1 Chapter 4. Section 4-3. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison...

1Chapter 4. Section 4-3. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman

MARIO F. TRIOLAMARIO F. TRIOLA EIGHTHEIGHTH

EDITIONEDITION

ELEMENTARY STATISTICSSection 4-3 Binomial Probability Distributions


DefinitionsBinomial Probability Distribution

1. The experiment must have 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.

4. The probabilities must remain constant for each trial.


Notation for Binomial Probability Distributions

P(x) = probability of getting exactly x success among n trials

n = fixed number of trials

x = specific number of successes in n trials

p = probability of success in one of n trials

q = probability of failure in one of n trials

(q = 1 - p )

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


P(x) = • px • qn-x (n - x )! x!

Binomial Probability Formula

n !

Method 1

P(x) = nCx • px • qn-x

for calculators with nCr function, r = x


P(x) = • px • qn-xn ! (n - x )! x!

Number of outcomes with

exactly x successes

among n trials

Probability of x successes

among n trials for any one

particular order

Binomial Probability Formula

Method 1 – Using a formula


Example: Find the probability of getting exactly 3 correct responses among 5 different requests from AT&T directory assistance. Assume in general, AT&T is correct 90% of the time.



This is a binomial experiment where:

n = 5

x = 3

p = 0.90

q = 0.10

Using the binomial probability formula to solve:

P(3) = 5C3 • 0.9 • 0.1 = 0.0729

Example: Find the probability of getting exactly 3 correct responses among 5 different requests from AT&T directory assistance. Assume in general, AT&T is correct 90% of the time.

3 2



0123456789

101112131415

0.2060.3430.2670.1290.0430.0100.0020.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+0.0+

P(x) n x

15 0123456789

101112131415

0.2060.3430.2670.1290.0430.0100.0020.0000.0000.0000.0000.0000.0000.0000.0000.000

P(x) x

Table A-1 Binomial Probability Distribution

For n = 15 and p = 0.10Method 2


Example: Using Table A-1 for n = 15 and p = 0.10, find the following:

a) The probability of exactly 3 successesb) The probability of at most 3 successes

a) P(3) = 0.129

b) P(at most 3) = P(0 or 1 or 2 or 3)

= P(2) or P(1) or P(2) or P(3)

= 0.206 + 0.343 + 0.267 + 0.129

= 0.945

Note = This method is limited because a table may not be available for every n and/or p.

Method 2 – Using a table


Probabilities with “Exact” successes

• Press 2nd, VARS (DISTR). • Select the option binompdf(). • Complete the entry binompdf(n, p, x)

to obtain P(x). – n is the number of trials– p is the probability of success–x is the EXACT number of

successes.

Method 3 – Using TI-83/4


• Example: What is the probability of getting exactly 2 heads when 4 tosses are made?

• Solution: –P(2) = binompdf(4, 0.5, 2)–P(2) = 0.375

Method 3 - Using TI-83/4

Probabilities with “Exact” successes


Probabilities with “At most” successes

• Example: What is the probability of getting at most 2 heads when 4 tosses are made?

• Express at most 2 as an inequality.– P( x ≤ 2) which means x = 0 or 1 or 2

• Solution: – P( x ≤ 2) = P(0) + P(1) + P(2)– P( x ≤ 2) = 0.0625 + 0.25 + 0.375 = 0.6875– Where the probabilities would computed using binompdf(4,0.5, 0) then binompdf(4,0.5, 1) etc…



Probabilities with “At most” successes

• Press 2nd, VARS, select the option binomcdf().

• Note: The “c” indicates this is a cumulative function. It adds all the probabilities from zero up to x number of successes.

• Complete the entry to obtain P(At most x) = binomcdf(n, p, x), where x is the MAXIMUM number of successes.



• Example: What is the probability of getting at most 2 heads when 4 tosses are made?

• Solution: – P( x ≤ 2) = binomcdf(4, 0.5, 2) = 0.6875.



Probabilities with “At least” successes

• When doing at least problems we must use the complement rule P(A) = 1 – P(not A)

• Complete the entry P(At least x) = 1 - binomcdf(n, p, x- 1),

where x is the MINIMUM number of successes.



• Example: What is the probability of getting at least 3 heads when 4 tosses are made?

• Solution:– P(x≥3) = 1 – P(x ≤ 2)– P(x≥3) = 1 - binomcdf(4, 0.5, 2) = 0.3125.

Note: This is the same as • P( x ≥ 3)= P(x=3)+ P(x=4)• P( x ≥ 3)= 0.25 + 0.0625 = 0.3125



Recap

• P(x) Ask to find the probability of EXACT number of successes.

– Formula: P(x) = nCx· px · qn-x

– Calculator: P(x) = binompdf(n,p,x)

• P(X x) Ask to find the probability of AT MOST a number of successes.

– Calculator: P(X x ) = binomcdf(n, p, x)

• P(X x) Ask to find the probability of AT LEAST a number of successes.

– Calculator: P(X x ) = 1 - binomcdf(n, p, x-1)

1 Chapter 4. Section 4-3. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison...

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