an urn contains 6 white and 9 blacks balls. if 4 balls are to be randomly selected without replacement, what is the probabilty that the first 2 selected are white and the last 2 black?
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The probability that the first one is white is 2/5. The probability that the second is white, given that the first was, is 2/5*5/14. The chance the third ball is black, given the first two were white, is 2/5*5/14*9/13. The chance the first two are white, then the next two are black is 2/5*5/14*9/13*2/3
About 18/273, which is a little less than 7%.
here ,first find the probability to get first two selected ball is white.
And then find the probability to get second two ball is black.
Here follow multiplication rule.
Required probability=P(w)*P(b)...find it