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Study the given information carefully and answer the question that follow:
A basket contains 4 red, 5 blue and 3 green marbles
If three marbles are picked at random what is the probability that at least one is blue ?
The total number of ways of picking three marbles is $$^{12}C_3 = 220$$
The total number of ways of picking three marbles such that none of them is blue (that is all of them are either red or green) is $$^7C_3 = 35$$
Hence, the number of favorable cases (atleast one is blue) is $$220 - 35 = 185$$
So, the required probability is $$\frac{185}{220} = \frac{37}{44}$$
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