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Binomial Distributions

Section 6.2: How Can We Find Probabilities When Each

Observation Has Two Possible Outcomes?

Binomial distributions are based on categorical data with only two possible responses.

Examples Did you pass the test Have a car Are you north or south of the equator Which court is a case going through (federal/state)

One response will be called a “success” the other a “failure”.

Binomial Distributions

Each trial must be independent. Asking one person can’t effect others answers.

The probability for each trial must be the same.

Notation: Let n = the number of trials Let p = the probability of success Let q = the probability of failure Let X be the random variable for the number of successes for n

trials.

Conditions:

Let X be the random variable for the number of successes for n trials.

Conditions:

P(5) would be the probability of getting 5 success in a given number of trials.

P(10) would be the probability of getting 10 success in a given number of trials.

P(0) would be the probability of getting no success in a given number of trials.

Example;

I’m going to guess the card you draw from the deck. (replace the cards between trials)

We are going to try this 3 times.

How likely is it I will get exactly one right?

How likely I get more then one right?

Lets try,

I’m going to guess the card you draw from the deck. We are going to try this 3 times. How likely is it I will get exactly one right? How likely I get more then one right?

Notation; n = p = q = X is the number of successes in three trials, either 0, 1, 2, or 3.

Goal; Find P(1) and P(2 or 3)

Lets try,

31/5251/52

List the sample space; FFF, SFF, FSF, FFS, SSF, SFS, FSS, SSS

Which correspond to one success, x = 1? SFF, FSF, FFS

What are their probabilities? P(SFF) = P(S and F and F) = P(S) x P(F) x P(F) = (1/52) x (51/52) x (51/52) =

0.0185

P(FSF) = P(F) x P(S) x P(F) = q x p x q = (51/52) x (1/52) x (51/52) = 0.0185

P(FFS) = q x q x p = (51/52) x (51/52) x (1/52) = 0.0185

There sum is (0.0185) + (0.0185) + (0.0185) = 0.0555 or 5.55%

Notice this is 3(p)(q²)

Finding probabilities; (old school)

3(p)(q²)

Parts:

The number of ways our goal could happen.

A p for each success we need

A q for each failure we want.

For two successes;It can happen three ways, SSF, SFS, FSS,

So, 3(p²)(q) = 3(1/52)²(51/52)

Three right guesses of your card can only happen one way, SSS.

So, 1(1/52)³(51/52)º =

In general we have,

Try three;

xnx qp )()(#P(x)

Number of ways that combination happens

For a small number of trials, like three, we can list the sample space relatively fast and see how many ways the successes and failures come up.

FFF, SFF, FSF, FFS, SSF, SFS, FSS, SSS

We could even do it for four;

FFFF, SFFF, FSFF, FFSF, FFFS, SSFF, SFSF, SFFS, FSSF, FSFS, FFSS, SSSF, SSFS, SFSS, FSSS, SSSS

Or how about five trials;

“Put list of the 32 combination here”

Listing the sample Space?

When you go survey people you are asking 30, 40, even 100 people, that’s 100 trials.Billons of trillions of combinations!!!

1,267,657,368,594,201,568,956,327,459,028

We need a tool!!

Listing the sample space?

How many ways can x successes be distributed over n trials?

Note: ! Refers to the factorial operation.

Combinatorics;

)!(!

!#

xnx

n

Factorials find the product of all counting numbers less then or equal to the number given.

What??So, 5! = 1 x 2 x 3 x 4 x 5 = 120

7! = 1 x 2 x 3 x 4 x 5 x 6 x 7 = 5040

The What?

So, for n=5 and x=3,

Thus of the 32 ways to combine 5 trials, ten of them have three successes.

Finding the Ways;

1012

120

2*6

120

]1*2][1*2*3[

1*2*3*4*5

!2!3

!5

)!35(!3

!5

Our formula;

xnx qp )(x)!-(nx!

n!P(x)

For;n trialsp probability of successq probability of failure (1-p)X number of successes

Use tech:

60% of Americans are overweight. If you gather 15 random people what is the likelihood 10 will be overweight?

Example;

101510 )40.0()60.0(10)!-(1510!

15!P(10)

xnx qp )(x)!-(nx!

n!P(x)

Questions?

Finding Means and Standard deviations for binomial distributions;

More formula;

np npq

So, given that 60% of Americans are overweight how many would you expect to find in a group of 15 random people?

μ = np = 15(0.60) = 9

With

We can use our normal distribution tools here under special conditions.

If npq is larger than 15 [ten really] we may use normal distribution to examine our probabilities of being in an interval.

Example of μ and σ;

9.16.3)40.0)(60.0(15

You play a game “many” times and find that you win 73% of the time. You want to know in a set of 50 games how many will you usually win.

On average you will win, μ = 50(0.73) = 36.5 games.

But, your standard deviation is 3.14

So you will usually fall between 36.5+2(3.14) and 36.5-2(3.14)

Therefore you should expect to win between 30.22 and 42.78 games.

If you wanted to find where ALL of your wins should fall use three S.D.s.

Example, (last one)

Questions?