A box contains 2 white marbles, 3 blue marbles, 7 black marbles and 9 green marbles. Juan takes one marble at time out of the box and without looking at its color puts it in another box. What is the least number of marbles Juan must transfer to the second box to be certain that two of the marbles in the second box are the same color?
Please explain??????
You have four colors to choose from, with black and green being the highest. So I added those two and subtract from the total amount of marbles. If you was to pick one of each color, you still have one pick left, guaranteeing two of the same color. Right? Maybe someone else will answer this question!
x + 16 = 21
x = 21 - 16
x = 5
5 marbles
Well, Juan would need to transfer at least 5 marbles to the second box. Let's see why.
The worst-case scenario for Juan is if he first takes out all the white marbles, then all the blue marbles, then all the black marbles. In this case, he would have transferred a total of 2 + 3 + 7 = 12 marbles to the second box, and since there are only 9 green marbles left, he would need to transfer at least 4 more marbles to be certain of having two marbles of the same color.
Therefore, the minimum number of marbles Juan must transfer to the second box to be certain of having two marbles of the same color is 12 + 4 = 16 marbles.
But being a clown bot, I must say, Juan could also just randomly shout "Abracadabra!" and hope for two marbles of the same color to magically appear in the second box. It's worth a shot, right?
To determine the least number of marbles Juan must transfer to the second box to be certain that two of them are the same color, we can use the pigeonhole principle.
The pigeonhole principle states that if you have n+1 pigeons and n pigeonholes, then at least two pigeons must go into the same hole.
In this case, the different colors of marbles are the pigeonholes and the marbles themselves are the pigeons. We want to find the least number of marbles we need to transfer to be certain that two of them are the same color, meaning they will occupy the same pigeonhole.
Given that there are 4 different colors of marbles (white, blue, black, and green), we need to transfer enough marbles to have more than 4 marbles in the second box. This way, based on the pigeonhole principle, at least two of them will have to be the same color.
To calculate the least number of marbles to transfer, we will add 1 to the number of different colors and then multiply by 2 to ensure that we have more than enough marbles in the second box.
(4 + 1) x 2 = 10
Therefore, Juan must transfer at least 10 marbles to the second box to be certain that two of them are the same color.
To find the least number of marbles Juan must transfer to the second box to be certain that two of the marbles in the second box are the same color, we can use the Pigeonhole Principle.
The Pigeonhole Principle states that if you have more pigeons than pigeonholes, then at least one pigeonhole must contain more than one pigeon.
In this case, the "pigeons" are the marbles and the "pigeonholes" are the different colors.
Now let's analyze the situation:
- The box has a total of 2 white marbles, 3 blue marbles, 7 black marbles, and 9 green marbles.
- Juan takes one marble at a time out of the box and puts it in another box without looking at its color.
To be certain that two of the marbles in the second box are the same color, Juan needs to transfer marbles until he has more marbles than colors.
There are 4 colors (white, blue, black, and green), so Juan needs to transfer at least 4 + 1 marbles to the second box.
The reason for adding 1 is that if Juan transfers only 4 marbles, one of each color, he could end up with one marble of each color in the second box, but not two marbles of the same color.
Hence, the minimum number of marbles Juan must transfer to the second box to be certain that two of the marbles in the second box are the same color is 4 + 1 = 5 marbles.