7.3. Binomial Distribution and Binomial Experiments http://www.ck12.org
(
n
k)
=
n!
k!(n−k)!P(X= 3 ) = 220 ( 0. 5 )^12 = 0. 0537
(
12
3
)
=
12!
3!( 12 − 3 )!
(
12
3
)
= 220
c) At least 10 heads means getting 10 heads, 11 heads or 12 heads.
(
n
k)
=
n!
k!(n−k)!(
12
10
)
=
12!
10!( 12 − 10 )!
(
12
10
)
= 66
(
n
k)
=
n!
k!(n−k)!(
12
11
)
=
12!
11!( 12 − 11 )!
(
12
11
)
= 12
(
n
k)
=
n!
k!(n−k)!(
12
12
)
=
12!
12!( 12 − 12 )!
(
12
12
)
= 1
P(X= 10 ) = 66 ( 0. 5 )^12 ≈ 0. 0161
P(X= 11 ) = 12 ( 0. 5 )^12 ≈ 0. 0029
P(X= 12 ) = 1 ( 0. 5 )^12 ≈ 0. 0002
At least 10 heads: 0. 0161 + 0. 0029 + 0. 0002 ≈ 0. 0192
Example:
Since the closing of the two main industries, a small town has experienced a decrease in the population of teenagers.
According to the statistics of the local high school, approximately 7.6% of teens ages 14−19 are no longer registered.
Suppose you randomly chose four people who were enrolled at the high school before the population dropped.
a) What is the probability that none of the four are registered?
b) What is the probability that at least one is not registered?
c) Create a probability distribution table for this situation.
Solution:
a)
P(X=k) =(
n
k)
Pk( 1 −p)n−kP(X= 0 )
(
4
0
)
( 0. 076 )^0 ( 1 − 0. 076 )^4 = 0. 7289
b)
P(X≥ 1 ) = 1 −[P(X= 0 )]
P(X≥ 1 )≈ 1 − 0. 7289 = 0. 2711
c) Using technology: