Question
Suppose there is an urn with 6 red balls and 4 yellow balls. We draw two balls from the urn, without replacement. (a) What is the probability that both balls are red? (b) What is the probability that both balls are yellow? (c) What is the probability of drawing one red and one yellow ball, in either order? (d) What is the probability that at least one of the balls is yellow?
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Answer to a math question Suppose there is an urn with 6 red balls and 4 yellow balls. We draw two balls from the urn, without replacement. (a) What is the probability that both balls are red? (b) What is the probability that both balls are yellow? (c) What is the probability of drawing one red and one yellow ball, in either order? (d) What is the probability that at least one of the balls is yellow?
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a) the probability that both balls are red = \frac{6}{6+4}\cdot\frac{\left(6-1\right)}{6+4-1}\:=\:\frac{6}{10}\cdot\frac{5}{9}\:=\:\:\frac{1}{3}
b) the probability that both balls are yellow = \frac{4}{6+4}\cdot\frac{\left(4-1\right)}{6+4-1}\:=\:\frac{4}{10}\cdot\frac{3}{9}\:=\:\:\frac{2}{15}
c) the probability of drawing one red and one yellow ball, in either order= \frac{6}{6+4}\cdot\frac{\left(4\right)}{6+4-1}\:=\:\frac{6}{10}\cdot\frac{4}{9}\:=\:\:\frac{4}{15}
d) the probability that at least one of the balls is yellow:
P\left(atleast\:1\:ball\:yellow\right)\:=\:1-\:P\left(atleat\:1\:ball\:yellow\right)\~\:=\:1-\:P\left(no\:ball\:is\:Yellow\right)
1-\:P\left(no\:Ball\:are\:yellow\right)\:=\:1-\:P\left(all\:are\:red\right)\:=\:1-\:\frac{1}{3}\:=\:\frac{2}{3}
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