Question

Jar A contains 8 red balls and 2 green balls. Jar B contains 6 red balls and 3 green balls. You randomly select one ball from Jar A and, without knowing its color, drop it into Jar B. You then randomly select one ball from Jar B.
What is the probability that the ball you select from Jar B will be red?

Answer 1

Before the change:
Jar A contains 8 red balls and 2 green balls.
Jar B contains 6 red balls and 3 green balls.

After the change:
Jar B contains 6 red balls, 3 green balls,
and one ball that has a 1/5 chance of being green and 4/5 chance of being red.

So, total balls in jar B = 10

While we consider a normal red ball to be 'one red ball' we can take the unknown ball to be '4/5 red ball'.

So, total red balls = 6 + 4/5 = 34/5

Probability of red ball = Red balls/Total balls
= 34/5 x 1/10 = 34/50
= 0.68

Answer 2

Another way to look at it:
Jar A has 8R and 2G, total=10
Jar B has 6R and 3G, total=9

We transfer 1 ball from A to B.

8 time out of 10, we transferred a red, so B contains 7R+3G, total=10
Probability of drawing a red
= 7/10 (8 times out of 10)
= 7/10 * (8/10)
=0.56

2 times out of 10, we transfer a green, so B contains 6R+4G, total=10
Probability of drawing a red
= 6/10 *(2/10)
=0.12

Total probability
=0.56+0.12
=0.68