Question 75992
Four balls are selected at random without replacement from an urn containing three white balls and five blue balls. Find the probability of two or three of the balls are white.
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P(2 balls white) = 3C2/8C2 = 6/28
P(3 balls white) = 3C3/8C3 = 1/[8*7*6/1*2*3]= 1/56
These are disjoint events so add to get:
12/56 + 1/56 = 13/56
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Cheers,
Stan H.