http://www.ck12.org Chapter 4. Discrete Probability Distribution
p(x) =
(
n
x
)
px( 1 −p)n−x
Substituting we have:p(x) =
( 3
x
)
(. 25 )x( 1 −. 25 )^3 −x
a. Forx=2,
p(x) =
(
3
2
)
(. 25 )x( 1 −. 25 )^3 −^2
= ( 3 )(. 25 )^2 ( 1 − 25 )^1
= 0. 14
The probability is 14% that exactly two out of three randomly selected calls will last longer than 3.8 minutes.
b. Here,x=0. We use the binomial probability formula,
p(x= 0 ) =
(
3
0
)
(. 25 )^0 ( 1 −. 25 )^3 −^0
=
3!
0!( 3 − 0 )!
(. 25 )^0 (. 75 )^3
= 0. 422
The probability is 42.2% that none of the three randomly selected calls will last longer than 3.8 minutes.
Example:
A car dealer knows that from past experience he can make a sale to 20% of the customers that he interacts with.
What is the probability that, in five randomly selected interactions, he will make a sale to
a. Exactly three customers?
b. At most one customer?
c. At least one customer?
d. Determine the probability distribution for the number of sales.
Solution:
The success here is making a sale to the customer. The probability that the seller makes a sale to any customer is
p=20%= 0 .2. The number of trials isn=5. The binomial probability formula for our case is
p(x) =
(
5
x
)
(. 25 )x(. 8 )^5 −x
a. Here we want the probability of exactly 3 sales,x=3:
p(x) =
(
5
3
)
(. 2 )^3 (. 8 )^5 −x= 0. 051
This means that the probability that the sales person makes exactly three sales in five attempts is 5.1%.