How many M&M bags with one color missing?

Question

A small bag of M&Ms from the vending machine in our office contains about 20 M&Ms. For this question, let's say exactly 20. They come in 5 colors: green, brown, yellow, blue and red. The distribution of colors appears to be random. If it is random, what is the percentage of bags with one color missing? Or has anyone information about M&Ms packing machines and knows that the distribution is not random??

Thanks Doug. It turns out that for my country there are only 11.1% green ones and 22,2% of red, blue, brown and yellow ones. But that can be only an average because this morning a bag contains 9 yellow M&M but only 2 red ones. Makes the question even trickier... :-)

Answer 1

If the distribution were equal (20% each) then the chances of an M&M being red would be 20%. The chance of it not being red would be 80%. The chance of 20 M&Ms not being red would be 80%^20, or 1.1529%. The chance of any one of 5 colors being absent would be 5 times that, or 5.7646%.

With the distribution in your country, it would be only 77.8%^20=0.665% for each of the colors except green, but green would have a probability of 88.9%^20=9.507%, so the total probability would be 4*0.665% + 9.507% = 12.147%.

Answer 2

NotEasilyFooled's answer is almost correct but...

There is a slight possibility that a bag misses more than one color. These possibilities have been counted double (or more) in his calculation. The correct answer will therefore be slightly less than his number. It is 5.618%.

How do you find this answer?

Let S = { gr, br, y, bl, rd } be the set of colors, and C a subset of S.
Define P(C) to be the probability that the colors in the M&M package are precisely those of C, and P*(C) the probability that they are a subset of C.

For instance, P*({gr,br,y}) is the probability that there are only green, brown and yellow M&Ms, but possibly some of these colors missing as well; P({gr,br,y}) is the probability that all these colors, and only them, are present. Obviously, P(C) <= P*(C). In fact,

P(C) = P*(C) - (sum of P*(D) over all subsets D of C)

Now P*(C) is easy to calculate; it is equal to
P*(C) = (0.20 * |C|)^20

if |C| = 1, P*(C) = 0.000000000001049%
if |C| = 2, P*(C) = 0.000001100%
if |C| = 3, P*(C) = 0.003656%
if |C| = 4, P*(C) = 1.153%
if |C| = 5, P*(C) = 100.0%

We will abbreviate this as P*(4) = 1.153%, etc. It is clear that P(1) = P*(1).

Suppose |C| = 2, for instance, C = {gr, rd}. The probability that precisely these two colors are present, and none is missing, is equal to

P({gr, rd}) = P*({gr, rd}) - P({gr}) - P({rd})

so P(2) = P*(2) - 2 P(1) = P*(2) - 2 P*(1)

Likewise,

P({gr, rd, y}) = P*({gr, rd, y}) - P({gr, rd}) - P({gr, y}) - P({rd, y}) - P({gr}) - P({y}) - P({rd})

so

P(3) = P*(3) - 3 P(2) - 3 P(1)
... = P*(3) - 3 [P*(2) - 2 P*(1)] - 3 P*(1)
... = P*(3) - 3 P*(2) + 3 P*(1)

In the same way we get

P(4) = P*(4) - 4 P(3) - 6 P(2) - 4 P(1)
... = P*(4) - 4 [P*(3) - 3 P*(2) + 3 P*(1)] - 6 [P*(2) - 2 P*(1)] - 4 P*(1)
... = P*(4) - 4 P*(3) + 6 P*(2) - 4 P*(1)

(Note for math fans: yep, you get the binomial coefficients with alternating sign.)

This is equal to 1.1383%. For any given set of four colors, this is the probability that precisely these colors are present in the M&Ms bag. Since there are five possible combinations of four colors, the final answer is 5 * 1.1383%.

As an exact fraction, the probability is

{4^20 - 4 * 3^20 + 6 * 2^20 - 4} / {5^19}
=
1,085,570,781,624 / 19,073,486,328,125

Answer 3

Assuming a perfectly random distribution:

The chance of a specific color not coming up in a bag is (4/5)^20. Since this can happen to five different colors, the chance of having a bag with not one of any color is 5*(4/5)^20.

This is equal to about 5.76 %, or 1 in about 17.3 bags.

Using the distribution above, not having a green is (8/9)^20 and not having another color is 4*(7/9)^20.

Added together, this is 12.11 % or about 1 in 8.26 bags.

Assuming this distribution is correct, the chance of receiving exactly 9 yellows is (2/9)^9*(7/9)^11*20nCr9 = 1.40%.

The chance of receiveing exactly 9 yellows AND exactly two reds is

1.40% * (2/7)^2*(5/7)^9*11nCr2 = 0.304% or 1 in 329 bags.

This just goes to show you that just about every bag is unique...but considering that there are 95 trillion ways you could pull 20 M&M's out of a bag, that makes sense.

Answer 4

http://www.m-ms.com/


Doug

Answer 5

hmmm, you have me wondering now! Sorry, I cant help on that one!