The question is on Math Probability and has five parts. According to Masterfoods, 12% of peanut M&M's are brown, 15% are yellow , 12% are red, 23% are orange and 15% are green.

Question

Round your answers to 4 decimal places where possible.

1. Compute the probability that a randomly selected peanut M&M is not orange.
2. Compute the probability that a randomly selected peanut M&M is brown or green.
3. Compute the probability that two randomly selected peanut M&Ms are both brown
4. If you randomly select six M&Ms, compute that probability that none of then are orange
5. 5. If you randomly select six peanut M&Ms, compute that probability that at least one of them is orange.
6. My name is Susan Davis
7. I am a student at Eastern Gateway Community College
8. My email is [email protected]

Answer 1

Although it's not explicitly stated (and the wording of the question is somewhat misleading), you'll need to assume you are selecting WITH replacement in order to answer parts 3, 4, and 5. Alternatively, if you're selecting WITHOUT replacement from a very large population (a reasonable assumption in this scenario), the probabilities computed WITH replacement will be close to the true probabilities.

1. P(Not Orange) = 1 - P(Orange).
2. Since "Brown" and "Green" are mutually exclusive, P(Brown OR Green) = P(Brown) + P(Green).
3. Selecting WITH replacement, P(2 of 2 are Brown) = P(Brown)*P(Brown) = P(Brown)**2
4. Selecting WITH replacement, P(6 of 6 are Not Orange) = P(Not Orange)**6
5. P(At least one Orange) = 1 - P(None are orange) = 1 - [Answer from #5]

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