There are just two ways to get the value of 11 when dice are thrown. one would be a 5 and 6 and the other is a 6 and 5.
10 ways.
The are more ways to roll a 6 (five ways) than to roll a 4 (three ways).
Assuming you mean a sum of 4, there are three ways this can be done: (1, 3), (2, 2) and (3, 1).
5 of 36. That's about 14% if you round up.
7/12 ≈ 0.583 Probability is number_of_ways_of_success/total_number_of_ways. There are 36 possible ways 2 dice can land when thrown. To be more than 6, the sum can be 7, 8, ..., 12 There is 1 way to achieve 12 (6 & 6); 2 ways to achieve 11 (5 & 6 and 6 & 5); ... 6 ways to achieve 7 (1 & 6, 2 & 5, ..., 6 & 1); A Total of 1 + 2 + ... + 6 = 21 ways of getting a sum greater than 6. Thus the probability of throwing a sum greater than 6 is 21/36 = 7/12.
Two ways- one on the first die, and two on the second die, and visa versa.
10 ways.
It is 11/18.
two ways one and three and a pair of twos
count the number of times you get number 1, when you roll a pair of dice.. cheerio
The are more ways to roll a 6 (five ways) than to roll a 4 (three ways).
There is 4 ways to roll a 9 with 2 dice, and 36 possible outcomes. So, the probability of rolling a sum of 9 with two dice is 4/36 or 1/9.
Assuming you mean a sum of 4, there are three ways this can be done: (1, 3), (2, 2) and (3, 1).
5 of 36. That's about 14% if you round up.
3
33,33333...%
7/12 ≈ 0.583 Probability is number_of_ways_of_success/total_number_of_ways. There are 36 possible ways 2 dice can land when thrown. To be more than 6, the sum can be 7, 8, ..., 12 There is 1 way to achieve 12 (6 & 6); 2 ways to achieve 11 (5 & 6 and 6 & 5); ... 6 ways to achieve 7 (1 & 6, 2 & 5, ..., 6 & 1); A Total of 1 + 2 + ... + 6 = 21 ways of getting a sum greater than 6. Thus the probability of throwing a sum greater than 6 is 21/36 = 7/12.