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Chapter 4 Introduction to Probability

 

4.2 Events and Their Probabilities

 

Event: An event is a collection of sample points

 

Probability of an Event: The probability of any event is equal to the sum of the probabilities of the sample points in the event

 

Probability Axioms

 

Given an event Ein a sample space Swhich is either finite with Nelements or countably infinite with N==inftyelements, then we can write

 

S=( union _(i==1)^NE_i),

and a quantity P(E_i), called the probability of event E_i, is defined such that

1. 0<=P(E_i)<=1.

2. P(S)==1.

3. Additivity: P(E_1 union E_2)==P(E_1)+P(E_2), where E_1and E_2are mutually exclusive.

4. Countable additivity: P( union _(i==1)^nE_i)==sum_(i==1)^(n)P(E_i)for n==1, 2, ..., Nwhere E_1, E_2, ...are mutually exclusive (i.e., E_1 intersection E_2==emptyset).

 

                Conditional Probability

 

The conditional probability of an event Aassuming that Bhas occurred, denoted P(A|B), equals

P(A|B)==(P(A intersection B))/(P(B)),

(1)

which can be proven directly using a Venn diagram. Multiplying through, this becomes

P(A|B)P(B)==P(A intersection B),

(2)

which can be generalized to

Rearranging (2) gives

P(B|A)==(P(B intersection A))/(P(A)).

(3)

Solving (3) for P(B intersection A)==P(A intersection B)and plugging in to (2) gives

P(A|B)==(P(A)P(B|A))/(P(B)).

 

 

                Bayes’ Theorem

 

                Let Aand B_jbe sets. Conditional probability requires that

P(A intersection B_j)==P(A)P(B_j|A),

(1)

where  intersection denotes intersection ("and"), and also that

P(A intersection B_j)==P(B_j intersection A)==P(B_j)P(A|B_j).

(2)

Therefore,

P(B_j|A)==(P(B_j)P(A|B_j))/(P(A)).

(3)

Now, let

S= union _(i==1)^NA_i,

(4)

so A_iis an event in Sand A_i intersection A_j==emptysetfor i!=j, then

A==A intersection S==A intersection ( union _(i==1)^NA_i)== union _(i==1)^N(A intersection A_i)

(5)

P(A)==P( union _(i==1)^N(A intersection A_i))==sum_(i==1)^NP(A intersection A_i).

(6)

But this can be written

P(A)==sum_(i==1)^NP(A_i)P(A|A_i),

(7)

so

P(A_i|A)==(P(A_i)P(A|A_i))/(sum_(j==1)^NP(A_j)P(A|A_j))

 

 

Activities

Conditional probability - Bayes' theorem type problems


A type of conditional probability problem involves finding the probability of an earlier event having occurred given that we know the outcome for a later event.

This type of problem and similar problems can be solved with the use of Bayes' theorem.

Rather than using Bayes' theorem, we will use the basic formula  and probabilities from a suitable tree diagram.



Consider the example in which two balls are removed from a bag containing three green and two red balls. The first ball is not replaced before the second is removed.

If the second ball is red, what is the probability that the first ball was green?

A = {first ball green} and B = {second ball red}
 

There is a simpler approach to this problem:
If the second ball is red, then the first ball must be one of the remaining four balls (three green and one red). Therefore the probability that it is green is 3 out of 4 i.e. 3/4.
 

Other problems do not lend themselves to this simpler treatment. Consider the same bag with the addition of one yellow ball. Again two balls are removed without replacement.

If the balls are different colours, what is the probability that one of the balls is green?

A = {one of the balls is green} and B = {different colours}




Archery problem - Rachel, Susan and Tiffany are shooting at a target in archery. With each arrow, the probability of hitting the centre is:

Rachel , Susan  and Tiffany 

What is the probability that:

Solution


 
 
 
 
 Homework Problems for Conditional Probability

1.    The probability that event A occurs is .63.   The probability that event B occurs is .45.   The probability that both events A and B occur is .18.    Find the following:
A)   P(A\B)
B)   P(B\A)

2.    The probability that event A occurs is .81.   The probability that event B occurs is .68.   The probability that both events A and B occur is .22.    Find the following:
A)   P(A\B)
B)   P(B\A)

3.    Two fair dice are rolled.   Find the probability that the sum rolled is at least ten, given:
A)   No information at all
B)   At least one die comes up six
C)   The same number appears on both dice

4.    A card is drawn at random from a standard deck of playing cards.   What is the probability that the card is less than a 7 given:
A)    No information at all
B)    The card is not a 2
C)    The card is a heart
D)    The card is a 3 or 4

5.    If a fair coin is flipped three times, what is the probability that it comes up tails at least once given:
A)    No information at all
B)    All three flips produce the same side
C)    It comes up tails at most once
D)    The third flip is heads
E)    It comes up heads at least once

6.   
In a small town the given chart below reflects the ethic and religious backgrounds of the townsmen.   Let C, B, I, and S denote the events of being Catholic, Baptist, Italian , and Spanish respectively.

Religion/Ethnic background

Italian

Spanish

Total

Catholic

145

122

267

Baptist

111

225

336

Total

256

347

603

Find the following:
A)    P(B)
B)    P(C\I)
C)    P(S\B)
D)    P(I\C)

 

7.    In a small town the given chart below reflects the sex and educational backgrounds of the townsmen.   Let M, W, P, B and H denote the events of being a Man, Woman, PHd degree, BA or MA degree and High School Diploma respectively.

 

PHD

BA or MA

High School Diploma

TOTAL

MEN

72

52

83

207

WOMEN

75

91

102

268

TOTAL

147

143

185

475

Suppose that one of these townsmen is selected at random.    What is the probability:
A)    The townsmen is a woman?
B)    The townsmen is a man?
C)    The townsmen has a High School Diploma?
D)    The townsmen is a man given that the townmen has a PHd?
E)    The townsmen has MA degree given that the person is a male?
F)    The townsmen is a woman given that the townmen has a BA?

8.    A family has three children.   Find the probability of having two boy, given that at most one of the children are girls.

9.    A family has three children.   Find the probability of having one boy, given that at least one of the children are boys.

 

Answers

Homework Solutions

1A)    .4
1B)    .29
2A)    .32
2B)    .27
3A)    1/3 or .33
3B)    .45
3C)    .17
4A)    .38
4B)    .33
4C)    .38
4D)    1
5A)    .88
5B)    .5
5C)    .75
5D)    .75
5E)    .86
6A)    .56
6B)    .57
6C)    .67
6D)    .54
7A)    P(W)= .56
7B)    P(M)= .44
7C)    P(H) = .39
7D)    P(M\P) = .49
7E)    P(B\M) = .25
7F)    P(W\B) = .64
8)    .75
9)    .43

 

 

Bayes’ Theorem

1) An auto insurance company charges younger drivers a higher premium than it does older drivers because younger drivers as a group tend to have more accidents.  The company has 3 age groups: Group A includes those under 25 years old, 22% of all its policyholders.  Group B includes those 25-39 years old, 43% of all its policyholders, Group C includes those 40 years old and older.  Company records show that in any given one-year period, 11% of its Group A policyholders have an accident.  The percentages for groups B and C are 3% and 2%, respectively.

a) What percent of the company’s policyholders are expected to have an accident during the next 12 months?

b)   Suppose Mr. X has just had a car accident.  If he is one of the company’s policyholders, what is the probability that he is under 25?

Say that this company not only classifies drivers by age, but in the case of drivers under 25 years old, it also notes whether they have had a driver’s education course.  One quarter of its policyholders under 25 have had a drivers education course and 5% of these have an accident in a one-year period. Of those under 25 who have not had a driver’s education course, 13% have an accident within a one-year period.  A 20-year-old woman takes out a policy with this company and within one year she as an accident.  What is the probability that she did not have a driver’s education course?

2) A medical research lab proposes a screening test for a disease.  To try out this test, it is given to 100 people, 60 of whom are known to have the disease and 40 of whom are known not to have the disease.  A positive test indicates the disease and a negative test indicates no disease.  Unfortunately, such medical tests can produce two kinds of errors:

1)   A false negative test:  For the 60 people who do have the disease, this screening indicates that 2 do not have it.

2)   A false positive test:  For the 40 people who do not have the disease, this screening test indicates that 10 do have it.

a)   Which of the false tests do you think is more serious and why?

b)   Incorporate the facts given above into a tree diagram.  (Be sure to convert the given integers into probabilities.)

c)   Suppose the test is given to a person whose disease status is unknown.  If the test result is negative, what is the probability that the person does not have the disease?