An urn contains four balls, 2 white and 2 black. Balls are drawn one by one but just the three of them and we do not return them into the urn. What is the probability that: A={first drawn ball is black}, B={black ball is drawn at least once} and C={one white ball is drawn and two black}?
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1$\begingroup$ Surely you must have some ideas, no? If you enumerate the balls as $W_1,W_2, B_1, B_2$ then you can just list all the possible triples. If you have no ideas at all, then this is one way to do it. $\endgroup$ – lulu Apr 13 '20 at 14:11
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A = first drawn ball is black = 1/2*1/3+(2*1/2*2/3*1/2) = 1/2 B = black ball is drawn at least once = will always happen because you choose 3 balls and we do not retirn them , and there are only 2 white balls C = one white ball is drawn and two black = 1/2
