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

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

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

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

If you repeatedly select random samples of five 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

Answer 1

here this is binomial with parameter n=5 and p=0.23

1)

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

P(X=4)=

(nCx)px(1−p)(n-x)    =

0.0108

2)

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

P(3<=X<=4)=

∑x=ab     (nCx)px(1−p)(n-x)    =

0.0829

3)

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

P(X<=4)=

∑x=0a     (nCx)px(1−p)(n-x)    =

0.9994

4)

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

P(X>=4)=1-P(X<=3)=

1-∑x=0x-1   (nCx)px(q)(n-x) =

0.0114

5)

expected to be blue =5*0.23 =1.15