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 think the answer is (C) 7. Because at worst case the student can draw 2 from each color and still he has to draw at least one more marble so that the third one of any color will constitute a color with three marbles.

What is OA?
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Nobody dies a virgin, life screws us all.

you all are right, it is 7

although, i really have hard time understanding why??? What if he keeps taking red ones... Or minimum no means that he's "lucky", keeps pulling different color each time

It should b R+G+B and then again R+G+B and then on 7th draw,he might pick either of R/G/B thus satisfying our condition.

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.