1. There are 8 brown balls, 4 orange balls and 5 black balls in a bag. Five balls are chosen at random. What is the probability of their being 2 brown balls, 1 orange ball and 2 black balls ?
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By: anil on 05 May 2019 01.39 am
Total number of balls in the bag = 8 + 4 + 5 = 17 P(S) = Total possible outcomes = Selecting 5 balls at random out of 17 => $$P(S) = C^{17}_5 = frac{17 imes 16 imes 15 imes 14 imes 13}{1 imes 2 imes 3 imes 4 imes 5}$$ = $$6188$$ P(E) = Favorable outcomes = Selecting 2 brown, 1 orange and 2 black balls. => $$P(E) = C^8_2 imes C^4_1 imes C^5_2$$ = $$frac{8 imes 7}{1 imes 2} imes 4 imes frac{5 imes 4}{1 imes 2}$$ = $$28 imes 4 imes 10 = 1120$$ $$ herefore$$ Required probability = $$frac{P(E)}{P(S)}$$ = $$frac{1120}{6188} = frac{280}{1547}$$
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