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An urn contains five red balls and three yellow balls. Four balls are drawn from the urn at random, without replacement. What is probability that the first two balls are red and the second two balls are yellow?

Thanks for your any help.

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    $\begingroup$ Find the probability that the first ball is red. Next, assume that the first ball removed was red and find the probability that the second ball is also red ... $\endgroup$ – Teresa Lisbon Nov 1 '18 at 9:43
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$\frac{5(redballs)}{8(allballs)}\times \frac{4(redballs)}{7(allballs)}\times \frac{3(yellowballs)}{6(allballs)} \times \frac{2(yellowballs)}{5(allballs)} = 0.0714$

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$$P(R_1\cap R_2\cap Y_3\cap Y_4)=P(R_1)P(R_2\mid R_1)P(Y_3\mid R_1\cap R_2)P(Y_4\mid R_1\cap R_2\cap Y_3)=\frac58\frac47\frac36\frac25$$

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