Aptitude Discussion

Q. |
An urn contains 6 red, 5 blue and 2 green marbles. If three marbles are picked at random, what is the probability that at least one is blue? |

✖ A. |
28/143 |

✖ B. |
115/197 |

✖ C. |
28/197 |

✔ D. |
115/143 |

**Solution:**

Option(**D**) is correct

P(None is blue), $=\dfrac{^8C_3}{^{13}C_3}$

$= \dfrac{56}{286}$

$= \dfrac{28}{143}$

P(At least one is blue), $ = 1- \dfrac{28}{143}$

$= \dfrac{115}{143}$

**Payel**

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*

That is true, but problem with your selection is that you are missing out on some selections.

For example you missed out P(2 blue, 1 red) or similar cases. And this is why you are not getting the corect result.

yes you are right . sorry for the mistake and thank you very much to clear it out. so the answer will be option (d).

**Payel**

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Total cases => $n(S)= ^{13}C_3 = 286$

Favorable Cases => $^5C_1 \times ^6C_2 + ^5C_1 \times ^2C_2 + ^5C_1 \times ^6C_1 \times ^2C_1 + ^5C_3$

$=5 \times 15 + 5 \times 1+ 5 \times 6 \times 2 + 10$

$=150$

Probability $= \dfrac{150}{286}$

I could not understand your point but it seems you are trying to add probabilities of instances where at least 1 blue marble is selected.

So, you have tried,

P(1 blue, 2 red) + P(1 blue, 2 green) + P(1 blue, 1 red, 1 green) + P(3 blue)

But what about the cases where 2 blue marbles can also be selected?

P.S. In this question it is simple to count the favorable cases, still you made a mistake. So negation method (described in the solution) is generally a preferred method.

in this question it is mentioned "at least one is blue" i.e. one or more than one blue marble can be selected.