Question

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 four 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 two of the four M&M’s are blue.



Compute the probability that two or three of the four M&M’s are blue.



Compute the probability that at most two of the four M&M’s are blue.



Compute the probability that at least two of the four M&M’s are blue.



If you repeatedly select random samples of four peanut M&M’s, on average how many do you expect to be blue? (Round your answer to two decimal places.)

blue M&M’s

With what standard deviation? (Round your answer to two decimal places.)

blue M&M’s

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Answer 1

1)

here this is binomial with parameter n=4 and p=0.23

probability that exactly two of the four M&M’s are blue :

P(X=2)=

(nCx)px(1−p)(n-x)    =

0.1882

2)

probability that two or three of the four M&M’s are blue:

P(2<=X<=3)=

∑x=ab     (nCx)px(1−p)(n-x)    =

0.2257

3)

probability that at most two of the four M&M’s are blue :

P(X<=2)=

∑x=0a     (nCx)px(1−p)(n-x)    =

0.9597

4)

probability that at least two of the four M&M’s are blue :

P(X>=2)=1-P(X<=1)=

1-∑x=0x-1   (nCx)px(q)(n-x) =

0.2285

5)

mean E(x)=μ=np=0.92

6)

standard deviation σ=√(np(1-p))=0.84