Binomial distribution

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%.)

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

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

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

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

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

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

Answer 1

a) (5C3)(.23^2)(.77^3)= .2415

b) .2415 + (5C4)(.23^4)(.77) = .2415 + .0108 = .2523

c) 1 - .2523 - (5C5)(.23^5) = .7471

d) .7471

e) 5(.23) = 1.15

f) sqrt(5(.23)(.77)) = .94