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.

compute the probability that exactly three of the four m& m’s are brown

compute the probability that two or three of the four m& m’s are brown.

compute the probability that at most three of the four m& m’s are brown.

compute the probability that at least three of the four m& m’s are brown.

if you repeatedly select random samples of four peanut m& m’s, on average how many do you expect to be brown? (in other words: what is the mean number of brown m& m’s in all samples of four m& m’s? ) round your answer to two decimal places.

with what standard deviation? (round your answer to two decimal places.)

Mean = 0.48

Step-by-step explanation:

Given that according to the Masterfoods, we have M&M manufactures different colours of peanuts.

ColourBrownYellowRed   Blue  OrangeGreen Total

Percent        12   15              12   23          23            15       100

We can sort this as brown 12% and non brown 88%

X no of brown in the selection of 4 peanuts is binomial since there are two outcomes and each is independent.

P(X=r) = 4Cr (0.12)^r (0.88)^(n-r)

E(x) = np =

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°

there are 70 more legs than eyes at the dog show (196 – 126 = 70). dogs have two more legs than eyes, and humans have the same number of legs as eyes. thus 70 is twice the number of dogs at the show, so the number of dogs is half of 70, which is 35. to get the number of humans, subtract 35 from 63: 63 – 35 = 28.

step-by-step explanation: