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

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

Answer 1

The way I like to think about these types of questions is to understand what they are asking. By asking you to compute that probability that at least one of them is yellow, they are essentially telling you to find the probability that none of them are yellow and subtract that from one.

We can see, from the question, 15% of the M&M's produced are yellow. This means that if we were to randomly select an M&M, the probability it would be yellow is 0.15. Subtracting this from 1 will give us the probability of selecting a non-yellow M&M (1-0.15=0.85).

Using these values, we can find the probability that none of the 4 M&M's we select will be yellow. p(none yellow)=0.85x0.85x0.85x0.85=0.52

Now, subtracting this value from 1 will give us the probability that at least one of the M&M's we select is yellow, 1-0.52=0.48.

So if we randomly select 4 M&M's, the probability we get at least one yellow is 0.48 (there is a 48% chance at least one is yellow).

Answer 2

The probability that at least one of them is yellow = 1 - probability than none of them are yellow.

Assuming the number of M and M s is large so that each selected M and M is an independent sample.

Probability selected M and M is not yellow is 1 - 0.15 = 0.85

Probability selected M and M is not yellow for all 4 selections is (0.85)^4

So probability at least one is yellow is 1 - (0.85)^4 = 0.48

Answer 3

This is 1 - P(all 4 are not yellow)

P(not yellow) = 1 - 15% = 85%

1 - P(all 4 are not yellow) = 1 - .85^4 = .4780