A salesman makes visits to customers Based on his past histo

A salesman makes visits to customers. Based on his past history, the probability he makes a sale on any visit is 0.15. It is reasonable to assume that customers\' decisions are independent of one another. If he makes 10 visits in a day, what is the chance he makes at least five sales?

Answer 0.0099 This answer is correct. The random variable X, the number of sales in a day, has the binomial distribution with n = 10 and p = 0.15. This answer was obtained using a TI calculator.

How do you get this answer? please show work

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) = P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)   
= ( 10 4 ) * 0.15^4 * ( 1- 0.15 ) ^6 + ( 10 3 ) * 0.15^3 * ( 1- 0.15 ) ^7 +
( 10 2 ) * 0.15^2 * ( 1- 0.15 ) ^8 + ( 10 1 ) * 0.15^1 * ( 1- 0.15 ) ^9 +
   ( 10 0 ) * 0.15^0 * ( 1- 0.15 ) ^10
= 0.9901

P( X > = 5 ) = 1 - P( X < 5) = 0.0099