Apr 20, 2023

Chapter 6: Chapter 6: Binomial Probability Binomial Probability

DistributionsDistributions

In Chapter 6:

6.1 Binomial Random Variables6.2 Calculating Binomial Probabilities6.3 Cumulative Probabilities6.4 Probability Calculators6.5 Expected Value and Variance of Binomial Random Variables6.6 Using the Binomial Distribution to Help Make Judgments

§6.1 Binomial Random Variables• Binomial = a family of discrete random

variables

• Binomial random variable ≡ the random number of successes in n independent Bernoulli trials (a Bernoulli trials has two possible outcomes: “success” or “failure”

• Binomials random variables have two parametersn number of trialsp probability of success of each trial

Binomial Example• Consider the random number of successful

treatments when treating four patients

• Suppose the probability of success in each instance is 75%

• The random number of successes can vary from 0 to 4

• The random number of successes is a binomial with parameters n = 4 and p = 0.75

• Notation: Let X ~b(n,p) represent a binomial random variable with parameters n and p. The illustration variable is X ~ b(4, .75)

§6.2 Calculating Binomial Probabilities

xnxxn qpCxX )Pr(

where

nCx ≡ the binomial coefficient (next slide)

p ≡ probability of success for each trial

q ≡ probability of failure = 1 – p

Formula for binomial probabilities:

Binomial Coefficient

)!(!

!

xnx

nCxn

where ! represents the factorial function, calculated:x! = x (x – 1) (x – 2) … 1For example, 4! = 4 3 2 1 = 24By definition 1! = 1 and 0! = 1

6)12)(12(

1234

)!2)(!2(

!4

)!24)(!2(

!424

C

For example:

Formula for the binomial coefficient:

Binomial Coefficient

)!(!

!

xnx

nCxn

The binomial coefficient is called the “choose function” because it tells you the number of ways you could choose x items out of n

nCx the number of ways to choose x items out of n

For example, 4C2 = 6 means there are six ways to choose two items out of four

Binomial Calculation – Example Recall the “Four patients example”. Four patients; probability of success of each treatment = .75. The number of success is the binomial random variable X ~ b(4,.75). Note q = 1 −.75 = .25. What is the probability of observing 0 successes under these circumstances?

0039.

0039.11

250750!4!0

!4

250750

)0(Pr

40

04004

..

..C

qpCX xnxxn

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

Area Under The Curve

Pr(

X =

2)

=.2

109 ×

1.

0

Recall the area under the curve (AUC) concept.

AUC = probability!

§6.3: Cumulative Probability

• Recall the cumulative probability concept

• Cumulative probability ≡ the probability of that value or less

• Pr(X x) • Corresponds to left

tail of pmf

Pr(X 2) on X ~b(4,.75)

Cumulative Probability Function

• Cumulative probability function lists cumulative probabilities for all possible outcome

• Example: The cumulative probability function for X~b(4, 0.75)Pr(X 0) = 0.0039

Pr(X 1) = 0.0508

Pr(X 2) = 0.2617

Pr(X 3) = 0.6836

Pr(X 4) = 1.0000

Pr(X 1) = Pr(X = 0) + Pr(X = 1) = .0039 + .0469 = 0.0508

Pr(X 2) = Pr(X = 0) + Pr(X = 1) + Pr(X = 2)

Pr(X 4) = Pr(X = 0) + Pr(X = 1) + Pr(X = 2) + Pr(X = 3)

Pr(X 4) = Pr(X = 0) + Pr(X = 1) + … + Pr(X = 4)

§6.5: Expected Value and Variance for Binomials

• The expected value (mean) μ of a binomial pmf is its “balancing point”

• The variance σ2 is its spread

• Shortcut formulas:

np npq2

Expected Value and Variance, Binomials, Illustration

For the “Four patients” pmf of 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 the “Four patients” example

• Note μ = 3, suggesting we should see 3 success on average

• Does the observation of 2 successes cast doubt on p = 0.75?

• No, because Pr(X 2) = 0.2617 is not too unusual

StaTable Probability Calculator• Calculates

probabilities for many types of random variables

• This figure shows probabilities for X~b(4,0.75)

• Available in Java, Windows, and Palm versions (download from website)

Pr(X = 2) = .2109

Pr(X ≤ 2) = .2617

x = 2

p = .75

n = 4