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 brown.

Answer 1

Start by thinking about what information you need, and what is extra. Since the question is asking about brown M&Ms, you can ignore most of it and just ask about the probability of brown or not brown. Start with a single pull, and calculate the probability of pulling a brown candy.

P(brown) = (number of brown possibilities) / (number of total possiblities) = 12% / 100% = 0.12

Whenever you know all the possible options, the total probability always adds up to 1, so

P(not brown) = 1 - 0.12 = 0.88

One way to do a probability problem is to count up all the possible ways you could pull n items, then count the number of those that meet your criteria. But there is a far far easier way.

This is because the question is not asking for the probability of pulling 1, 2, 3, or 4 brown M&Ms. It just cares about the probability of pulling at least 1 brown. The only way this could not happen is if you only pulled not brown candies. So we can calculate that, and then just subtract it from 1 (remember, the probabilities have add up to 1).

So now we get:

P(not brown first candy) * P(not brown second candy) * P(not brown third candy) * P(not brown fourth candy)

All the probabilities are the same: 0.88, and we don't care about the order, so we can write this as

P(not brown)⁴ = (0.88)⁴ = .5997

So the probability that you pull at least 1 brown candy is one minus the probability that you pull no brown candies:

1 - .5997 = .4003, or about 40%.