in a single throw of two fair dice,find the probability
that the production of the product of the numbers on the dice
`
(1) between 8 and 16.
The word "between" is ambiguous.
Here are the rolls between 8 to 16, inclusive of products of
8 and 16.
1. (2,4) -> 2×4 = 8
2. (2,5) -> 2×5 = 10
3. (2,6) -> 2×6 = 12
4. (3,3) -> 3×3 = 9
5. (3,4) -> 3×4 = 12
6. (3,5) -> 3×5 = 15
7. (4,2) -> 4×2 = 8
8. (4,3) -> 4×3 = 12
9. (4,4) -> 4×4 = 16
10. (5,2) -> 5×2 = 10
11. (5,3) -> 5×3 = 15
12. (6,2) -> 6×2 = 12
There are 36 possible rolls.
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
So the probability is 12 out of 36,
or 12/36 which reduces to 1/3, if "between" is inclusive.
But if "between" means STRICTLY between, then it does
not include products of 8 or 16, and the possible rolls
would be:
1. (2,5) -> 2×5 = 10
2. (2,6) -> 2×6 = 12
3. (3,3) -> 3×3 = 9
4. (3,4) -> 3×4 = 12
5. (3,5) -> 3×5 = 15
6. (4,3) -> 4×3 = 12
7. (5,2) -> 5×2 = 10
8. (5,3) -> 5×3 = 15
9. (6,2) -> 6×2 = 12
If this is the case, then the probability is 9 out of 36,
or 9/36 which reduces to 1/4, if "between" leaves out 8
and 16.
So you'll have to decide which of those two answers is
the one you're looking for. You might ask your teacher.
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(2) divisible by 4.
1. (1,4) -> 1×4 = 4
2. (2,2) -> 2×2 = 4
3. (2,4) -> 2×4 = 8
4. (2,6) -> 2×6 = 12
5. (3,4) -> 3×4 = 12
6. (4,1) -> 4×1 = 4
7. (4,2) -> 4×2 = 8
8. (4,3) -> 4×3 = 12
9. (4,4) -> 4×4 = 16
10. (4,5) -> 4×5 = 20
11. (4,6) -> 4×6 = 24
12. (5,4) -> 5×4 = 20
13. (6,2) -> 6×2 = 12
14. (6,4) -> 6×4 = 24
15. (6,6) -> 6×6 = 36
The probability is 15 out of 36, or 15/36 which
reduces to 5/12.
Edwin
