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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]
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 red or green.
3.Compute the probability that three randomly selected peanut M&M’s are all blue.
4.If you randomly select four peanut M&M’s, compute that probability that none of them are green.
5.If you randomly select four peanut M&M’s, compute that probability that at least one of them is green.

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Answer:

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By taking the quotients between the percentages, we will get:

  • 1) 0.770
  • 2)0.270
  • 3) 0.012
  • 4) 0.522
  • 5) 0.478

How to get the probabilities?

1) The probability of not getting an orange one is equal to the quotient between the percentage of not orange ones, and 100%.

We know that 23% are orange, then 100% - 23% = 77% are not orange, so the probability is:

P = 77%/100% = 0.770

2) 12% are red, 15% are green, so 27% are either red or green, the probability is:

P = 27%/100% = 0.270

3) To get 3 blues we can estimate the probability as:

P = (23%/100%)^3 = 0.23^3 = 0.012

But this would be incorrect, as we draw blue peanuts, the probability of drawing another is reduced (because now there are less on the package).

4) Same procedure as before, 15% are green, so 85% are not green, so to draw 4 non-green ones the probability is:

P = (85%/100%)^4 = 0.522

5) The probability of getting at least one green in 4 daws is 1 less what we got above.

P = 1 - 0.522 = 0.478

If you want to learn more about probability, you can read:

https://brainly.com/question/251701