Answer the following below using binomial distribution.

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 five of the six M&M’s are orange. 

Compute the probability that four or five of the six M&M’s are orange. 

Compute the probability that at most five of the six M&M’s are orange. 

Compute the probability that at least five of the six M&M’s are orange.

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

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

Answer 1

P(5)= C(6,5).23^5(1-.23)^1 = .0030

P(4)+P(5) = C(6,4).23^4(1-.23)^2+C(6,5).23^5(1-.23)^1 = .0279

1- P(6) = 1- .23^6 = .9999

P(5)+P(6) = C(6,5).23^5(1-.23)^1 + .23^6 = .0031

Alternatively, use

=BINOM.DIST(5,6,0.23,0)

=BINOM.DIST(5,6,0.23,0)+BINOM.DIST(4,6,0.23,0)

=BINOM.DIST(5,6,0.23,1)

=1-BINOM.DIST(4,6,0.23,1)

The final number in parentheses is 1 for CDF and 0 for pmf.

np = 6 * .23 = 1.38

sqrt(np(1-p)) = sqrt(6(.23)(1-.23)) = 1.03