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.

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 blue.

Compute the probability that a randomly selected peanut M&M is green or red.

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

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

Answer 1

Hello Nora!

I will answer each one, the questions will be bold.

Compute the probability that a randomly selected peanut M&M is not blue

P (blue) = 23%

P (blue)^C = 1 - P(blue) = 100% - 23% = 77%

Compute the probability that a randomly selected peanut M&M is green or red.

P (green) = 15%

P (red) = 12%

P (green or red) = P(green) + P(red) = 27%

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

P (yellow) = 15%

(P(yellow))³ = 0.338%

Please note here that I considered there was replacement in the selection process.

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

P(brown) = 12%

P(brown)^C = 1 - P(brown) = 100% - 12% = 88%

(P(brown)^C)⁶ = 46.440%