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Section 5-3Binomial Probability

Distributions

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Key Concept

This section presents a basic definition of a binomial distribution along with notation, and it presents methods for finding probability values.

Binomial probability distributions allow us to deal with circumstances in which the outcomes belong to two relevant categories such as acceptable/defective or survived/died.

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DefinitionsA binomial probability distribution results from a procedure that meets all the following requirements:

1. The procedure has a fixed number of trials.

2. The trials must be independent. (The outcome of any individual trial doesn’t affect the probabilities in the

other trials.)

3. Each trial must have all outcomes classified into two categories (commonly referred to as success and failure).4. The probability of a success remains the same in all trials.

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What is the chance that LeBron James makes 18 or fewer of his 70 three-point attempts?

Dealing out 5 cards and counting the spades.

Flipping a coin until you get a heads.

Binomial?

Yes, meets the requirements of the Binomial Distribution.

No, trials are not independent; probability changes as each card is dealt.

No, no fixed number of trials.

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Notation for Binomial Probability Distributions

S and F (success and failure) denote two possible categories of all outcomes; p and q will denote the probabilities of S and F, respectively, so

P(S) = p (p = probability of success)

P(F) = 1 – p = q (q = probability of failure)

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Notation (cont.)

n denotes the number of fixed trials. x denotes a specific number of successes in n

trials, so x can be any whole number between 0 and n, inclusive.

p denotes the probability of success in one of the n trials.

q denotes the probability of failure in one of the n trials.

P(x) denotes the probability of getting exactly x successes among the n trials.

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Important Hints

Be sure that x and p both refer to the same category being called a success.

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Methods for Finding Probabilities

We will now discuss three methods for finding the probabilities corresponding to the random variable x in a binomial distribution.

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Method 1: Using the Binomial Probability Formula

P(x) = • px • qn-x (n – x )!x!

n !

for x = 0, 1, 2, . . ., nwhere

n = number of trials

x = number of successes among n trials

p = probability of success in any one trial

q = probability of failure in any one trial (q = 1 – p)

Look Familiar??

. . . can be written as nCx

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Rationale for the Binomial Probability Formula

P(x) = • px • qn-xn ! (n – x )!x!

The number of outcomes with

exactly x successes

among n trials

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Binomial Probability Formula

P(x) = • px • qn-xn ! (n – x )!x!

Number of outcomes with

exactly x successes

among n trials

The probability of x successes among n trials

for any one particular order

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Recall the Distribution for flipping a Coin twice

Is this a binomial distribution? Yes!

What did our random variable x represent? The number of heads: x 0 1 2

Could we use the binomial formula to find the probability of getting two heads when a coin is flipped twice? Try it!

25.)5(.)5.0()!22(!2

!2)( 222

xpn=2 p=.5 x=2

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Define the random variable of interest asx = the number of laptops among these

12

The binomial random variable x counts the number of laptops purchased. The purchase of a laptop is considered a success and is denoted by S. The probability distribution of x is given by 12.,..,2,1,0)4.0()6.0()!12(!

!12)( 12

xxx

xp xx

Sixty percent of all computers sold by a large computer retailer are laptops and 40% are desktop models. The type of computer purchased by each of the next 12 customers will be recorded.

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What is the probability that exactly four of the next 12 computers sold are laptops?

If many groups of 12 purchases are examined, about 4.2% of them include exactly four laptops.

042.0

4.06.0!8!4!12

)4()4(

84

xPp

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What is the probability that between four and seven (inclusive) are laptops?

547.0227.0177.0101.0042.0

4.06.0!5!7!12...4.06.0

!8!4!12

)7()6()5()4()74(

5784

ppppxP

These calculations can become very tedious. We will examine how to use

Appendix A Table A-1 to perform these calculations.

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What is the probability that between four and seven (exclusive) are laptops?

278.0177.0101.0

4.06.0!6!6!124.06.0

!7!5!12

)6()5()74(

6675

ppxP

Notice that the probability depends on whether < or ≤ appears. This is

typical of discrete random variables.

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Method 2: UsingTable A-1 in Appendix A

Part of Table A-1 is shown below. With n = 12 and p = 0.80 in the binomial distribution, the probabilities of 4, 5, 6, and 7 successes are 0.001, 0.003, 0.016, and 0.053 respectively.

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To find p(x) for any particular value of x,

1.Locate the part of the table corresponding to the value of n2.Move down to the row labeled with the value of x.3.Go across to the column headed by the specified value of p.

The desired probability is at the intersection of the designated x row and p column.

Using Appendix A Table A-1 to Compute Binomial Probabilities

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What is the probability that between four and seven (inclusive) are laptops? Let’s use the appendix…

547.0227.0177.0101.0042.0

)7()6()5()4()74(

ppppxP

We will use Appendix A Table A-1 to find the values of these probabilities.

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Method 3: Using TechnologySTATDISK, Minitab, Excel and the TI-83 Plus calculator can all be used to find binomial probabilities.

TI-83 Plus calculator

We will explore a few ways the calculator can be used to find binomial probabilities…

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CalculatorCalculator

• Binomialpdf(n,p,x) – this calculates the probability of a single binomial P(x = k)

• Binomialcdf(n,p,x) – this calculates the cumulative probabilities from P(0) to P(k) OR P(X < k)

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Recall the Distribution for flipping a Coin twice

Use your calculator to find the probability of getting two heads when a coin is flipped twice? n=2 p=.5 x=2 binompdf(2, .5, 2) = .25

Use your calculator to fine the probability of getting at least one head in the flip of two coins? n=2 p=.5 x=0 1-binomcdf(2, .5, 0) = .75

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Three-Pointers

Find the probability that LeBron makes exactly 18 three-pointers in 70 attempts, assuming his ability to make a three-pointer is .315? n=70 p=.315 x=18 binompdf(70, .315, 18) = .0618What is the probability he will make at least 25? n=70 p=.315 x=24 P(x≥25) =1-binomcdf(70, .315, 24) =.2612

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Strategy for Finding Binomial Probabilities

Use computer software or a TI-83 Plus calculator if available.

If neither software nor the TI-83 Plus calculator is available, use Table A-1, if possible.

If neither software nor the TI-83 Plus calculator is available and the probabilities can’t be found using Table A-1, use the binomial probability formula.