ardz24 wrote:A bag contains a total of four billiard balls. The billiard balls are numbered as follows: 2, 3, 5 and 7.

Scenario 1: Two billiard balls are pulled out of the bag. What is the probability that the sum of the numbers appearing on the balls is odd?
Scenario 2: A billiard ball is pulled out of the bag then placed back in. Another billiard ball is pulled out of the bag (which may or may not be identical to the first) then placed back in. What is the probability that the sum of the two balls is odd?

I'm confused how to set up the formulas here. Can any experts help?

Scenario 1: Two billiard balls are pulled out of the bag. What is the probability that the sum of the numbers appearing on the balls is odd?

A sum of odd can be formed if a ball numbered 2 is drawn. So, the samples are (2, 3); (2 5); (2, 7): a total 3 samples out of 4C2 = 4.3/1.2 = 6 samples.

Probability = 3/6 = 1/2

Scenario 2: A billiard ball is pulled out of the bag then placed back in. Another billiard ball is pulled out of the bag (which may or may not be identical to the first) then placed back in. What is the probability that the sum of the two balls is odd?

Case 1: A ball number 2 is drawn and one ball from (3, 5, 7) is drawn

Probability = (Probability of drawing a numbered 2) * (Probability of drawing a numbered 3, 5 or 7) = (1/4) * (3/4) = 3/16

Case 2: A ball from (3, 5, 7) is drawn and then a ball number 2 is drawn

Probability = (Probability of drawing a numbered 3, 5 or 7) * (Probability of drawing a numbered 2) = (3/4) * (1/4) = 3/16

Total probability = 3/16 + 3/18 = 3/8.

Hope this helps!

-Jay
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