Hi All,

We're told that an urn contains 10 balls, numbered from 1 to 10 and we're told that 2 balls are selected at random WITH replacement. We're asked for the probability that the SUM of the 2 numbers on the balls will be EVEN. This question can be solved in a couple of different ways, including as a straight-probability question.

To start, there are two ways to get a sum that is EVEN:

(Odd on the first) and (Odd on the second)
(Even on the first) and (Even on the second)

Since there are an equal number of Odd and Even numbers AND we are replacing the first ball before we pull the second, the probability of pulling one Odd or one Even is the same each time: 1/2

(Odd) and (Odd) = (1/2)(1/2) = 1/4
(Even) and (Even) = (1/2)(1/2) = 1/4
Total probability of ending with a sum that is Even on two balls is 1/4 + 1/4 = 2/4 = 1/2

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

BTGmoderatorLU wrote:Source: Veritas Prep

An urn contains 10 balls, numbered from 1 to 10. If 2 balls are selected at random with replacement from the urn, what is the probability that the sum of the 2 numbers on the balls will be even?

A. 25%
B. 37.5%
C. 50%
D. 62.5%
E. 75%

The OA is C

The sum of the 2 numbers will be even if the 2 numbers chosen are both even or both odd. The probability of choosing 2 even numbers is 1/2 x 1/2 = 1/4. Likewise, the probability of choosing 2 odd numbers is 1/2 x 1/2 = 1/4. Therefore, the probability of choosing 2 numbers whose sum is even is 1/4 + 1/4 = 1/2.

Answer: C