SOLUTION: a bag contains 5 white and 8 red balls. Two drawings of 3 balls are made such that;
(a) the balls are replaced before the second trial.
(b) the balls are not replaced before
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-> SOLUTION: a bag contains 5 white and 8 red balls. Two drawings of 3 balls are made such that;
(a) the balls are replaced before the second trial.
(b) the balls are not replaced before
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Question 1129622: a bag contains 5 white and 8 red balls. Two drawings of 3 balls are made such that;
(a) the balls are replaced before the second trial.
(b) the balls are not replaced before the second trial.
Find the probability that the first drawing will give 3 white balls and the second 3 red balls in each case. Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! the first giving 3 white balls is (5/13)(5/13)(5/13) and then (8/13)(8/13)(8/13) for the second, if that is what is meant in the question
The second is (5/13)(4/12)(3/11) and then (8/10)(7/9(6/8); without replacement, the numerator and denominator change after each draw.
