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Instructions
An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles.
Question 53
If four marbles are picked at random, what is the probability that one is green, two are blue and one is red ?
Solution
Number of ways of selecting 1 green, 2 blue and 1 red = $$^2C_1*^4C_2*^6C_1 = 2*6*6 = 72$$
Probability of selecting 1 green, 2 blue and 2 red = $$\frac{72}{^{15}C_4}$$ = $$\frac{24}{455}$$
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