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Three green balls are mixed with five yellow balls in a single tub. (For each color, there is one ball with a black dot on it). One ball is picked, then another ball is picked without replacement. What is the probability that the first ball in green without the dot, but the second ball is yellow with the dot?

 Nov 13, 2018

you may as well treat the dotted balls as separate colors.


so your initial distribution is (g,gdot,y,ydot) = (2,1,4,1)


\(P[(g,ydot)] = \dfrac{2}{8}\dfrac{1}{7} = \dfrac{1}{28}\)

 Nov 13, 2018
edited by Rom  Nov 13, 2018

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