Question 1165451: 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]
Compute the probability that a randomly selected peanut M&M is not orange.
Compute the probability that a randomly selected peanut M&M is brown or orange.
Compute the probability that two randomly selected peanut M&M’s are both green.
If you randomly select six peanut M&M’s, compute that probability that none of them are red.
If you randomly select six peanut M&M’s, compute that probability that at least one of them is red.
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Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! a. 23% are orange so 0.77 probability not orange.
b. brown or orange is 0.35 probability
c. both green would be (15/100)*14/99=0.0212
d. none is red when 6 chosen
denominator is 100C6, number of ways to choose 6 out of a 100.
88 out of 100 are not red
so 88C6 are ways to pick one of those.
it is 88C6/100C6=0.4546
way 2
88/100 probability first is not red*87/99*86/98*85/97*84/96*83/95=0.4546
e. at least one of them is red is the complement, since it is a dichotomy, and that probability is 0.5454.
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