Bowl I contains eight red balls and six blue balls. Bowl II is empty. Four balls are selected at random, without replacement, and transferred from bowl I to bowl II. One ball is then selected at random from bowl II. Calculate the conditional probability that two red balls and two blue balls were transferred from bowl I to bowl II, given that the ball selected from bowl II is blue.

Step-by-step explanation:

The original possibility of getting blue balls out would be 6/14 and for red balls it would be 8/14. With only 4 balls taken the fractions change to a denominator of 4. There would then be blue - 2/4 (rounded down) and red - 2/4 (rounded up). in total both balls could be 2 of each colour although there is more chance of there being more blue balls as the probability was slightly higher!