I have a bag of M&M peanuts. red white and blue. equal numbers of each color. I reach into the bag and grab 3 . what is the chance of grabbing 1 of each color? what is the formula for calculating that probability?
There are 3x3x3 = 27 choices and 6 ways to permute RWB so the probablility is 6/27.
Or:
The chance of the 2nd pick being different from the first is (2N)/(3N-1) and the chance of the 3rd pick being different from the first and the second is (N)/(3N-2).
The chance that both these events occur is the product of their individual chances of occurring.
As N increases the chance that both these events occur approaches 2/9 = 6/27.
Here’s my take on it. It’s been while since I took probability, so take it with a grain of salt.
To be considered a “success”, you have to take three candies out of the bag, and they have to all be different colors.
When you take the first candy out, it will contribute to the success of the “experiment” no matter which color it is. The probability of here then is 100 percent, or 1.
When you take the 2nd out, the only condition is that it cannot be the same color of the 1st. So, the probability of this is (2/3 X) divided by (X-1). The minus 1 is because of the first one removed. (X = total amount of M&Ms)
When taking the last one out, it has to be of the remaining color that hasn’t been chosen yet. The probability of this is (1/3 X) divided by (X-2). The minus 2 is because two have been previously removed.
To get the total probability, you multiply all three. So the answer is
1 times ((2/3 X) divided by (X-1)) times ((1/3 X) divided by (X-2))
where X = total amount of M&Ms.
The above can be simplified, but I’ll leave that as an exercise to the reader…
It depends how many M&M’s are in the bag. For example, if there are only 3 M&M’s in the bag, then the probability of pulling one of each out is 1. But to generalize, assume there are n of each color in the bag. There are (3n)(3n-1)(3n-2)/321= (27n[sup]3[/sup]-27n[sup]2[/sup]+6n)/6= ((3n)[sup]2/sup+2n)/2 different 3 M&M combos. There are n[sup]3[/sup] ways to pick out a “red-white-blue” combo. So… the probability would be
2n[sup]3[/sup]/((3n)[sup]2/sup+2n)
i got it by doing
chances 1st one is good = 3/3
changes 2nd one is good 2/3
chances 3rd one is good 1/3
total = 6/27 = 2/9
but what is the expression for the calculation for any number of colors N?
is it:
(n-1)!/ n**(n-1)
justinh, you’re forgetting that the number of M&M’s goes down by one at each step, so, for example, when you draw the second M&M, the probability of drawing a different color won’t be exactly 2/3. However, as the number of M&M’s in the bags increases, the probability will have a limiting value of 2/9.
If we change the problem to one where we have a bag with n different colors, m of each color, then the probability is:
[(mn-n)! * (n!) * m] / (mn)!