What is the probability that both balls are red given that the first ball you drew was red?

Question

Assume that each bag contains 5 balls. Bag a contains 3 red and 2 white, while bag b contains 1 red, 3 white, and 1 blue. You randomly select one ball from bag a, note the color, and place the ball in bag b. You then select a ball from bag b at random and make note of its color.
(1) What is the probability that both balls are red?

(2) What is the probability that both balls are red given that the first ball you drew was red?

Answer 1

In bag A, there are 5 balls, of which 3 are red, and 2 are white.

So, the chance of drawing a red ball is 3 of 5, or 60% (decimal 0.6).

 

We then place the ball from bag A into bag B, giving 6 balls.

 

If the ball from bag A was not red, then the chance of drawing a red ball will be 1 of 6, or 16.67% (decimal 0.1667)

 

If the ball from bag A was red, then the chance of drawing a red ball will be 2 of 6, or 33.3% (decimal 0.333)

 

Q1. The probability that both balls are red is the product of the first selection times the product of the second selection. Which would be, 0.60 x 0.166 = 0.099, or ≅ 10%

 

Q2. Given that the first ball was red, and then placed into bag B, the probability will be the product of the first selection times the second selection. Which would be 0.60 x 0.333 = 0.1998, ≅ 20%