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 on of each color?
Assuming that he gets a different one each time = 3, so the fourth one will be the same as one of the first 3.
To ensure that he has taken out at least one of each color, he will have to take out a maximum of 3 marbles from his pocket.
To ensure that the boy has taken out at least one of each color, we need to figure out the worst-case scenario. This means we assume he always picks the same color marble until he has already taken out one of each color.
Let's break it down step by step:
1. First, the boy can pick any marble without restriction.
2. For the second marble, he needs to avoid picking the same color as the first marble. So, he has two options: either pick a different color or pick another marble of the same color but not the same as the first marble.
3. From the third marble onwards, he needs to avoid picking the same color as the first two marbles. So, he again has two options: either pick a different color or pick another marble of the same color but not the same as the first two marbles.
Based on these steps, we can calculate the minimum number of marbles the boy needs to pick to ensure he has taken out at least one of each color:
First marble: Any color, so 1 possibility.
Second marble: 2 possibilities (different color or same color but different from the first).
Third marble: 2 possibilities (different color or same color but different from the first two).
In general, for the nth marble, the boy will have n possibilities: either pick a different color or pick another marble of the same color but different from the previous (n-1) marbles.
In this case, since the boy has a total of 3 + 4 + 4 = 11 marbles, he will have to take out at least 11 marbles to ensure that he has taken out at least one of each color.