Quantitative Aptitude Ques 1035

Question: An urn contains 3 red and 4 green marbles. If three marbles are picked at random, then what is the probability that atleast two see green?

Options:

A) $ \frac{20}{35} $

B) $ \frac{22}{35} $

C) $ \frac{14}{19} $

D) $ \frac{5}{7} $

Show Answer

Answer:

Correct Answer: B

Solution:

  • $ P,(R)=\frac{3}{7}, $ $ P,(G)=\frac{4}{7} $ P (selecting 3 marbles out of 7) $ ={}^{7}C _3 $ P (atleast two are green) $ =\frac{{}^{4}C _2\times {}^{3}C _1+{}^{4}C _3}{{}^{7}C _3} $ $ =\frac{\frac{4\times 3}{2\times 1}\times 3+\frac{4\times 3\times 2}{3\times 2\times 1}}{\frac{7\times 6\times 5}{3\times 2\times 1}}=\frac{18+4}{35}=\frac{22}{35} $
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