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 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.)

Answer 1

Answer:

Mean = 0.48

Step-by-step explanation:

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

Colour Brown Yellow Red   Blue  Orange Green 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)

$P(X=3)=0.006082\\P(X=2 or 3)= P(3)+P(2) = 0.006082+0.06691=0.0729\\P(x\leq 3) = 0.99979\\P(X\geq 4) = 0.000207$

E(x) = np = $4(0.12) = 0.48$