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