In his pocket, a boy has 3 red marbles, 4 blue marbles, and

I agree...only logic will help

take out 4G + 4B + 1R = 9 marbles

What about scenario of choosing 3 red and 4 blue and still seeing distinct color.
In other words why can't we have 8 marbles drawn and still see three distinct colors? 3red+4 blue=7 marbles and on 8th draw should be green since remaining should be all green marbles?

Can someone please explain why?

Thanks in advance.

The word "ensure" in the question basically is asking in the worst case scenario. So yes technically, the scenario outlined above is true as well, it is not the worst possible outcome.

In order to guarantee the outcome of one of each color, you need to take into account all of the balls in the two largest groups which equals 8.

Hope that helped.

What is the difference between this question and the one below.Please explain its confusing:

There are 15 black chips and 5 white chips in a jar. What is the least number of chips we should pick to guarantee that we have 2 chips of the same color?
A. 3
B. 5
C. 6
D. 16
E. 19

Worst case scenario would be if the first two chips we pick will be of the different colors. But the next chip must match with either of two, so 3 is the answer.

Answer: A.

9 (4+4+1) worst case

In Such cases, Go for the worst case scenario and hence here we can assume that balls drawn first is (Blue or Green) = 4 and then a single ball to make sure that the guy has atleast one ball of each color which is +1.

Hence the answer is 4+4+1 = 9

In his pocket, a boy has 3 red marbles, 4 blue marbles, and 4 green marbles. How many will he have to take out of his pocket to ensure that he has taken out at least one of each color?

We need to maximise the possibility of worst case scenario's.

Consider that first 4 draws can be from green and next 4 can be from blue marbles, even then not all color is represented, but the next card has to be Blue and thus all colors are represented.

So 8 unsucessful draws (4B + 4G) + 1 Blue = 9

Ans: D

In his pocket, a boy has 3 red marbles, 4 blue marbles, and 4 green marbles. How many will he have to take out of his pocket to ensure that he has taken out at least one of each color?

A. 3
B. 7
C. 8
D. 9
He can start with 3 red, 4 blue and then 1 green but that is not the worst case scenario, the worst case scenario is that he starts with 4 green, 4 blue and finally 1 red, making it 9 marbles.
E. 11

In his pocket, a boy has 3 red marbles, 4 blue marbles, and 4 green marbles. How many will he have to take out of his pocket to ensure that he has taken out at least one of each color?

To ensure that the boy has taken out at least one marble of each color, we need to consider the worst-case scenario where he takes out the maximum number of marbles of the same color before getting one of each color.

In this case, the boy would need to take out all the marbles of two colors (red and blue or red and green) before he can be certain to have at least one of each color.

To calculate the minimum number of marbles he needs to take out, we add up the number of marbles of the two colors he has the most of, which are red and blue:

3 red marbles + 4 blue marbles = 7 marbles.

Therefore, the boy would need to take out at least 7 marbles from his pocket to ensure that he has taken out at least one marble of each color.

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