(36 + 29) + 75 = 140
36
29+3 = 32 and 29+7 = 36
There are 36 permutations of two dice. Only one of them has a sum of two. Then probability, then, of rolling a sum of more than two is 35 in 36, or about 0.9722.
The probability is 29/36.
(36 + 29) + 75 = 140
36
29+3 = 32 and 29+7 = 36
There are 36 permutations of two dice. Only one of them has a sum of two. Then probability, then, of rolling a sum of more than two is 35 in 36, or about 0.9722.
There are 3 pairs of prime numbers with sum of 36. These are: (31, 5), (29, 7), (23, 13), & (19, 17) Hope this helps! :D
The probability is 29/36.
29 and 36
It is (36+28)-15 = 49
Because the sum of its proper divisors is greater than 36.
This is easier to solve by looking at the reverse problem, what is the probability of the sum being 11 or more. Out of the 6*6 = 36 outcomes, three (5,6), (6,5) and (6,6) satisfy this event. So the probability of getting a sum of 11 or more is 3/36. So the probability of less than 11 is 1-Pr(>=11) = 1 - 3/36 = 33/36 = 11/12 or 0.91667
7 & 29, 13 & 23, 17 & 19...
There 36 possibilities, and out of the 36, there are 26 possibilities of getting a sum less than 9. 26/36 = 2/3