Box A has 3 blue and 7 red balls in it, box B has 1 blue and 1 red. A) Three bal

Question

Box A has 3 blue and 7 red balls in it, box B has 1 blue and 1 red.

A) Three balls are selected without replacement from box A and one ball is selected from ball B. What is the chance that at least one of these four balls is blue?

B) Five balls are drawn without replacement from box A. What is the chance that exactly two balls are blue or the first two balls drawn are the same color?

C) A ball is randomly selected from box A and put into box B. Then a ball is randomly selected from the three balls in box B. What is the chance that the ball put from box A into box B is blue, given that the ball drawn from box B is blue?

Explanation / Answer

A) P(at least one of these four balls is blue) = 1 - P(all balls are red)

= 1 - (7C3/10C3)x(1/2)

= 1 - 0.1458

= 0.8542

B) P(exactly two balls are blue or the first two balls drawn are the same color) = P(exactly 2 balls are blue) + P(first 2 balls are red) + P(first 2 balls are blue) - P(first 2 balls are blue and rest of the three are red)

= (3C2x7C3/10C5) + (7/10 x 6/9) + (3/10 x 2/9) - (3/10 x 2/9 x 7/8 x 6/7 x 5/6)

= 0.4167 + 0.4667 + 0.0667 - 0.0417

= 0.9084

C) P(ball put from box A into box B is blue |  the ball drawn from box B is blue)

= P(ball put from box A into box B is blue and the ball drawn from box B is blue) / P(ball drawn from box B is blue)

= (3/10 x 2/3)/[(3/10 x (2/3) + (7/10)(1/3)]

= 0.4615

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