| 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? |
A) \[\frac{20}{35}\]
B) \[\frac{22}{35}\]
C) \[\frac{14}{19}\]
D) \[\frac{5}{7}\]
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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