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]

(a) Compute the probability that a randomly selected peanut M&M is not yellow.

(b) Compute the probability that a randomly selected peanut M&M is orange or yellow.

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

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

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

(a) 0.85

(b) 0.38

(c) 0.002

(d) 0.986

(e) 0.12

Step-by-step explanation:

Brown = 12%,

Yellow = 15%

Red = 12%

Blue = 23%

Orange = 23%

Green = 15%

(a) Probability that a randomly selected peanut M&M is not yellow.

P (Yellow) = 15/100 = 0.15

Therefore: P (Not Yellow)

=1-0.15 = 0.85

(b) Probability that a randomly selected peanut M&M is orange or yellow.

P (Orange OR Yellow)

= P (Orange) + P (Yellow)

=23/100 + 15/100

=38/100 = 0.38

(c) Probability that three randomly selected peanut M&M's are all red.

P (Red and Red and Red) = 12/100 X 12/100 X 12/100 = 0.001728 = 0.002

(d) Probability that neither of them are red.

P (neither of them are red) = 1 - P (both are red) = 1 - (12/100 X 12/100)

= 1-0.0144 = 0.9856 = 0.986

(e) Probability that at least one of them is red.

P (RB or RY or RR or Rb or RO or RG)

= (0.12 X 0.12) + (0.12 X 0.15) + (0.12 X 0.12) + (0.12 X 0.23) + (0.12 X 0.23) + (0.12 X 0.15)

= 0.12