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A box contains some red balls and some blue balls. There are four more blue balls than red balls. A ball is removed at random,...Asked by Anonymous
A box contains some red balls and some blue balls. There are four more blue balls than red balls. A ball is removed at random, replaced and a second ball randomly removed. The probability that the two balls are different colours is 21/50. How many balls of each colour are in the box?
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Answered by
Candlelight
Math - Probability
Let the number of red balls in the box be x.
Then the number of blue balls in the box is x + 4.
Hence the total number of balls in the box is 2x + 4.
P(R) = x/(2x + 4)
P(B) = (x + 4)/(2x + 4)
Thus P(2 different colours, with replacement):
= P(R, then B) + P(B, then R)
= 2 [x/(2x + 4)][(x + 4)/(2x + 4)] = 21/50
Solve the equation to find your answer.
Let the number of red balls in the box be x.
Then the number of blue balls in the box is x + 4.
Hence the total number of balls in the box is 2x + 4.
P(R) = x/(2x + 4)
P(B) = (x + 4)/(2x + 4)
Thus P(2 different colours, with replacement):
= P(R, then B) + P(B, then R)
= 2 [x/(2x + 4)][(x + 4)/(2x + 4)] = 21/50
Solve the equation to find your answer.
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