2006-10-24 19:44:27 · 13 answers · asked by shaneeeeeee 1 in Education & Reference ➔ Higher Education (University +)
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
All the possible combinations of the two rolls. Look at the ones that contain the number "4". Out of 36, 11 of them contain 4.
11/36
Good Luck.
This is right by the way guranteed.
2006-10-24 20:00:18 · answer #1 · answered by Bekk 2 · 1⤊ 0⤋
Here are all the combinations:
1 2 3 4 5 6
1 11 12 13 14 15 16
2 21 22 23 24 25 26
3 31 32 33 34 35 36
4 41 42 43 44 45 46
5 51 52 53 54 55 56
6 61 62 63 64 65 66
Just count up all the instances that contain at least one 4. There are 11, so the odds are 11/36. (or nearly 1/3)
2006-10-24 20:14:27 · answer #2 · answered by supensa 6 · 3⤊ 0⤋
There is a 5/6 chance of not getting a "4" on each roll.
So the chance of NOT getting a "4" on two successive rolls is
5/6 x 5/6 = 25/36.
Therefore the chance of getting a "4" at least once is 1 minus 25/36 = 9/36 = 1/4.
There is a (1/4) 25% chance of getting at least one "4".
2006-10-24 20:51:21 · answer #3 · answered by Zam 2 · 0⤊ 2⤋
Assuming both throws are independent...
3 possibilities:
first throw 4 AND second throw not 4 1/6*5/6 OR +
first throw not 4 AND second throw 4 5/6*1/6 OR +
first throw 4, AND second throw 4 1/6*1/6
so answer = 5/36+5/36+1/36= 11/36.
2006-10-24 20:22:51 · answer #4 · answered by ekonomix 5 · 2⤊ 0⤋
1 IN 6
2006-10-24 19:53:19 · answer #5 · answered by cork 7 · 0⤊ 2⤋
1/6 is the probability of getting a four if you throw a dice once, because the number of outcomes is 6 and there is only one four, so it is 1/6. But you will throw the dice twice, so (1/6)(1/6) = 1/36
I hope this helps!
2006-10-24 20:26:10 · answer #6 · answered by Kevin Y 2 · 0⤊ 3⤋
The chances of throwing a particular number on one die is 1 in 6
So, if you have two dice (or one die and two throws!), it is (1/6 + 1/6) = 2/6 = 1/3. Of course, that is hoping the die is not weighted!
2006-10-24 20:12:19 · answer #7 · answered by Anonymous · 0⤊ 3⤋
11/36 is the answer.
1/6 + (5/6)*(1/6) = 11/36
the chance of getting on the first roll + (the chance of not getting on the first roll) x (the chance of getting on the secon roll)
2006-10-24 19:49:12 · answer #8 · answered by lataliano 3 · 3⤊ 0⤋
6 to the power of 4
6 X 6 X 6 X 6 = 1 in 1,296 chances
multiply by 4 for the exact #4 dice rolls
1 in 5,184 chances for it to happen.
2006-10-24 19:48:39 · answer #9 · answered by Anonymous · 0⤊ 3⤋
1/36
2006-10-24 20:03:57 · answer #10 · answered by Naveed 2 · 0⤊ 2⤋