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 three decimal places, for example: 0.123] Compute the probability that a randomly selected peanut M&M is not yellow. Compute the probability that a randomly selected peanut M&M is orange or yellow. Compute the probability that three randomly selected peanut M&M's are all red. If you randomly select two peanut M&M's, compute that probability that neither of them are red. If you randomly select two peanut M&M's, compute that probability that at least one of them is red.

Answer 1

Answer:

a) P(Y') = 0.850

b) P(O) + P(Y) = 0.380

c) 0.001728 = 0.002 to 3 d.p

d) 0.7744 = 0.774 to 3 d.p

e) 0.120

Step-by-step explanation:

Probability of brown M&M's = P(Br) = 0.12

Probability of yellow M&M's = P(Y) = 0.15

Probability of red M&M's = P(R) = 0.12

Probability of blue M&M's = P(Bl) = 0.23

Probability of orange M&M's = P(O) = 0.23

Probability of green M&M's = P(G) = 0.15

Total probability = 1

a) The probability that a randomly selected peanut M&M is not yellow = 1 - P(Y) = 1 - 0.15 = 0.85

b) The probability that a randomly selected peanut M&M is orange or yellow = P(O) + P(Y) = 0.23 + 0.15 = 0.38

c) The probability that three randomly selected peanut M&M's are all red = P(R) × P(R) × P(R) = 0.12 × 0.12 × 0.12 = 0.001728

d) If you randomly select two peanut M&M's, compute that probability that neither of them are red

This probability = P(R') × P(R')

P(R') = 1 - P(R) = 1 - 0.12 = 0.88

P(R') × P(R') = 0.88 × 0.88 = 0.7744

e) If you randomly select two peanut M&M's, compute that probability that at least one of them is red

This probability = [P(R) × P(R')] + [P(R) + P(R)] = (0.12 × 0.88) + (0.12 × 0.12) = 0.1056 + 0.0144 = 0.120