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

Question

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

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 4 decimal places where possible)

Answer 1

Compute the probability that none are blue, then subtract the probability from 1.

Assume we have a very large bag to select the M and M's from so that each selection is independent (as we were not given number of M and Ms from which we were selecting).

Since probability of getting a blue M and M in a single selection is 0.23 (23% in decimal), probability of not getting a blue M and M is 1 - 0.23 = 0.77.

Probability of not getting a blue M and M three times in a row is 0.77 * 0.77 * 0.77 = 0.4565

So probability of getting at least 1 blue M and M is 1 - 04565 = 0.5435.