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 five 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 five M&M’s are brown.

Compute the probability that two or three of the five M&M’s are brown.

Compute the probability that at most two of the five M&M’s are brown.

Compute the probability that at least two of the five M&M’s are brown.

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

With what standard deviation? (Round your answer to two decimal places.)  brown M&M’s

Answer 1

it is a binomial probability distribution,

and probability is given by

P(X=x) = C(n,x)*px*(1-p)(n-x)

Sample size , n = 5

Probability of brown, p =0.12

a)

P ( X = 2) = C(5,2)*0.12^2*0.88^3=0.0981

b)

P(X=3 or X=2) = P(X=3) + P(x=2)

P ( X = 3) = C(5,3)*0.12^3*0.88^2=0.0134

P(X=3 or X=2) = P(X=3) + P(x=2) = 0.0981+0.0134 = 0.1115

c)

P(X≤2) = P(x=0) + P(X=1) + P(X=2) = 0.9857

d)

P(X≥2) = 1 - P(x=0) - P(X=1) = 0.1125

e)

Mean = np = 5*0.12 = 0.60

Variance = np(1-p) = 0.5280

Standard deviation = √variance = 0.73