A certain box holds 3 green balls, 2 white balls, and 1 blue

Question

A certain box holds 3 green balls, 2 white balls, and 1 blue ball. Three balls are taken at random without repetition. What is the probability of having the blue ball among the taken ones?

stolyar · Answer

The only thing that I can add is that it is 1/2.

Let us reformulate the question: what is the probability of NOT having the blue ball among the taken ones?

evensflow · Answer

Select all three which are not blue..

So, 5C3/6C3.

Opps!! it was a typo. I corrected the mistake.

RK73 · Answer

Stop! i stuck!

let's do it slowly - 
when we choose three balls, we have 6c3 possibilities.
from these choices we have 3c1 outcomes with blue one.
where I am wrong?

AkamaiBrah · Answer

There are 6c3 or 20 possible combinations of 3 balls. 

To calculate the number of possible arrangement of 3 balls which include a blue ball assume that the arrangement has a blue ball. Now calculate the number of ways that 2 balls can be combined of the 5 left to accompany the blue one. This is 5c2 = 10. Hence, the probability is 1/2. 

Another way to do this is say: 
Suppose I pull the balls out 1 by 1 (w/o replacement). For the 3 balls to have the blue one, it must be the 1st, 2nd, or 3rd ball. 

Pr(1st) = 1/6
pr(2nd) = 5/6*1/5 = 1/6
pr(3rd) = 5/6*4/5*1/4 = 1/6

These are mutually exclusive events, so we can add them together to get a total probability of 1/2.

Bunuel · Answer

Testing...............................................

A certain box holds 3 green balls, 2 white balls, and 1 blue

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