2007-08-20 05:28:21 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Find the probably of getting a 2, then subtract this from 1, to find the probably of NOT getting one.
Probability of 2 on first die = 1/6
Probability of 2 on second die = 1/6
So out of 36 possibilities, you have 6/36 + 6/36 possibilities of getting a two, subtracting 1 for the fact that they count the double 2's twice, so its 11/36 of getting at least one 2.
So the probability of getting a 2 is 11/36, therefore the probably of not getting a 2 on either die is 1-11/36 = 25/36
[Edit] In response to the Reverend:
35/36 is the probably of not getting two 2's at the same time. 25/36 is the probability of not getting a 2 on either one.
2007-08-20 05:39:18 · answer #1 · answered by Jon G 4 · 0⤊ 0⤋
On 2d6 (Two 6 sided dice the most common type) there are 36 Combinations of number that can show up. As "2" is only able to be rolled on ONE of these combinations "1" and "1" then the odds of not getting a total of "2" is 35:36.
Now if your question is not having a 2 on either
showing "face" of 2d6 then the answer will be different because a full 1/3rd off all possible roils or 12/36 has the potential to have a "2" on a showing face the probability is 1:3
2007-08-20 12:56:39 · answer #2 · answered by topdawgco97 4 · 0⤊ 0⤋
The probability of getting a 1 on 1 of the dice is 1/6. So the probability of getting two 1's is (1/6)*(1/6) or 1/36.
The probability of not getting a 1 is just 1 minus the probability of getting two 1's or 1 - 1/36 = 35/36
2007-08-20 12:41:16 · answer #3 · answered by Captain Mephisto 7 · 0⤊ 1⤋
Chance of getting a 2 = (1/6)(1/6) = 1/36
Chance of not getting a 2 = 1 - 1/36 = 36/36 - 1/36 = 35/36
2007-08-20 12:42:03 · answer #4 · answered by kellenraid 6 · 0⤊ 1⤋
Let's suppose Die A and Die B
probability of getting a 2 on both A&B = 1/36
probability of getting a 2 only on die A= 1/6*5/6 = 5/36
probability of getting a 2 on only die B= 5/6*1/6 = 5/36
Then add up all the probabilities and you get 11/36
Finally since this i ssthe probability of getting ATLEAST one 2, do 1 - 11/36 and this gives you:
25/36.
2007-08-20 12:42:19 · answer #5 · answered by meg 1 · 0⤊ 1⤋
"Throwing 2 dices once" equals "Throwing 1 dice twice"
P(getting 2) = 1/6
P(NOT getting 2) = 1 - 1/6 = 5/6
P(NOT getting 2 in two consec. throws) =
P(NOT getting 2) * P(NOT getting 2) = (5/6)^2 = 0.694
2007-08-20 13:04:24 · answer #6 · answered by juanito l 1 · 0⤊ 0⤋
35/36.
2007-08-20 12:37:57 · answer #7 · answered by Anonymous · 0⤊ 1⤋
the answer is .. 25/36
2007-08-20 12:39:41 · answer #8 · answered by nambi23 1 · 0⤊ 1⤋
On getting double 2s or one 2?? Please be more clear. For one 2 1/6.
2007-08-20 12:36:04 · answer #9 · answered by Anonymous · 0⤊ 0⤋