Question 1166393: 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. You randomly select six peanut M&M’s from an extra-large bag of the candies. (Round all probabilities below to four decimal places; i.e. your answer should look like 0.1234, not 0.1234444 or 12.34%.)
Compute the probability that exactly five of the six M&M’s are red.
Compute the probability that four or five of the six M&M’s are red.
Compute the probability that at most five of the six M&M’s are red.
Compute the probability that at least five of the six M&M’s are red.
If you repeatedly select random samples of six peanut M&M’s, on average how many do you expect to be red? (Round your answer to two decimal places.)
With what standard deviation? (Round your answer to two decimal places.)
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! p(red)=0.12
5 of 6 is 6(# ways it can happen)*0.12^5*0.88=0.0001
4 of 6 are red is 6C4*0.12^4*0.88^2=0.0024
so 4 or 5 are red has prob 0.0025
at most 5 is 1-all=1-.12^6=1.0000 (rounded)
at least 5 of the 6 are red are the same as exactly 5, since 6 is <0.0001
np is the mean and this would be 6*0.12=0.72 expected value for red.
sd: variance is np(1-p)=0.72*0.88=0.6336
sqrt(V)= sd=0.7960 or 0.80
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