *
2 x 6 = 12 *
3 x 3 = 9
3 x 4 = 12 *
3 x 5 = 15 *
3 x 6 = 18 *
4 x 4 = 16 *
4 x 5 = 20 *
4 x 6 = 24
5 x 5 = 25
5 x 6 = 30
6 x 6 = 36

There are 21 overall possibilities of which 7 are between 10 and 20 inclusive. So the answer should be 7/21 = 1/3

[shu132]

1 x 1 = 1
1 x 2 = 2
1 x 3 = 3
1 x 4 = 4
1 x 5 = 5
1 x 6 = 6
2 x 2 = 4
2 x 3 = 6
2 x 4 = 8
2 x 5 = 10 *
2 x 6 = 12 *
3 x 3 = 9
3 x 4 = 12 *
3 x 5 = 15 *
3 x 6 = 18 *
4 x 4 = 16 *
4 x 5 = 20 *
4 x 6 = 24
5 x 5 = 25
5 x 6 = 30
6 x 6 = 36

There are 21 overall possibilities of which 7 are between 10 and 20 inclusive. So the answer should be 7/21 = 1/3

No there are definitely 36 possibilities.. remember that there are 2 dice, and therefore (2,5) is not the same as (5,2).. so the possible outcomes are:

(1,1) --> (1,6)
(2,1) --> (2,6)
(3,1) --> (3,6)
(4,1) --> (4,6)
(5,1) --> (5,6)
(6,1) --> (6,6)

[Logophobic]

I think you have answered your own question. Look at the 36 possibilities, and notice that (4, 4) occurs only once. The 13 qualifying rolls are: 2x5, 2x6, 3x4, 3x5, 3x6, 4x3, 4x4, 4x5, 5x2, 5x3, 5x4, 6x2, and 6x3.