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?

a)

b) 80.42%

c) 87.24%

Step-by-step explanation:

a)

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

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

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

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

1. 3. can also be expressed as (22/7)

2. -4.5

3. 3.6

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a = 3.14

b = 18

c = 18/5 or 3.6 or 3 3/5

step-by-step explanation: