2007-08-17 10:22:04 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Required outcomes are:-
P(sum of 2) requires (1 , 1) = 1 / 36 ,
P(sum of 3) requires (1 , 2) , (2 , 1) = 2 / 36
P(sum of 12) requires (6,6) = 1 / 36
P(2 or 3 or 12) = 4/36 = 1/9
2007-08-21 09:56:08 · answer #1 · answered by Como 7 · 1⤊ 0⤋
4/36 = 1/9
1+1 = 2
1+2 = 3
2+1 = 3
6+6 = 12
2007-08-17 10:27:05 · answer #2 · answered by civil_av8r 7 · 1⤊ 0⤋
There is only one way to get a total of 2, namely get 1 on the first die and 1 on the second. There are a total of 6 times 6, or 36 different totals. Therefore, Pr(2) = 1/36.
There are two ways to get 3: either a 1 on the first die and 2 on the second or 2 on first and 1 on second. Thus, Pr(3) =
2/36 = 1/18. We find
Pr(2) = Pr(12) = 1/36
Pr(3) = Pr(11) = 2/36 = 1/18
Pr(4) = Pr(10) = 3/36 = 1/12
Pr(5) = Pr(9) = 4/36 = 1/9
Pr(6) = Pr(8) = 5/36
Pr(7) = 6.
2007-08-17 10:52:32 · answer #3 · answered by Tony 7 · 1⤊ 0⤋
There is only 1 chance in 36 of getting a 2 or a 12 on a dice roll
There is a 2 in 36 chance of getting a 3 (either 1,2 or 2,1).
Therefore there is four chances in 36 (1+2+1) of getting one of these results, or 1/9 if you reduce the fraction.
2007-08-17 11:04:01 · answer #4 · answered by tinfoil666 3 · 0⤊ 1⤋
7 to the power 13
2007-08-17 10:32:39 · answer #5 · answered by peter 2 · 0⤊ 1⤋