Building a

Binomial Table

The table is a tool for building a binomial distribution.

Given the number of binomial trials, n, and the probability of getting a success in one of those trials, p, we use the table to calculate all total probabilities for X = 0, 1, ..., n.

prob(X) = \dfrac{n!}{X!(n-X)!} p^X (1-p)^{n-X}
prob(X)=n!X!(n−X)!pX(1−p)n−Xprob(X) = \dfrac{n!}{X!(n-X)!} p^X (1-p)^{n-X}
A = {_n}C_X = \dfrac{n!}{X!(n-X)!}
A=nCX=n!X!(n−X)!A = {_n}C_X = \dfrac{n!}{X!(n-X)!}

To form the table, we break the formula for the total probability into two parts: Part A and Part B.

Part A is simply a combination that gives the number of ways we can get X successes out of n trials.

Part B is the probability of getting X successes from n trials in just one of those ways.

B = p^X (1-p)^{n-X}
B=pX(1−p)n−XB = p^X (1-p)^{n-X}
prob(X) = \dfrac{n!}{X!(n-X)!} p^X (1-p)^{n-X}
prob(X)=n!X!(n−X)!pX(1−p)n−Xprob(X) = \dfrac{n!}{X!(n-X)!} p^X (1-p)^{n-X}

To get the total probability, prob(X), we multiply the probability of getting X successes in just one way by the total number of ways.

The result, A x B, is the probability of getting X successes out of n trials in all ways.

n = 3, p = 0.45

X =

A =

B =

prob(X) =

0

1

2

3

A Sample Table With Formulas

_3C_0
3C0_3C_0
_3C_1
3C1_3C_1
_3C_2
3C2_3C_2
_3C_3
3C3_3C_3
.45^0 (1-.45)^{3-0}
.450(1−.45)3−0.45^0 (1-.45)^{3-0}
.45^1 (1-.45)^{3-1}
.451(1−.45)3−1.45^1 (1-.45)^{3-1}
.45^2 (1-.45)^{3-2}
.452(1−.45)3−2.45^2 (1-.45)^{3-2}
.45^3 (1-.45)^{3-3}
.453(1−.45)3−3.45^3 (1-.45)^{3-3}
A \times B
A×BA \times B
A \times B
A×BA \times B
A \times B
A×BA \times B
A \times B
A×BA \times B

n = 3, p = 0.45

X =

A =

B =

prob(X) =

0

1

2

3

A Sample Table With Final Values

0.16637

0.16637

1

3

3

1

0.13612

0.40837

0.11137

0.09112

0.33412

0.09112

n = 3, p = 0.45

X =

A =

B =

prob(X) =

0

1

2

3

Don't Forget to Check Your Calculations

0.16637

0.16637

1

3

3

1

0.13612

0.40837

0.11137

0.09112

0.33412

0.09112

The values in Row A should add up to 

2^n:
2n:2^n:
2^n = 2^3 = 8
2n=23=82^n = 2^3 = 8
1 + 3 + 3 + 1 = 8 = check!
1+3+3+1=8=check!1 + 3 + 3 + 1 = 8 = check!

n = 3, p = 0.45

X =

A =

B =

prob(X) =

0

1

2

3

Don't Forget to Check Your Calculations

0.16637

0.16637

1

3

3

1

0.13612

0.40837

0.11137

0.09112

0.33412

0.09112

The values in Row prob(X) should add up to 1:

0.16637 + 0.40837 + 0.33412 + 0.09112 = 0.99998 = \text{close enough}
0.16637+0.40837+0.33412+0.09112=0.99998=close enough0.16637 + 0.40837 + 0.33412 + 0.09112 = 0.99998 = \text{close enough}

Fin

1. Probability - Building a Binomial Table

By smilinjoe

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1. Probability - Building a Binomial Table

a.k.a., Building a binomial distribution.

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