Question

The proportion of brown M&M’s in a milk chocolate packet is approximately 14% (Madison, 2013). Suppose a package of M&M’s typically contains 52 M&M’s.

A.) State the random variable.

B.) Argue that this is a binomial experiment

Find the probability that

C.) Six M&M’s are brown.

D.) Twenty-five M&M’s are brown.

E.) All of the M&M’s are brown.

F.) Would it be unusual for a package to have only brown M&M’s? If this were to happen, what would you think is the reason

Answer 1

Ans:

a)Let X be the number of brown M &M 's.

X has binomial distrbution with n=52 and p=0.14

P(x=k)=52C_k*0.14^k*(1-0.14)^52-k

b)X is binomial,as there are only two possible outcomes i.e. brown or not brown,also probability of being brown is fixed i.e. p=0.14 and trials are independent.

c)

P(x=6)=52C₆*0.14⁶*(1-0.14)⁴⁶=0.1487

d)

P(x=25)=52C₂₅*0.14²⁵*(1-0.14)²⁷=0.0000

e)

P(x=52)=0.14⁵²=0.0000

f)Yes,it is unusual for a package to have only brown M &M's,as probability of all being brown is less than 0.05