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 brown or red.

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

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

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

Kindly check explanation

Step-by-step explanation:

Given the following :

P(brown) = 12% = 0.12

P(Yellow) = 15% = 0.15

P(Red) = 12% = 0.12

P(blue) = 23% = 0.23

P(orange) = 23% = 0.23

P(green) = 15% = 0.15

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

P(not yellow) = P(Yellow)' = 1 - P(Yellow) = 1 - 0.15 = 0.85

B.) Compute the probability that a randomly selected peanut M&M is brown or red.

P(Brown) or P(Red) :

0.12 + 0.12 = 0.24

C.) Compute the probability that three randomly selected peanut M&M’s are all brown.

P(brown) * P(brown) * P(brown)

0.12 * 0.12 * 0.12 =0.001728

D.) If you randomly select three peanut M&M’s, compute that probability that none of them are blue.

P(3 blue)' = 1 - P(3 blue)

P(3 blue) = 0.23 * 0.23 * 0.23 = 0.012167

1 - P(3 blue) = 1 - 0.012167 = 0.987833

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

P(1 blue) + p(2 blue) + p(3 blue)

(0.23) + (0.23*0.23) + (0.23*0.23*0.23)

0.23 + 0.0529 + 0.012167

= 0.295067

one set of data has more variation than the other. hope this !