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 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 two of the six M&M’s are brown.

Compute the probability that two or three of the six M&M’s are brown.

Compute the probability that at most two of the six M&M’s are brown.

Compute the probability that at least two of the six M&M’s are brown.

Answer 1

here this is binomial with parameter n=6 and p=0.12

1)

probability =

P(X=2)=

(nCx)px(1−p)(n-x)    =

0.1295

2)

probability =

P(2<=X<=3)=

∑x=x1x2     (nCx)px(1−p)(n-x)    =

0.1531

3)

probability =

P(X<=2)=

∑x=0x     (nCx)px(1−p)(n-x)    =

0.9739

4)

probability =

P(X>=2)=

1-P(X<=1)=

1-∑x=0x-1     (nCx)px(1−p)(n-x)    =

0.1556