A student is asked to pick marbles from a bag that contains 18 marbles. Eight marbles are red, 4 are green, and 6 are black. What is the minimum number of marbles that a blindfolded student would have to draw from the bag to be certain of having at least three marbles of the same color?

A. 5
B. 6
C. 7
D. 13
E. 14

i get 7...he takes out 2 of each..so the 7th one will be of the 3rd color

What you are saying is right for the best case. However, in order to be certain, worst case has to be considered.

The answer is 7.

There is no way we can have 7 marbles without picking at least 3 of the same color, infact we cannot write 7 as the combination of only 2 marbles of the same color:

\(2G+2B+2R+1?=7\) marbles

The last one that is missing must be G or B or R so we have at LEAST 3 of a color.
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Took a while to understand the question. But here's a simplified version of the question stem for those who are struggling to understand why 17 is not the correct answer.

The question simply asks after how many tries when will the student have atleast three marbles of the same colour.
If he is lucky then he can pick 3 of same colour (red, green or black) in the first three tries. This is the best case.

In the worst case, he draws them in the below order:

RRGGBBR (on the 7th he picks a red making it atleast three of same colour)
Likewise, there can be several other orders in which he may draw them i.e. GGBBRRG, BBRRGGB, BRBRGGB, BRGBGRB and so on.

Essentially, when the student draws 7 marbles there would be atleast 3 marbles of the same colour.