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An urn contains 3 red, 4 black, 5 pink, and 7 green marbles. If five marbles are drawn at random, what is the probability that two are black and three are green?
35/1938
1/1938
1/210
1/1162
Five marbles are drawn at random from an urn of 19 marbles in 19C5 ways, then n(T) = 19C5 = 19!/5!×14! = 11628 Let E be the event that two marbles are black and three are green, then n(E)= 4C2 × 7C3 = 210 Thus , required probability = n(E)/n(T) = 210/11628 = 35/1938
By: bhavesh kumar singh ProfileResourcesReport error
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