# 6: Binomial Probability Distributions

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```							              Chapter 6:
Binomial Probability
Distributions
June 12
In Chapter 6:

6.1 Binomial Random Variables
6.2 Calculating Binomial Probabilities
6.3 Cumulative Probabilities
6.4 Probability Calculators
6.5 Expected Value and Variance ..
6.6 Using the Binomial Distribution to Help
Make Judgments
Binomial Random Variables
• Bernoulli trial ≡ a random event with two
possible outcomes (“success” or “failure”)
• Binomial random variable ≡ the random
number of successes in n independent
Bernoulli trials, each trial with the same
probability of success
• Binomials have two parameters:
n  number of trials
p  probability of success of each trial
Binomials (cont.)
• Only two outcomes are possible (success
and failure)
• The outcome of each trial does not
depend on the previous trial
(independence)
• The probability for success p is the same
for each trial
• Trials are repeated a specified number of
times n
Calculating Binomial
Probabilities by hand
Formula:
n x
Pr( X  x) n C x p q   x

where

nCx   ≡ the binomial coefficient (next slide)
p ≡ probability of success for a single trial
q ≡ probability of failure for single trial = 1 – p
Binomial Coefficient
Formula for the binomial coefficient:
n!
n Cx 
x!(n  x)!
where ! represents the factorial function:
x! = x  (x – 1)  (x – 2)  …  1
For example, 4! = 4  3  2  1 = 24
By definition 1! = 1 and 0! = 1
For example:
4!         4!      4  3  2 1
4 C2                                      6
(2!)( 4  2)! (2!)( 2)! (2 1)( 2 1)
Binomial Coefficient
n!
n Cx 
x!(n  x)!
The binomial coefficient tells you the number of
ways you could choose x items out of n

nCx    the number of ways to x items out of n
For example, 4C2 = 6
Therefore, there are 6 ways to choose 2 items
out of 4.
Binomial Calculation – Example
“Four patients example”: X ~ b(4,.75).
Note q = 1 −.75 = .25.
What is the probability of 0 successes?
Pr( X  0)  n C x p x q n  x
 4 C0  0.750  0.254 0
4!
          0.750  0.254
0!4!
 1 1  .0039
 .0039
X~b(4,0.75), continued
Pr(X = 1) = 4C1 · 0.751 · 0.254–1
= 4 · 0.75 · 0.0156
= 0.0469

Pr(X = 2) = 4C2 · 0.752 · 0.254–2
= 6 · 0.5625 · 0.0625
= 0.2106
X~b(4, 0.75) continued

Pr(X = 3) = 4C3 · 0.753 · 0.254–3
= 4 · 0.4219 · 0.25
= 0.4219

Pr(X = 4) = 4C4 · 0.754 · 0.254–4
= 1 · 0.3164 · 1
= 0.3164
pmf for X~b(4, 0.75)
Tabular and graphical forms
x        Pr(X = x)

0          0.0039

1          0.0469

2          0.2109

3          0.4210

4          0.3164
Pr(X = 2)
=.2109 × 1.0
AUC = probability!
Cumulative Probability
= Pr(X  x) = Left “Tail”
This figure illustrates Pr(X  2) on X ~b(4,.75)
Cumulative Probability Function
Cumulative probability function (cdf) =
cumulative probabilities for all outcome
Example: cdf for X~b(4, 0.75)
Pr(X  0) = 0.0039
Pr(X  1) = 0.0508   Pr(X = 0) + Pr(X = 1)
Pr(X  2) = 0.2617   Pr(X = 0) + Pr(X = 1) + Pr(X = 2)

Pr(X  3) = 0.6836   Pr(X = 0) + Pr(X = 1) + … + Pr(X = 3)

Pr(X  4) = 1.0000   Pr(X = 0) + Pr(X = 1) + … + Pr(X = 4)
Calculating Binomial Probabilities
with the StaTable Utility
StaTable is a free computer program that
calculates probabilities for many types of random
variables, including binomials
StaTable Binomial Calculator

Number of successes x

Binomial parameter p
Binomial parameter n

Calculates Pr(X = x)

Calculates Pr(X ≤ x)
StaTable Probability Calculator
StaTable
Exact and
cumulative                       x=2

probability of “2” for          p = .75
X~b(n = 4, p = .75)
n=4

Pr(X = 2) = .2109

Pr(X ≤ 2) = .2617
§6.5: Expected Value and
Variance for Binomials
• Expected value μ
• Variance σ2
• Shortcut formulas:

  np                npq
2
Expected Value and Variance,
Binomials, Illustration
For X~b(4,.75)
μ = n∙p
= (4)(.75) = 3
σ2 = n∙p∙q
= (4)(.75)(.25)
= 0.75
§6.6 Using the Binomial
• Suppose we observe 2
successes in a “Four
patients” experiment?
• Assume X~b(4, .75)
• 3 success are
expected
• Does the observation
of 2 successes cast   Pr(X  2) = 0.2617.
doubt on p = 0.75?    What does this infer?

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