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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?









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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}$

$= \dfrac{115}{143}$

(5) Comment(s)


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


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).


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

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

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


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.