Let X1, X2, X3, X4, X5 and X6 denote the numbers of blue, brown, green, orange, red, and yellow M&M candies, respectively, in a sample of size n. According to the M&M Web site, the color proportions are p1=0.24, p2=0.13, p3=0.16, p4 = 0.20, p5 =0.13, and p6 =0.14. a) If n = 12, what is the probability that there are exactly two M&Ms of each color? b) For n = 20, what is the probability that there are at most five orange candies? (Hint: Treat an orange candy as a success and any other color as a failure) c) In a sample of 20 M&Ms, what is the probability that the total number of candies that are blue, green, or orange is at least 10?

Since the probability of taking a color does not depend on the previous color obtained, the events are independent, so the probability of getting exactly two of each color is

[tex]\large (0.24)^2*(0.13)^2*(0.16)^2*(0.20)^2*(0.13)^2*(0.14)^2= 3.3018*10^{-10}[/tex]

b)

We can model this with a binomial distribution treating an orange candy as a “success” and any other color as a “failure”.

So the probability of “success” p equals 0.20 and the probability of “failure” q equals 0.80 and the probability that there are at most 5 orange candies out of 20 equals

[tex]\large \sum_{k=0}^{5}\binom{20}{k}(0.2)^k(0.8)^{20-k}=\binom{20}{0}(0.2)^0(0.8)^{20}+\binom{20}{1}(0.2)^1(0.8)^{19}+\binom{20}{2}(0.2)^2(0.8)^{18}+\\\binom{20}{3}(0.2)^3(0.8)^{17}+\binom{20}{4}(0.2)^0(0.8)^{16}+\binom{20}{5}(0.2)^5(0.8)^{15}[/tex]

this can be computed either by hand or with the aid of a computer to obtain that probability that there are at most five orange candies is 0.8042 0r 80.42%

c)

In this case we use once more the binomial distribution but this time “success” is getting blue, green or orange that has a probability of 0.24 + 0.16 + 0.20 = 0.60 and “failure” has a probability of 0.4.

Now we are looking for

[tex]\large \sum_{k=10}^{20}\binom{20}{k}(0.6)^k(0.4)^{20-k}[/tex]

again by hand or with the computer help, this probability equals 0.8724 0r 87.24%