According to Masterfoods

Question

According to Masterfoods, the company that manufactures M&M’s, 12% of peanut M&M’s are brown, 15% are yellow, 12% are red, 23% are blue, 23% are orange and 15% are green. You randomly select four peanut M&M’s from an extra-large bag of the candies. (Round all probabilities below to four decimal places; i.e. your answer should look like 0.1234, not 0.1234444 or 12.34%.)

Compute the probability that exactly two of the four M&M’s are orange.

Compute the probability that two or three of the four M&M’s are orange.

Compute the probability that at most two of the four M&M’s are orange.

Compute the probability that at least two of the four M&M’s are orange.

If you repeatedly select random samples of four peanut M&M’s, on average how many do you expect to be orange? (Round your answer to two decimal places.)

 orange M&M’s

With what standard deviation? (Round your answer to two decimal places.)

 orange M&M’s

Answer 1

Several of these questions can be answered simply through the application of the binomial distribution.

Let p = probability of (fraction of) M&Ms of that color, and

q = 1-p probability of being any of the others.

P(r,n,p) = binomial probability of getting exactly r of the desired color in a sample of n M&Ms.

p(r,n,p) = C(r,n) p^r (1-p)^n-r

where

C(r,n) = n! / (r!) (n-r)! combinations, the number of ways of getting r things from a group of n.

and the rest of P(r,n,p) is the probability of any single sequence that

has r out of n selected.

So, for the probability that two (r=2) of four (n=4) in a sample being orange (p=0.23) rather than another color (0.77),

we have

P(r,n,p) = [4!/(2!)(2!)] (0.23)^2 (0.77)^2 = 0.1882,

recalling that n! = (n)(n-1)(n-2)... "n factorial."

I'll leave the rest of the questions as "exercises for the students."

A key is that you can calculate for one color at a time (yes/no) at p and lump the other probabilities as 1-p.