According to masterfoods....

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 brown 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 brown candy is the third M&M selected.

Compute the probability that the first brown candy is the third or fourth M&M selected.

Compute the probability that the first brown candy is among the first three M&M’s selected.

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

Answer 1

For this problem, q = 1 - p.

Excel has a negative binomial distribution function. With one success this becomes the geometric distribution

P(X = 3)= q^2p = .88²(.12) = 0.0929

Alternatively, in Excel, NEGBINOM.DIST(2,1,0.12,0) (the 0 means exactly this solution)

P(X = 3 or X = 4) = q^2p+ q^3p = q^2p(1 + q) = .88²(.12)(1.88) = 0.1747

In Excel, NEGBINOM.DIST(2,1,0.12,0) + NEGBINOM.DIST(3,1,0.12,0)

1 - q^3 = 1 - .88^3 (1 - all 3 not being brown) = 0.3185

Alternatively, NEGBINOM.DIST(2,1,0.12,1) (note that the final value is 1 here; this is cumulative - up to 2 failures)

Expected value is 1/p = 1/.12 = 25/3 = 8.3333 ; since this is not a probability, I assumed that we still want 4 numbers after the decimal point (note that this is the number of trials until a success, not the number of failures until a success; be careful which version of the geometric distribution that you use)