(a) Box 1 contains 3 red balls and 5 green balls. Box 2 contains 5 red balls and
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Question
(a)
Box 1 contains 3 red balls and 5 green balls. Box 2 contains 5 red balls and 3 green balls. A ball is randomly selected from Box 1 and placed in box 2. Then a ball is again randomly selected from Box 2.
What is the probability that red balls are selected from both boxes?
(b)
Box 1 contains 3 red balls and 5 green balls. Box 2 contains 5 red balls and 3 green balls. A ball is randomly selected from Box 1 and placed in box 2. Then a ball is again randomly selected from Box 2 and placed in box 1.
What is the probability that after the end of the process, box 1 ends with 3 red balls and 5 green balls, the same number of red and green balls as in the beginning of the process?
(Hint: This question is from chapter 2 - Probability)
Explanation / Answer
(a) Total no of balls in Box 1 = 3 Red + 5 Green = 8
Probability of picking a red ball = 3/8. Now we place this red ball in box 2. Therefore the red ball count (5 red) increases by 1 = 6 Red balls and total balls = 6 red + 3 green = 9 balls.
Therefore probability of picking a red ball from box 2 = 6/9 = 2/3.
Therefore the required probability = 3/8 * 2/3 = 1/4
(b) For this part we have to calculate the cases where
(1) a red ball is picked from box 1 and put into box2, and then a red ball is picked from box 2 is replaced into box 1. (We have calculated this to be 1/4 in part a)
(2) a green ball is picked from box 1 and put into box2, and then a green ball is picked from box 2 is replaced into box 1.
Probability of picking a green ball from box 1 = 5/8. Now box 2 will have a total of 9 balls of which 4 are green (1 got added from box 1) . Therefore probability of picking a ball from box 2 = 4/9. The required probability in this case is 5/8 * 4/9 = 5/18
Therefore the required probability = 1/4 + 5/18 = 19/36
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