According to a recent survey 74 of all customers will return

According to a recent survey, 74% of all customers will return to the same grocery store. Suppose 12 customers are selected at random, what is the probabilty that:

A) Exactly five of the customers will return

B) All twelve will return

C) At least seven will return

D) At least one will return

E) How many customers would be expected to return to the same store?

Solution

Binomial Distribution

PMF of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
Where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial

a)
P( X = 5 ) = ( 12 5 ) * ( 0.74^5) * ( 1 - 0.74 )^7
= 0.0141

b)
P( X = 12 ) = ( 12 12 ) * ( 0.74^12) * ( 1 - 0.74 )^0
= 0.027

c)

P( X < 7) = P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 12 6 ) * 0.74^6 * ( 1- 0.74 ) ^6 + ( 12 5 ) * 0.74^5 * ( 1- 0.74 ) ^7 + ( 12 4 ) * 0.74^4 * ( 1- 0.74 ) ^8 + ( 12 3 ) * 0.74^3 * ( 1- 0.74 ) ^9 + ( 12 2 ) * 0.74^2 * ( 1- 0.74 ) ^10 + ( 12 1 ) * 0.74^1 * ( 1- 0.74 ) ^11 + ( 12 0 ) * 0.74^0 * ( 1- 0.74 ) ^12
= 0.0646

P( X > = 7 ) = 1 - P( X < 7) = 0.9354

d)
P( X < 1) = P(X=0)
= ( 12 0 ) * 0.74^0 * ( 1- 0.74 ) ^12
= 0
P( X > = 1 ) = 1 - P( X < 1) = 1

e)
Expected to return store = np = 12 * 0.74 = 8.88