Binomial Theorem
Suppose you flip a coin 20 times.
Binomial Distribution
Suppose you flip a coin 20 times.
How many heads would you expect to flip?
Binomial Distribution
Suppose you flip a coin 20 times.
How many heads would you expect to flip?
Remember that.
Binomial Distribution
Suppose you flip a coin 20 times.
How many heads would you expect to flip?
What is the probability of flipping exactly ten heads in twenty tries?
Binomial Distribution
Suppose you flip a coin 20 times.
How many heads would you expect to flip?
What is the probability of flipping exactly ten heads in twenty tries?
HHHHHHHHHHTTTTTTTTTT
20!
(0.5)10(0.5)10 ≈ 0.176
10!10!
Binomial Distribution
Let’s see how theory and reality compare.
Flip a coin 20 times and record the number of heads.
Binomial Distribution
Let’s see how theory and reality compare.
In theory we have:
Number of Heads
Probability of Heads
0
P (0H) =
20!
20
0
20!0! (0.5) (0.5)
1
P (1H) =
20!
19
1
19!1! (0.5) (0.5)
2
P (2H) =
20!
18
2
18!2! (0.5) (0.5)
3
4
5
6
7
8
9
10
11
...
≈ 9.5 × 10−7
≈ 1.9 × 10−5
≈ 0.00018
P (3H) ≈ 0.0011
P (4H) ≈ 0.0046
P (5H) ≈ 0.0148
P (6H) ≈ 0.0370
P (7H) ≈ 0.0739
P (8H) ≈ 0.1201
P (9H) ≈ 0.1602
P (10H) ≈ 0.1762
P (11H) ≈ 0.1602
...
Binomial Distribution
0.20
0.18
0.16
0.14
Probability
Let’s see how theory and reality compare.
In theory we have:
Heads Probability of Heads
0
P (0H) ≈ 9.5 × 10−7
1
P (1H) ≈ 1.9 × 10−5
2
P (2H) ≈ 0.00018
3
P (3H) ≈ 0.0011
4
P (4H) ≈ 0.0046
5
P (5H) ≈ 0.0148
6
P (6H) ≈ 0.0370
7
P (7H) ≈ 0.0739
8
P (8H) ≈ 0.1201
9
P (9H) ≈ 0.1602
10
P (10H) ≈ 0.1762
11
P (11H) ≈ 0.1602
...
...
0.12
0.10
0.08
0.06
0.04
0.02
0
1 2 3 4
5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Number of Heads
Binomial Distribution
0.20
0.18
0.16
0.14
Probability
Let’s see how theory and reality compare.
In theory we have:
Heads Probability of Heads
0
P (0H) ≈ 9.5 × 10−7
1
P (1H) ≈ 1.9 × 10−5
2
P (2H) ≈ 0.00018
3
P (3H) ≈ 0.0011
4
P (4H) ≈ 0.0046
5
P (5H) ≈ 0.0148
6
P (6H) ≈ 0.0370
7
P (7H) ≈ 0.0739
8
P (8H) ≈ 0.1201
9
P (9H) ≈ 0.1602
10
P (10H) ≈ 0.1762
11
P (11H) ≈ 0.1602
...
...
0.12
0.10
0.08
0.06
0.04
0.02
0
1 2 3 4
5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Number of Heads
Note that the most likely outcomes are those centered around the number we “expected” to flip.
Binomial Distribution
Now consider a familiar case as an example with asymmetric probabilities of winning or losing.
Binomial Distribution
The Trojans win about 60% of the time while the Bulldogs win about 40% of the time.
If the two teams play a five game series, what are the possible outcomes (wins and losses)
and their respective probabilities of occurring?
Binomial Distribution
The Trojans win about 60% of the time while the Bulldogs win about 40% of the time.
If the two teams play a five game series, what are the possible outcomes (wins and losses)
and their respective probabilities of occurring?
The Trojans could win either no games, 1 game, 2 games, 3 games, 4 games, or all 5 games.
These are all the possible outcomes and are contained in the expression (0.4 + 0.6)5.
Binomial Distribution
The Trojans win about 60% of the time while the Bulldogs win about 40% of the time.
If the two teams play a five game series, what are the possible outcomes (wins and losses)
and their respective probabilities of occurring?
(0.4 + 0.6)5 =
1
1
1
1
1
1
1
1
1
1
8
3
6
28
6
10
15
21
1
4
10
20
35
56
1
3
4
5
7
2
5
15
35
70
1
1
6
21
56
1
7
28
1
8
1
1 · (0.4)5 + 5 · (0.6)1(0.4)4 + 10 · (0.6)2(0.4)3 + 10 · (0.6)3(0.4)2 + 5 · (0.6)4(0.4)1 + 1 · (0.6)5
P (0 Wins) + P (1 Win)
+
P (2 Wins)
+
P (3 Wins)
+
P (4 Wins)
+ P (5 Wins)
Binomial Distribution
The Trojans win about 60% of the time while the Bulldogs win about 40% of the time.
If the two teams play a five game series, what are the possible outcomes (wins and losses)
and their respective probabilities of occurring?
(0.4 + 0.6)5 =
1
1
1
1
1
1
1
1
1
8
3
5
7
6
15
1
4
10
20
35
56
1
3
10
21
28
2
4
6
1
1
5
15
35
70
1
6
21
56
1
7
28
1
8
1
1 · (0.4)5 + 5 · (0.6)1(0.4)4 + 10 · (0.6)2(0.4)3 + 10 · (0.6)3(0.4)2 + 5 · (0.6)4(0.4)1 + 1 · (0.6)5
5!
0!5!
5!
1!4!
5!
2!3!
5!
3!2!
5!
4!1!
5!
5!0!
Binomial Distribution
The Trojans win about 60% of the time while the Bulldogs win about 40% of the time.
If the two teams play a five game series, what are the possible outcomes (wins and losses)
and their respective probabilities of occurring?
(0.4 + 0.6)5 =
1
1
1
1
1
1
1
1
1
8
3
5
7
6
15
1
4
10
20
35
56
1
3
10
21
28
2
4
6
1
1
5
15
35
70
1
6
21
56
1
7
28
1
8
1
1 · (0.4)5 + 5 · (0.6)1(0.4)4 + 10 · (0.6)2(0.4)3 + 10 · (0.6)3(0.4)2 + 5 · (0.6)4(0.4)1 + 1 · (0.6)5
5!
0!5!
5!
1!4!
5!
2!3!
5!
3!2!
5!
4!1!
5!
5!0!
Binomial Distribution
The probability distribution of the Trojans’ possible wins (in a five game series with the Bulldogs) is
shown in the table and histogram below.
X is the number of times the Trojans win,
P (X) is the corresponding probability of that number of wins.
X
0
1
2
3
4
5
P (X = x)
P (X = 0) ≈ 0.01
P (X = 1) ≈ 0.08
P (X = 2) ≈ 0.23
P (X = 3) ≈ 0.35
P (X = 4) ≈ 0.26
P (X = 5) ≈ 0.08
Probability Distribution Function (table)
Binomial Distribution
The probability distribution of the Trojans’ possible wins (in a five game series with the Bulldogs) is
shown in the table and histogram below.
X is the number of times the Trojans win,
P (X) is the corresponding probability of that number of wins.
X
0
1
2
3
4
5
P (X = x)
P (X = 0) ≈ 0.01
P (X = 1) ≈ 0.08
P (X = 2) ≈ 0.23
P (X = 3) ≈ 0.35
P (X = 4) ≈ 0.26
P (X = 5) ≈ 0.08
0.3
0.2
0.1
1
2
3
4
5
Probability Distribution Function (table and graph)
Binomial Distribution
The probability distribution of the Trojans’ possible wins (in a five game series with the Bulldogs) is
shown in the table and histogram below.
Note that the most likely outcome is the one we might “expect” since 60% of 5 is 3.
X
0
1
2
3
4
5
P (X = x)
P (X = 0) ≈ 0.01
P (X = 1) ≈ 0.08
P (X = 2) ≈ 0.23
P (X = 3) ≈ 0.35
P (X = 4) ≈ 0.26
P (X = 5) ≈ 0.08
0.3
0.2
0.1
1
2
3
4
5
In fact the results of our two examples give us some insight into the way the probabilities of events like
these are distributed.
The mean, µ, of a binomial distribution is given by µ = np where n is the number of trials (think
games played) and p is the probability of success in any one trial (winning a game).
In fact the results of our two examples give us some insight into the way the probabilities of events like
these are distributed.
The mean, µ, of a binomial distribution is given by µ = np where n is the number of trials (think
games played) and p is the probability of success in any one trial (winning a game).
The spread
p of the number of successes near the mean is measured by the standard deviation and given
by: σ = np(1 − p).
In fact the results of our two examples give us some insight into the way the probabilities of events like
these are distributed.
In general any event that involves only two outcomes – a win or loss (a yes or no, H or T) is said to be
distributed binomially. If the probability of any one success is p and the random variable X measures
the number of successes over n tries then we write
X ∼ B(n, p)
µ = np
p
σ = np(1 − p)
So what does this mean to the team playing these games?
The long term outcome from repeated five game series, or expected value, is the sum of the products
of the probability and the number of games won:
X
0
1
2
3
4
5
P (X = x)
x · P (X = x)
P (X = 0) ≈ 0.01
0
P (X = 1) ≈ 0.08
0.08
P (X = 2) ≈ 0.23
0.46
P (X = 3) ≈ 0.35
1.05
P (X = 4) ≈ 0.26
1.04
P (X = 5) ≈ 0.08
0.4
0.3
0.2
0.1
1
X
2
3
4
5
The sum
x · P (X = x) ≈ 3 so on average we expect the Trojans to win three games out of a five
game series with the Bulldogs (or any other team they have a 60-40 win/loss ratio with).
The long term outcome from repeated five game series, or expected value, is the sum of the products
of the probability and the number of games won:
X
0
1
2
3
4
5
P (X = x)
x · P (X = x)
P (X = 0) ≈ 0.01
0
P (X = 1) ≈ 0.08
0.08
P (X = 2) ≈ 0.23
0.46
P (X = 3) ≈ 0.35
1.05
P (X = 4) ≈ 0.26
1.04
P (X = 5) ≈ 0.08
0.4
0.3
0.2
0.1
1
X
2
3
4
5
The sum
x · P (X = x) ≈ 3 so on average we expect the Trojans to win three games out of a five
game series with the Bulldogs (or any other team they have a 60-40 win/loss ratio with).
Need more?
X
The sum
x · P (X = x) ≈ 3 so on average we expect the Trojans to win three games out of a five
game series with the Bulldogs (or any other team they have a 60-40 win/loss ratio with).
X
0
1
2
3
4
5
P (X = x)
x · P (X = x)
P (X = 0) ≈ 0.01
0
P (X = 1) ≈ 0.08
0.08
P (X = 2) ≈ 0.23
0.46
P (X = 3) ≈ 0.35
1.05
P (X = 4) ≈ 0.26
1.04
P (X = 5) ≈ 0.08
0.4
Imagine playing 100 of these five game series.
X
The sum
x · P (X = x) ≈ 3 so on average we expect the Trojans to win three games out of a five
game series with the Bulldogs (or any other team they have a 60-40 win/loss ratio with).
X
0
1
2
3
4
5
P (X = x)
x · P (X = x)
P (X = 0) ≈ 0.01
0
P (X = 1) ≈ 0.08
0.08
P (X = 2) ≈ 0.23
0.46
P (X = 3) ≈ 0.35
1.05
P (X = 4) ≈ 0.26
1.04
P (X = 5) ≈ 0.08
0.4
Imagine playing 100 of these five game series.
According to the probability table your team will have one series (P (X = 0) = 0.01) where it
wins 0 games;
X
The sum
x · P (X = x) ≈ 3.03 so on average we expect the Trojans to win three games out of a
five game series with the Bulldogs (or any other team they have a 60-40 win/loss ratio with).
X
0
1
2
3
4
5
P (X = x)
x · P (X = x)
P (X = 0) ≈ 0.01
0
P (X = 1) ≈ 0.08
0.08
P (X = 2) ≈ 0.23
0.46
P (X = 3) ≈ 0.35
1.05
P (X = 4) ≈ 0.26
1.04
P (X = 5) ≈ 0.08
0.4
Imagine playing 100 of these five game series.
According to the probability table your team will have one series (P (X = 0) = 0.01) where it
wins 0 games;
It will have 8 series (in the 100) where it wins one game;
X
The sum
x · P (X = x) ≈ 3.03 so on average we expect the Trojans to win three games out of a
five game series with the Bulldogs (or any other team they have a 60-40 win/loss ratio with).
X
0
1
2
3
4
5
P (X = x)
x · P (X = x)
P (X = 0) ≈ 0.01
0
P (X = 1) ≈ 0.08
0.08
P (X = 2) ≈ 0.23
0.46
P (X = 3) ≈ 0.35
1.05
P (X = 4) ≈ 0.26
1.04
P (X = 5) ≈ 0.08
0.4
Imagine playing 100 of these five game series.
According to the probability table your team will have one series (P (X = 0) = 0.01) where
it wins 0 games;
It will have 8 series (in the 100) where it wins one game;
It will have 23 series (in the 100) where it wins two games;
It will have 35 series (in the 100) where it wins three games;
It will have 26 series (in the 100) where it wins four games;
It will have 8 series (in the 100) where it wins five games;
X
The sum
x · P (X = x) ≈ 3.03 so on average we expect the Trojans to win three games out of a
five game series with the Bulldogs (or any other team they have a 60-40 win/loss ratio with).
X
0
1
2
3
4
5
P (X = x)
x · P (X = x)
P (X = 0) ≈ 0.01
0
P (X = 1) ≈ 0.08
0.08
P (X = 2) ≈ 0.23
0.46
P (X = 3) ≈ 0.35
1.05
P (X = 4) ≈ 0.26
1.04
P (X = 5) ≈ 0.08
0.4
Imagine playing 100 of these five game series.
According to the probability table your team will have one series (P (X = 0) = 0.01) where
it wins 0 games;
It will have 8 series (in the 100) where it wins one game;
It will have 23 series (in the 100) where it wins two games;
It will have 35 series (in the 100) where it wins three games;
It will have 26 series (in the 100) where it wins four games;
It will have 8 series (in the 100) where it wins five games;
So out out of 100 five game series, or 500 games, you win a total of 0 + 8(1) + 23(2) + 35(3) +
26(4) + 8(5) = 303 games for an average of 303/500 or about 3 out of every five games.
Calculator for Binomial Distributions
Recall the flipping heads activity.
If X is the number of heads flipped, find the probability of flipping at most 12 heads: P (X ≤ 12).
Calculator for Binomial Distributions
Recall the flipping heads activity.
If X is the number of heads flipped, find the probability of flipping at most 12 heads: P (X ≤ 12).
P (X = x)
9.5 × 10−7
1.9 × 10−5
0.00018
0.0011
0.0046
0.0148
0.0370
0.0739
0.1201
0.1602
0.1762
0.1602
0.1201
0.0739
0.0370
0.0148
0.0046
0.0011
0.00018
1.9 × 10−5
9.5 × 10−7
0.20
0.18
0.16
0.14
Probability
X
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
0.12
0.10
0.08
0.06
0.04
0.02
0
1 2 3 4
5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Number of Heads
P (X ≤ 12) = P (X = 0) + P (X = 1) + P (X = 2) + · · · + P (X = 12)
= binomcdf(20, 0.5, 12) ≈ 0.8684
Calculator for Binomial Distributions
Recall the flipping heads activity.
If X is the number of heads flipped, find the probability of flipping at least 12 heads: P (X ≥ 12).
Calculator for Binomial Distributions
Recall the flipping heads activity.
If X is the number of heads flipped, find the probability of flipping at least 12 heads: P (X ≥ 12).
P (X = x)
9.5 × 10−7
1.9 × 10−5
0.00018
0.0011
0.0046
0.0148
0.0370
0.0739
0.1201
0.1602
0.1762
0.1602
0.1201
0.0739
0.0370
0.0148
0.0046
0.0011
0.00018
1.9 × 10−5
9.5 × 10−7
0.20
0.18
0.16
0.14
Probability
X
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
0.12
0.10
0.08
0.06
0.04
0.02
0
1 2 3 4
5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Number of Heads
P (X ≥ 12) = P (X = 12) + P (X = 13) + P (X = 14) + · · · + P (X = 20)
= 1−binomcdf(20, 0.5, 11) ≈ 0.2517
Calculator for Binomial Distributions
Recall the flipping heads activity.
If X is the number of heads flipped, find the probability of flipping between 8 and 14 heads:
P (8 ≤ X ≤ 14).
Calculator for Binomial Distributions
Recall the flipping heads activity.
If X is the number of heads flipped, find the probability of flipping between 8 and 14 heads:
P (8 ≤ X ≤ 14).
P (X = x)
9.5 × 10−7
1.9 × 10−5
0.00018
0.0011
0.0046
0.0148
0.0370
0.0739
0.1201
0.1602
0.1762
0.1602
0.1201
0.0739
0.0370
0.0148
0.0046
0.0011
0.00018
1.9 × 10−5
9.5 × 10−7
0.20
0.18
0.16
0.14
Probability
X
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
0.12
0.10
0.08
0.06
0.04
0.02
0
1 2 3 4
5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Number of Heads
P (X ≤ 12) = P (X = 8) + P (X = 9) + P (X = 10) + · · · + P (X = 14)
= binomcdf(20, 0.5, 14) – binomcdf(20, 0.5, 7) ≈ 0.8477