Question 1200285: 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 all answers to 3 places after the decimal point, if necessary.)
(a) Compute the probability that a randomly selected peanut M&M is not brown.
P(not brown) =
(b) Compute the probability that a randomly selected peanut M&M is yellow or red.
P(yellow or red) =
(c) Compute the probability that three randomly selected peanut M&M’s are all green.
P(all green) =
(d) If you randomly select six peanut M&M’s, compute the probability that none of them are yellow.
P(none are yellow) =
(e) If you randomly select six peanut M&M’s, compute the probability that at least one of them is yellow.
P(at least one is yellow) =
Answer by math_tutor2020(3817)   (Show Source):
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Answers:
(a) 0.88
(b) 0.27
(c) 0.003
(d) 0.377
(e) 0.623
Explanation:
(a) 12% are brown, so 100%-12% = 88% are not brown.
This percentage converts to 0.88
(b) 15% are yellow and 12% are red.
That's a total of 15+12 = 27% which converts to 0.27
(c) 15% are green which converts to 0.15
Getting 3 green in a row has probability (0.15)^3 = 0.003375 = 0.003 when rounding to three decimal places.
(d) 15% are yellow, so 85% are not yellow.
Getting 6 non-yellow candies in a row has probability (0.85)^6 = 0.377149515625 which rounds to 0.377
(e) Subtract the previous result from 1 to get 1-0.377 = 0.623
This works because P(at least one yellow)+P(none are yellow) = 1
The events are complementary.
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