Two numbers are randomly selected without replacement from the first 12 positive integers. Find the probability that the sum of these two numbers is 10.
a) 7/144
b) 1/24
c) 1/18
d) 5/144
e) 1/16
+128406 The denominators that you have indicate that this is probably with replacement
The number of possible sums without replacement are 12 * 11 = 132
Sums of 10 =
1 9
9 1
2 8
8 2
7 3
3 7
4 6
6 4
So.......the probability that the numbers sum to 10 without replacement is just 8 / 132 = 2 / 33
However.....with replacement we have 12 * 12 = 144 possible sums
1 9
9 1
2 8
8 2
7 3
3 7
4 6
6 4
5 5
So....the probability that we have a sum of 10 with replacement is just 9 / 144 = 1 / 16
