A student is asked to pick marbles from a bag that contains

The worst case scenario would be if we remove 2 marbles of each color, so 6 marbles, and we still won't have 3 marbles of the same color. The 7th marble will be red, green or black, thus after 7 draws we are guaranteed to have at least three marbles of the same color.

Answer: C.

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Hope it helps.
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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?

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

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

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

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.

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.

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.

7 (worst case(2+2+2) +1)

Bunuel why wouldnt we have 8+6+3 as the ans? The person can also pick all red balls continuously and then all blacks and then the last three of green?

We need the minimum number of marbles to guarantee that we have at least three marbles of the same color.

So, 8 + 6 + 3 = 17 cannot be an answer because we can guarantee the desired outcome in 7 (17 is not minimum).

You cannot guarantee the desired outcome in less than 7 picks because as I said in the solution above we can have a case when we pick 2 marbles of each color, so 6 marbles, and we still won't have 3 marbles of the same color.

But for 7 draws, there does not exist a case when after the 7th draw we don't have at least three marbles of the same color. That's why 7 is an answer.


Check other Worst Case Scenario Questions from our Special Questions Directory to understand the concept better.

Hope it helps.
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