A bag contains 6 red and 4 black balls.docx
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The document describes a probability problem involving two bags of balls. The first bag contains 6 red balls and 4 black balls, and the second bag initially contains 3 red and 8 black balls. A ball is drawn from the first bag and placed in the second bag, then a ball is drawn from the second bag. The probability of the second drawn ball being red is calculated as 3/10.
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A bag contains 6 red and 4 black balls.docx
- 1. A bag contains 6 red and 4 black balls. Another bag contains 3 red and 8 black balls. A ball is drawn from the first bag and is placed in the second bag. A ball is then drawn from the second bag. What is the probability that the ball drawn from the second bag is red ball. Solution Bag-I Bag-II 6 red balls 3 red balls 4 black balls 8 black balls R1 = Red ball from bag I R2 = Red ball from bag II B1 = Black ball from bag I B2 = Black ball from bag II If a red ball is placed in bag-II from bag-I Bag-II 4 red balls 8 black balls Therefore (R2) = 4 12
- 2. If a black ball is placed in bag-II from bag-I Bag-II 3 red balls 9 black balls Therefore (R2) = 3 12 () = (R1 R2) (B1 R2) () = (R1) P(R2) + (B1) (R2) () = 6 10 4 12 + 4 10 3 12 = 36 120 = 3 10