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
