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S160 #9
The Binomial Distribution, Part 1
JC Wang
February 16, 2016
The Binomial Distribution
Computing Binomial Probabilities
Outline
1
The Binomial Distribution
Binomial Random Variables
2
Computing Binomial Probabilities
Using Formula
JC Wang (WMU)
S160 #9
S160, Lecture 9
2 / 11
The Binomial Distribution
Computing Binomial Probabilities
Binomial Process and Binomial Random Variable
A sequence of (fixed) n observations is called a binomial process if
each observation results in exactly one of two possible outcomes
(conveniently called success and failure)
P(success) = p and P(failure) = q = 1 − p for all observations
observations are independent
X = total number of successes among the n observations is a
binomial random variable with parameters n and p and is denoted
X ∼ binomial(n, p)
JC Wang (WMU)
S160 #9
S160, Lecture 9
3 / 11
The Binomial Distribution
Computing Binomial Probabilities
Binomial Process and Binomial Random Variable
A sequence of (fixed) n observations is called a binomial process if
each observation results in exactly one of two possible outcomes
(conveniently called success and failure)
P(success) = p and P(failure) = q = 1 − p for all observations
observations are independent
X = total number of successes among the n observations is a
binomial random variable with parameters n and p and is denoted
X ∼ binomial(n, p)
JC Wang (WMU)
S160 #9
S160, Lecture 9
3 / 11
The Binomial Distribution
Computing Binomial Probabilities
Binomial Process and Binomial Random Variable
A sequence of (fixed) n observations is called a binomial process if
each observation results in exactly one of two possible outcomes
(conveniently called success and failure)
P(success) = p and P(failure) = q = 1 − p for all observations
observations are independent
X = total number of successes among the n observations is a
binomial random variable with parameters n and p and is denoted
X ∼ binomial(n, p)
JC Wang (WMU)
S160 #9
S160, Lecture 9
3 / 11
The Binomial Distribution
Computing Binomial Probabilities
Examples of Binomial Random Variables
1
A 5-question multiple-choice quiz has 5 choices on each question.
X = number of correct answers (success = correct) in the quiz by
guessing all. Then X ∼ binomial(n = 5, p = 15 = 0.2).
2
Past experience: 40% phone respondents agree to be interviewed
(success = a respondent agrees to be interviewed) for market
research survey. Of 50 reached by Reliable Research, X
respondents agree to be interviewed. Then
X ∼ binomial(n = 50, p = 0.40).
3
Historical data shows that 20% of TV buyers at TV World purchase
extended warranty (success = a buyer purchases extended
warranty). X extended warranties were sold along with the 300 TV
sets sold last quarter. Then X ∼ binomial(n = 300, p = 0.20).
JC Wang (WMU)
S160 #9
S160, Lecture 9
4 / 11
The Binomial Distribution
Computing Binomial Probabilities
Examples of Binomial Random Variables
1
A 5-question multiple-choice quiz has 5 choices on each question.
X = number of correct answers (success = correct) in the quiz by
guessing all. Then X ∼ binomial(n = 5, p = 15 = 0.2).
2
Past experience: 40% phone respondents agree to be interviewed
(success = a respondent agrees to be interviewed) for market
research survey. Of 50 reached by Reliable Research, X
respondents agree to be interviewed. Then
X ∼ binomial(n = 50, p = 0.40).
3
Historical data shows that 20% of TV buyers at TV World purchase
extended warranty (success = a buyer purchases extended
warranty). X extended warranties were sold along with the 300 TV
sets sold last quarter. Then X ∼ binomial(n = 300, p = 0.20).
JC Wang (WMU)
S160 #9
S160, Lecture 9
4 / 11
The Binomial Distribution
Computing Binomial Probabilities
Examples of Binomial Random Variables
1
A 5-question multiple-choice quiz has 5 choices on each question.
X = number of correct answers (success = correct) in the quiz by
guessing all. Then X ∼ binomial(n = 5, p = 15 = 0.2).
2
Past experience: 40% phone respondents agree to be interviewed
(success = a respondent agrees to be interviewed) for market
research survey. Of 50 reached by Reliable Research, X
respondents agree to be interviewed. Then
X ∼ binomial(n = 50, p = 0.40).
3
Historical data shows that 20% of TV buyers at TV World purchase
extended warranty (success = a buyer purchases extended
warranty). X extended warranties were sold along with the 300 TV
sets sold last quarter. Then X ∼ binomial(n = 300, p = 0.20).
JC Wang (WMU)
S160 #9
S160, Lecture 9
4 / 11
The Binomial Distribution
Computing Binomial Probabilities
Examples of Binomial Random Variables
1
A 5-question multiple-choice quiz has 5 choices on each question.
X = number of correct answers (success = correct) in the quiz by
guessing all. Then X ∼ binomial(n = 5, p = 15 = 0.2).
2
Past experience: 40% phone respondents agree to be interviewed
(success = a respondent agrees to be interviewed) for market
research survey. Of 50 reached by Reliable Research, X
respondents agree to be interviewed. Then
X ∼ binomial(n = 50, p = 0.40).
3
Historical data shows that 20% of TV buyers at TV World purchase
extended warranty (success = a buyer purchases extended
warranty). X extended warranties were sold along with the 300 TV
sets sold last quarter. Then X ∼ binomial(n = 300, p = 0.20).
JC Wang (WMU)
S160 #9
S160, Lecture 9
4 / 11
The Binomial Distribution
Computing Binomial Probabilities
iClicker Question 9.1
The probability that a defective item is observed at a production line is
0.02. A quality engineer, working at the production line, inspect an
item. What is the chance that the item is found to be non-defective?
A. 0.02
B. 1
C. 0.98
D. −0.02
E. none of the previous
JC Wang (WMU)
S160 #9
S160, Lecture 9
5 / 11
The Binomial Distribution
Computing Binomial Probabilities
Outline
1
The Binomial Distribution
Binomial Random Variables
2
Computing Binomial Probabilities
Using Formula
JC Wang (WMU)
S160 #9
S160, Lecture 9
6 / 11
The Binomial Distribution
Computing Binomial Probabilities
Using Formula to Compute Binomial Probability
X ∼ binomial(n, p)
P(X = j) =
n!
pj q n−j , j = 0, 1, 2, . . . , n.
j!(n − j)!
where n! = n × (n − 1) × · · · × 2 × 1 and 0! = 1.
Multiple-choice quiz: X ∼ binomial(5, 0.2), for example
P(X = 2) =
=
=
JC Wang (WMU)
5!
5!
.22 .85−2 =
.22 .83
2!(5 − 2)!
2! × 3!
5×4×3×2×1
.22 .83
(2 × 1) × (3 × 2 × 1)
5×4 2 3
.2 .8 = .2048
2×1
S160 #9
S160, Lecture 9
7 / 11
The Binomial Distribution
Computing Binomial Probabilities
TV+More Example, continued
Recall that historical data shows that 20% (i.e., p = 0.2) of TV buyers
at TV World purchase extended warranty. If (n =) 10 TV sets were sold
in one day, what is the probability that (j =) 3 extended warranties
were sold?
Now, X , the number of extended waranties sold along with 10 TV sets
has X ∼ binomial(10, .2) distribution and hence
10!
(0.2)3 (1 − 0.2)10−3
3!(10 − 3)!
10 × 9 × 8 × 7!
=
(0.2)3 (0.8)7
3 × 2 × 1 × 7!
= 0.2013.
P(X = 3) =
JC Wang (WMU)
S160 #9
S160, Lecture 9
8 / 11
The Binomial Distribution
Computing Binomial Probabilities
Using Formula
olympics swimmer example
A swimmer competes in three events in the Summer Olympics. The
swimmer’s winning/losing one event is independent of her result in any
other event. If the probability of winning any one event is 0.45, what is
the chance that she wins two or three events?
X ∼ binomial(3, 0.45)
3!
3!
.452 .551 +
.453 .550 = .334125 + .091125 = .42525.
2!1!
3!0!
|
{z
} |
{z
}
P(X = 2)
JC Wang (WMU)
P(X = 3)
S160 #9
S160, Lecture 9
9 / 11
The Binomial Distribution
Computing Binomial Probabilities
The ‘Language’ of Probability
Note first that X , the number of successes, can only assume
values 0, 1, ..., n.
‘only 2’ or ‘exactly 2’: P(X = 2)
‘at most 3’ or ‘no more than 3’ or ‘3 or less’: P(X ≤ 3) = P(X =
0, 1, 2, or 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3).
‘at least 5’ or ‘no less than 5’ or ‘5 or more’ if n = 10:
P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X =
8) + P(X = 9) + P(X = 10)
‘at least 8’ or ‘no less than 8’ or ‘8 or more’ if n = 10:
P(X ≥ 8) = P(X = 8) + P(X = 9) + P(X = 10)
etc.
JC Wang (WMU)
S160 #9
S160, Lecture 9
10 / 11
The Binomial Distribution
Computing Binomial Probabilities
The ‘Language’ of Probability
Note first that X , the number of successes, can only assume
values 0, 1, ..., n.
‘only 2’ or ‘exactly 2’: P(X = 2)
‘at most 3’ or ‘no more than 3’ or ‘3 or less’: P(X ≤ 3) = P(X =
0, 1, 2, or 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3).
‘at least 5’ or ‘no less than 5’ or ‘5 or more’ if n = 10:
P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X =
8) + P(X = 9) + P(X = 10)
‘at least 8’ or ‘no less than 8’ or ‘8 or more’ if n = 10:
P(X ≥ 8) = P(X = 8) + P(X = 9) + P(X = 10)
etc.
JC Wang (WMU)
S160 #9
S160, Lecture 9
10 / 11
The Binomial Distribution
Computing Binomial Probabilities
The ‘Language’ of Probability
Note first that X , the number of successes, can only assume
values 0, 1, ..., n.
‘only 2’ or ‘exactly 2’: P(X = 2)
‘at most 3’ or ‘no more than 3’ or ‘3 or less’: P(X ≤ 3) = P(X =
0, 1, 2, or 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3).
‘at least 5’ or ‘no less than 5’ or ‘5 or more’ if n = 10:
P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X =
8) + P(X = 9) + P(X = 10)
‘at least 8’ or ‘no less than 8’ or ‘8 or more’ if n = 10:
P(X ≥ 8) = P(X = 8) + P(X = 9) + P(X = 10)
etc.
JC Wang (WMU)
S160 #9
S160, Lecture 9
10 / 11
The Binomial Distribution
Computing Binomial Probabilities
The ‘Language’ of Probability
Note first that X , the number of successes, can only assume
values 0, 1, ..., n.
‘only 2’ or ‘exactly 2’: P(X = 2)
‘at most 3’ or ‘no more than 3’ or ‘3 or less’: P(X ≤ 3) = P(X =
0, 1, 2, or 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3).
‘at least 5’ or ‘no less than 5’ or ‘5 or more’ if n = 10:
P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X =
8) + P(X = 9) + P(X = 10)
‘at least 8’ or ‘no less than 8’ or ‘8 or more’ if n = 10:
P(X ≥ 8) = P(X = 8) + P(X = 9) + P(X = 10)
etc.
JC Wang (WMU)
S160 #9
S160, Lecture 9
10 / 11
The Binomial Distribution
Computing Binomial Probabilities
The ‘Language’ of Probability
Note first that X , the number of successes, can only assume
values 0, 1, ..., n.
‘only 2’ or ‘exactly 2’: P(X = 2)
‘at most 3’ or ‘no more than 3’ or ‘3 or less’: P(X ≤ 3) = P(X =
0, 1, 2, or 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3).
‘at least 5’ or ‘no less than 5’ or ‘5 or more’ if n = 10:
P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X =
8) + P(X = 9) + P(X = 10)
‘at least 8’ or ‘no less than 8’ or ‘8 or more’ if n = 10:
P(X ≥ 8) = P(X = 8) + P(X = 9) + P(X = 10)
etc.
JC Wang (WMU)
S160 #9
S160, Lecture 9
10 / 11
The Binomial Distribution
Computing Binomial Probabilities
The ‘Language’ of Probability
Note first that X , the number of successes, can only assume
values 0, 1, ..., n.
‘only 2’ or ‘exactly 2’: P(X = 2)
‘at most 3’ or ‘no more than 3’ or ‘3 or less’: P(X ≤ 3) = P(X =
0, 1, 2, or 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3).
‘at least 5’ or ‘no less than 5’ or ‘5 or more’ if n = 10:
P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X =
8) + P(X = 9) + P(X = 10)
‘at least 8’ or ‘no less than 8’ or ‘8 or more’ if n = 10:
P(X ≥ 8) = P(X = 8) + P(X = 9) + P(X = 10)
etc.
JC Wang (WMU)
S160 #9
S160, Lecture 9
10 / 11
The Binomial Distribution
Computing Binomial Probabilities
The ‘Language’ of Probability
Note first that X , the number of successes, can only assume
values 0, 1, ..., n.
‘only 2’ or ‘exactly 2’: P(X = 2)
‘at most 3’ or ‘no more than 3’ or ‘3 or less’: P(X ≤ 3) = P(X =
0, 1, 2, or 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3).
‘at least 5’ or ‘no less than 5’ or ‘5 or more’ if n = 10:
P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X =
8) + P(X = 9) + P(X = 10)
‘at least 8’ or ‘no less than 8’ or ‘8 or more’ if n = 10:
P(X ≥ 8) = P(X = 8) + P(X = 9) + P(X = 10)
etc.
JC Wang (WMU)
S160 #9
S160, Lecture 9
10 / 11
The Binomial Distribution
Computing Binomial Probabilities
iClicker Question 9.2
The probability that a defective item is observed at a production line is
0.02. A quality engineer, working at the production line, inspect the
next 4 items. What is the set of possible number of defectives?
A. {1,2,3,4}
B. {0,1,2,3,4}
C. {1,2}
D. {3,4}
E. none of the previous
JC Wang (WMU)
S160 #9
S160, Lecture 9
11 / 11