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!

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step-by-step explanation:

2/5+1/2=9/10

6/1-9/10=5 1/10

y=mx+b is the final answer!   : )