dabaobao wrote:
First 15 positive integers are written on a board. If two numbers are selected one by one from the board at random (the numbers are not necessarily different), what is the probability that the sum of these numbers is odd?
A) 49/225
B) 56/225
C) 98/225
D) 105/225
E) 112/225
Veritas Prep Official Solution
Numbers: 1, 2, 3, 4, …, 13, 14, 15
When will the sum of two of these numbers be odd? When one number is odd and the other is even.
P(Sum is Odd) = P(First number is Odd)*P(Second number is even) + P(First number is Even)*P(Second number is Odd)
P(Sum is Odd) = (8/15)*(7/15) + (7/15)*(8/15) = 112/225
Explanation
On first selection, we can pick any number so the probability is 1. The second selection depends on what you selected in the first pick. If you selected an odd number in the first pick, the probability of selecting an even number is 7/15. If you selected an even number in the first pick, the probability of selecting an odd number is 8/15. So what do you do? Do you use 7/15 or 8/15 with 1? You cannot say so you must take individual cases.
Case 1: Select an odd number and then an even number: (8/15) * (7/15)
Case 2: Select an even number and then an odd number: (7/15) * (8/15)
The total cases considered here are 15*15 (select first number in 15 ways and select the second number in 15 ways since the second number can be the same as the first number). In 8*7 ways, you will select an odd number and then an even number. In 7*8 ways, you will select an even number and then an odd number. In both the cases, the sum will be odd. This gives us a probability of (56+56)/225 = 112/225
The total probability of 1 is obtained as follows:
1 = P(first number odd, second number even) + P(first number even, second number odd) + P(first number odd, second number odd) + P(first number even, second number even)
= 56/225 + 56/225 + 64/225 + 49/225 = 1
We are only interested in the 56/225 + 56/225 part.