Compute the probability

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 peanut M&M’s from an extra-large bag looking for a blue candy. (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 the first blue candy is the sixth M&M selected.

Compute the probability that the first blue candy is the sixth or seventh M&M selected.

Compute the probability that the first blue candy is among the first six M&M’s selected.

If every student in a large Statistics class selects peanut M&M’s at random until they get a blue candy, on average how many M&M’s will the students need to select? (Round your answer to two decimal places.)

_____ blue M&M’s

Answer 1

P(blue) = .23

a)The trial of the first success is the geometric distribution.

Thus, P(first success in trial 6) = (1-.23)^(6-1)(.23) = (.77)^5(.23) = 0.0623

b) P(first sucess in trials 6 or 7) = (.77)^5(.23) + (.77)^6(.23) = (.77)^5(1 + (.77))(.23) = (.77)^5(1.77).23 = 0.1102

c) P (at least one candy in first 6 trials) = 1 - P(all non-blue in 6 trials) = 1 - .77^6 = 0.7916

d) The mean of the geometric distribution is 1/p = 1/.23 = 4.35