Tina randomly selects two distinct numbers from the set { 1, 2, 3, 4,

Let c be number picked by Sergio, and a and b be numbers picked by Tina (a and b have to be distinct)

So, question asks for probability that c > a + b.

No. of ways of selecting a, b and c:
a and b can be the following:
1 + 2 = 3, 1 + 3 = 4, 1 + 4 = 5, 1 + 5 = 6
2 + 3 = 5, 2 + 4 = 6, 2 + 5 = 7
3 + 4 = 7, 3 + 5 = 8
4 + 5 = 9

So total 10 ways.

Ways of selecting c = 10
Total ways of selecting a, b, and c = 10*10 = 100 (which forms the denominator of our probability)

No. of ways for given outcome:
c > a + b, minimum value of a + b is 3,l since a and b cannot be both 1 which would have given us a sum of 2.
So, min value of c is 4.

If a + b = 3: possible a,b = (1,2) only, but c can be any of the 7 nos from 4 to 10, so total ways = 1*7 = 7
If a + b = 4: possible a,b = (1,3) only, since (2,2) is not allowed, c can be from 5 to 10, so total ways = 1*6 = 6
If a + b = 5: possible a,b = (1,4) or (2,3), c can be from 6 to 10, so total ways = 2*5 = 10
If a + b = 6: possible a,b = (1,5) or (2,4), c can be from 7 to 10, so total ways = 2*4 = 8
If a + b = 7: possible a,b = (2,5) or (3,4), c can be from 8 to 10, so total ways = 2*3 = 6
If a + b = 8: possible a,b = (3,5) only, c can be from 9 to 10, so total ways = 1*2 = 2
If a + b = 9: possible a,b = (4,5) only, c can be only 10, so total ways = 1*1 = 1
Total ways = 7 + 6 + 10 + 8 + 6 + 2 + 1 = 40

So probability = 40/100 = 2/5

Option A