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)

Compute the probability that exactly three of the five M&M’s are red.

Compute the probability that three or four of the five M&M’s are red.

Compute the probability that at most three of the five M&M’s are red.

Compute the probability that at least three of the five M&M’s are red.

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

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

Answer 1

here radom variable number of red M&M follows bi8nomial distribution with paramter p=0.12 and n=5

a)

probability that exactly three of the five M&M’s are red =P(X=3)==0.0134

b)

probability that three or four of the five M&M’s are red=P(X=3)+P(X=4)

=+ =0.0134+0.0009=0.0143

c)

probability that at most three of the five M&M’s are red =P(X<=3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)

=

=0.5277+0.3598+0.0981+0.0134=0.9991

d)probability that at least three of the five M&M’s are red =P(X>=3)=1-P(X<=2)

=1-(P(X=0)+P(X=1)+P(X=2))=1-(0.5277+0.3598+0.0981)=0.0143

e)

expected to be red =np=5*0.12=0.60

f) standard deviation =sqrt(np(1-p))=0.73