SOLUTION: Please help me solve this question. The Web site for M&M™ candies claimed that 24% of plain M&M candies are blue, 20% are orange, 16% green, 14% yellow, and 13% each red and

Question 1171365: Please help me solve this question.

The Web site for M&M™ candies claimed that 24% of plain M&M candies are blue, 20% are orange, 16% green, 14% yellow, and 13% each red and brown.
(a) Pick one M&M at random from a package.
Describe the sample space.
What is the probability that the one you pick is blue or red?
What is the probability that the one you pick is not green?
(b) You pick three M&M’s in a row randomly from three separate packages.
Describe the sample space for the outcomes of your three choices.
What is the probability that every M&M is blue?
What is the probability that the third M&M is red?
What is the probability that at least one is blue?

Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
for every 100, 24 are blue, 20 are orange, 16 green, 14 yellow, 13 red, and 13 brown.
Sample space is what is picked {blue, orange, green, yellow, red, brown}
prob pick blue or red is 37/100, the sum of both minus getting both at the same time, which is 0.
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not green is 1-p(green)=0.84
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now the sample space is {blue 1, blue 2, blue 3, orange 1, orange 2, orange 3...., brown 3}
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every one blue is (24/100)^3=0.013824
the third is red is 0.13, the same for any given package
at least one is blue is 1- none is blue=1-.76^3=0.561.
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