Statistics and 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. (Round your answers to 4 decimal places where possible)

a. Compute the probability that a randomly selected peanut M&M is not orange.

b. Compute the probability that a randomly selected peanut M&M is brown or green.

c. Compute the probability that three randomly selected peanut M&M’s are all red.

d. If you randomly select three peanut M&M’s, compute that probability that none of them are orange.

e. If you randomly select three peanut M&M’s, compute that probability that at least one of them is orange.

Answer 1

P(Brown) = 0.12 P(Yellow) = 0.15 P(Red) = 0.12

P(Blue) = 0.23 P(Orange) = 0.23 P(Green) = 0.15

a.) Use the complement rule: P(Not Orange) = 1 - P(Orange) = 1 - 0.23 = 0.77

The probability that a randomly selected peanut M&M is not orange is 0.77.

b.) Use the addition rule for probability:

P(Brown or Green) = P(Brown) + P(Green) - P(Brown and Green) = 0.12 + 0.15 - 0 = 0.27

The probability that a randomly selected peanut M&M is either brown or green is 0.27. Keep in mind that this is mutually exclusive because no peanut M&M is both brown and green.

c.) Use the multiplication rule for probability: P(Red)*P(Red)*P(Red) = (0.12)³ = 0.001728

The probability that 3 randomly selected peanut M&M's are all red is 0.001728.

d.) Use the multiplication rule for probability: [P(Not Orange)]³ = (0.77)³ ≈ 0.4565

The probability that 3 randomly selected peanut M&M's are not orange is about 0.4565.

e.) Use the complement rule:

P(at least one M&M is orange) = 1 - P(3 M&M's are not orange) = 1 - (0.77)³ = 0.5435

The probability that 3 randomly selected peanut M&M's where at least one of them is orange is about 0.5435.