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 peanut M&M’s from an extra-large bag looking for a orange candy. (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 the first orange candy is the fourth M&M selected.

Compute the probability that the first orange candy is the fourth or fifth M&M selected.

Compute the probability that the first orange candy is among the first four M&M’s selected.

If every student in a large Statistics class selects peanut M&M’s at random until they get a orange candy, on average how many M&M’s will the students need to select? (Round your answer to two decimal places.) orange M&M’s

Answer 1

this is Geometric distribution with parameter p=0.23

1) probability that the first orange candy is the fourth M&M selected :

P(X=4)=

(1-p)x-1p=(1-0.23)^3*0.23 =

0.1050

2)

probability that the first orange candy is the fourth or fifth M&M selected :

P(3<=X<=4)=

(1-p)x1-1-(1-p)x2=

0.2414

3)

probability that the first orange candy is among the first four M&M’s selected :

P(X<=4)=

1-(1-p)x=

0.6485

4)

expected number =1/p=1/023 =4.35