A box contains 6 red and 1 blue balls. A second box contains 1 red and 4blue balls. One ball is removed from each box at random without replacement. All the remaining balls are put into the third box. What is the probability that a ball drawn randomly from the third box is blue?

What are the different balls which can be left in a box?

Box 1:
a) Pick blue, 6 red remains. Probability of this event = 1/7

b) Pick red, 5 red and 1 blue remains. Probability = 6/7

Box 2:
a) Pick blue, 1 red and 3 blue remains. Probability = 4/5

b) Pick red, 4 blue remains. Probability = 1/5

Which now means that you can have

I) 1 a) and 2 a) giving a total of 7 red and 3 blue. Probability = 1/7 * 4/5 = 4/35

ii) 1 a) and 2 b) giving a total of 6 red and 4 blue. Probability = 1/7 * 1/5 = 1/35

iii) 1 b) and 2 a) giving a total of 6 red and 4 blue. Probability = 6/7 * 4/5 = 24/35

iv) 1 b) and 2 b) giving a total of 5 red and 5 blue. Probability = 6/7 * 1/5 = 6/35

This can be condensed into:
I. P(7 red + 3 blue) = 4/35
II. P(6 red + 4 blue) = 1/35 + 24/35 = 25/35
III. P(5 red + 5 blue) = 6/35

So, now P(blue is picked from 3rd box) = P(blue in case I.) + P(blue in case II.) + P(blue in case III.)

This should be easy now. I hope it helped! :)

Post the answer that you got!