Question

You sell two types of sandwiches. Suppose 60% of all customers will choose chicken, the rest will choose beef.

a) Let X=number of customers until the first chicken is sold. What is the mean and variance of X?

b) what is the average number of chicken sandwiches sold in 10 customers?

c) what is the probability of selling 2 chicken sandwiches to the next 10 customers?

Answer 1

first chicken = p

first beef second chicken = (1-p)p

first beef . third chicken = (1-p)^2p

nth chicken (1-p) ^(n-1) p

P(x=n) = (1-p) ^(n-1) p

so modeling this as a negative binomial

where r is the number of successes(chicken in our case)

n is the number of trails

p is the probabilty of chicken .6

so using standard result for negative binomial

with r=1

mean =1/(.6) = .6/.4 = 1.67

variance= (1-p)/(p)^2 = .4/(.6^2) =1.11

b)

average number of chicken sandwiched = E(x)= np = 10*.6 = 6

c)

p selling 2 chicken in 10 customer s

10c2 (.6^2) (.4^8) = 0.01061683