Chapter 5

Probability Theory

• Rule 1 : The probability P(A) of any event A satisfies 0 < P(A) < 1

• Rule 2 : If S is the sample space in a probability model, then P(S) = 1

• Rule 3 : The complement of any event A is the event that A will not occur, written as Ac . The complement rule states that P(AC) = 1 - P(A)

• Rule 4 : (Addition rule for disjoint events) If A and B are disjoint, P(A or B) = P(A) + P(B)

5.1 General Probability Rules

• Rule 5 : (Multiplication rule) If A and B are independent events, P(A and B) = P(A) P(B)

We have already met and used five rules in §4.2

General Addition Rule

For any two events A and B,

P(A or B) = P(A) + P(B)-P(A & B)

Example

Two dice are thrown, what is the probability of

1) getting a total of 12 or an absolute difference of 4?

B: A Total of 12, C An absolute difference of 4P(B or C)=P(B)+P(C)=(1/36)+(4/36)=5/36

2) getting at least one “6” on both dice or an absolute difference of 4?

D: At least one 6P(C or D)=P(C)+P(D)-P(C&D) =(4/36)+(11/36)-(2/36)=13/36

3) getting an absolute difference of 4 or odd total?P(C or D)=P(C)+P(D)-P(C&D) =(4/36)+(18/36)-0=22/36

Caffeine ExampleCommon sources of caffeine are coffee, tea and cola

drinks. Suppose that:

55% of adults drink coffee,

25% of adults drink tea,

45% of adults drink cola

15% drink both coffee and tea,

5% drink all three beverages,

25% drink both coffee and cola

And 5% drink only tea

(a) Draw a Venn diagram marked with this notation.

(b) What percent of adults drink only cola?

(c) What percent drink none of these beverages?

The Binomial Setting:1) There is a fixed number n of observations

2) The n observations are independent

3) Each observation falls into just one of two categories, which, for convenience, we call “success” and “failure”.4) The probability of a success, call it p, is the same for each observation

5.2 The Binomial Distributions

Example: Toss a fair coin 10 times, and count the number ofheads which will appear. The random variable, X, is the number of heads that appear. The probability of a successis p = 0.5

This experiment satisfies the four requirements for the Binomial Setting, so this is a Binomial Distribution.

Binomial Distributions

The distribution of the count X of successes in the binomialsetting is called the binomial distribution with parametersn and p.

The parameter n is the number of observations

The parameter p is is the probability of a success on anyone observation.

The possible values of X are the whole numbers from 0to n.

We say that X is B(n,p)

Binomial DistributionsExample: Imagine we have a big bunch of transistors, sayroughly 10,000. We will say that 1000 of them are bad.

Pull an SRS of size 10 without replacement.

Let X be the amount of transistors which are bad in the SRS.

Q: Is this the binomial setting?

A: No. The first transistor has a 1000/10,000 = 0.1 chance ofbeing bad. What about the second?

The second has either a 1000/9,999 = 0.10001 or 999/9,999 chance = 0.09991 chance of being badWe can consider this a B(10,0.1)

ExampleIn each of the following cases, decide whether or not a binomial setting is the appropriate model, and give yourreasons.

(a) Fifty students are taught about binomial distributions by a television program. After completing their study, all students take the same exam. The number of students who pass is counted.

(b) A chemist repeats a solubility test 10 times on the same substance. Each test is conducted at temperature 10 higher than the previous test. She counts the number of times that the substance dissolves completely.

Example (cont.)

In each of the following cases, decide whether or not a

binomial setting is the appropriate model, and give your

reasons.

(c) A student studies binomial distributions using a computer-assisted instruction. After the initial instruction is completed, the computer presents 10 problems. The student solves each problem and enters the answer; the computer gives additional instruction between problems if the student’s answer is wrong. The number of problems that the student solves correctly is counted.

Binomial Probability Distribution Function

n = sample sizep = probability of ‘success’ k = number of ‘successes’ in sample (X = 0, 1, 2, ..., n)

knk ppk

nkXP

1)(

)!(!!

knkn

k

n

Example 1

Toss 1 Coin 4 times in a row. Note # Tails. What’s the Probability of 3 tails?

This is a binomial setting with n=4, p=0.5 i.e. B(4,0.5)

25.0

5.015.03

4)3(

1)(

343

XP

ppk

nkXP knk

Using TI-83 Calculator under distributions menuBinompdf(4,0.5,3)=0.25

Example 1

Evans is concerned about a low retention rate for employees. On the basis of past experience, management has seen a turnover of 10% of the hourly employees annually. Thus, for any hourly employees chosen at random, management estimates a probability of 0.1 that the person will not be with the company next year. Choosing 3 hourly employees a random, what is the probability that 1 of them will leave the company this year? Let: p = .10, n = 3, x = 1

• Using the Binomial Probability Function

f xn

x n xp px n x( )

!!( )!

( )( )

1f xn

x n xp px n x( )

!!( )!

( )( )

1

f ( )!

!( )!( . ) ( . )1

31 3 1

0 1 0 91 2

f ( )!

!( )!( . ) ( . )1

31 3 1

0 1 0 91 2

= (3)(0.1)(0.81)= (3)(0.1)(0.81)

= .243= .243

Binomial Mean & Standard Deviation

A: If a count X has the B(n,p) distribution, then :

X = np X = np (1-p)

Example: Refer back to Example 2

Expected ValueExpected Value EE((xx) = 3(.1) = .3 employees out ) = 3(.1) = .3 employees out

of 3of 3VarianceVariance

Var(x) = 3(.1)(.9) = .27Var(x) = 3(.1)(.9) = .27Standard DeviationStandard Deviation

employees 52.)9)(.1(.3)(SD x employees 52.)9)(.1(.3)(SD x

The Normal approximation to Binomial distributions

• Suppose that a count X has the Binomial distribution with n trials and success probability p. When n is large, the distribution of X is approximately Normal,

• As a rule of thumb, we will use the Normal approximation when n and p satisfy

p1np ,npN

.10p1n and 10np

Examples• Read Pages 329-330• Exercise 5.41, p. 334

• The chance that Jodi will score 70% or lower is about 12.5%

.1251.015.1ZP70XP and 33.4 ,75N )a

) 187.5, 6.85 and 175 1.82 0.0344.b N P X P Z

• Slightly over 3% chance that Jodi will score 70% or lower.

5.3 The Poisson Distributions

• The Poisson Setting:1. The number of successes that occur in any

unit of measure is independent of the number of successes that occur in any non-overlapping unit of measure.

2. The probability that a success will occur in a unit of measure is the same for all units of equal size and is proportional to the size of the unit.

3. The probability that 2 or more successes will occur in a unit approaches 0 as the size of the unit becomes smaller.

Poisson Probability Function

( )!

keP X k

k

( )!

keP X k

k

where:where:

P(X=k) P(X=k) = probability of = probability of xx occurrences in an occurrences in an intervalinterval

µ = mean number of occurrences in an intervalµ = mean number of occurrences in an intervalee = 2.71828 = 2.71828

Using the Poisson Probability Function

Patients arrive at the emergency room of Mercy Hospital at the average rate of 6 per hour on weekend evenings. What is the probability of 4 arrivals in 30 minutes on a weekend evening?

= 6/hour = 3/half-hour, x = 4

4 33 (2.71828)( 4) .1680

4!P X

4 33 (2.71828)

( 4) .16804!

P X

Example 5.15

Poisson Mean & Standard Deviation

• Mean =

• Variance =

• Standard Deviation =