Math 365, Elementary Statistics

Lesson 3 : Probability

Introductionback to top

Probability to a statistician is the probability of the occurrence of an event. To an ordinary person it is the quantified chance of occurrence of that event.

Some of the early theory of probability originated in gambling and later theories developed in bioscience. We get very tempted when we see somebody win $1 million in a lottery, but lottery operators design their games and machines in such a way that they will make more money than they give, in the long run.


3.1 Basic Concept of Probabilityback to top

We are all familiar with simple probabilistic statements. If you toss a coin the probability that the HEAD will show up is 1 out of 2. If you roll a die the probability of getting the face 5 is 1 out of 6. The probability of having an accident on a particular busy street on a particular day is 1 out of 100. (When a child says "probably we should invite Aaron for my birthday," however, the "probably" may have little to do with mathematics of probability, but shows the awareness of the concept of probability at a basic human level.)

When we toss a coin for a large number of times we find that essentially half the time the head shows up. As we continue to toss, we see that the ratio of the number of Heads to the number of tosses remains close to and moves around 1/2. So we say that if we toss a coin, the probability that the head will show up is .50. On the other hand, if this ratio remains close to and moves around .49 then we will say the probability of heads is .49. To understand the concept of probability empirically, we visit a flash animation of a coin tossing experiment.

We observe the accidents on a street over a long period of time and observe that on about one in a hundred days there is an accident. The longer we continue to observe, we see that the ratio of the number of days there is an accident to the number of days observed remains close to one to one hundred. So we say that probability of an accident on a day on that street is 1 percent.

These examples explain the basic notion of probability. The probability of an event is understood as the "relative frequency," the ratio of occurrences of the EVENT to the total number of times the EXPERIMENT is repeated.

3.2 Sets and Subsets, Statistical Experiments, Sample Space, Events, Probabilityback to top

This section provides basic definitions that we will need for the rest of the course.

Sets and Subsets

Definition. By a set S we mean a collection of objects. The objects in this set S are also called elements of the set. A set E is said to be a subset of a set S if each element of E is also an element of S. We write

E ⊆ S

to mean that E is a subset of S. Obviously, a subset E of S is a smaller collection than or equal to S.

The following are some examples. We also explain the usage of braces to describe a set.

  1. Let D = the collection of all 52 cards in a deck. Then D is a set. Let E be the collection of all the hearts in this deck. Then E is a subset of D. In brace notation

    E={x in D : X is a Heart }

    This is read "the set of x in D such that x is a heart"

  2. Let T be the collection of all those who filed a tax return to the IRS for the year 1999. Then T is a set. Let L be the collection of those whose Adjusted Gross Income in the return was less or equal to $30,000. Then L is a subset of T. Let C be the collection of those who declared capital gains income. Then C is a subset of T. We write
    L ⊆ T
    C ⊆ T
    In brace notation
    L = {x ∈ T : the Adjusted Gross Income of x is less or equal to $30,000}.

    The symbol ∈ means "an element of"
    x ∈ T means x is an element of T

  3. Let N be the collection of all integers, and let E be the collection of even integers. Then N, E are set and
    E ⊆ N
    In brace notation
    N = {n : n is an integer}
    E = {n ∈ N : n is even}.  

  4. Let R be the set of all (real) numbers. Let I be the set of all numbers between 0 and 1, not equal to 0,1. Then R,I are sets and I is a subset of R. In brace notation

    R = {x : x is a real number}
    I = {x ∈ R : 0  <  x  <   1}.   

  5. S = {1,7,13,17,19} is a set.

  6. Let S be the collection of you and your siblings, B be the collection of your brothers, and F be the collection of your sisters. Then S,B,F are sets and we have

    F ⊆ S
     B ⊆ S.

Statistical Experiments and Sample Space

Definitions.

  1. A statistical experiment is a procedure that produces exactly one out of many possible outcomes. All the possible outcomes are known, but which outcome will result when you perform the experiment is not known.
  2. Given an experiment, the set of all possible outcomes is called the sample space.
  3. Given an experiment, an outcome of the experiment is called a sample point. So, the sample space consists of sample points.

Examples. The following are examples of some experiments and their sample spaces.

  1. Suppose your experiment is tossing a coin. The outcomes are H (heads) and T (tails). So, the sample space is S = {H,T}.

  2. Suppose your experiment is tossing a coin twice. The sample points (or outcomes) are HH,HT,TH,TT and the sample space is S = {HH,HT,TH.TT}.

  3. Your experiment is rolling a die. The outcomes are 1,2,3,4,5,6 and the sample space is S = {1,2,3,4,5,6}.

  4. Suppose that your experiment is rolling a die twice. Then the sample space is

S= left bracket (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) right bracket
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

In brace notation, we can write

S = {(i,j) : i = 1,2,3,4,5,6  and j = 1,2,3,4,5,6}.


5. Suppose your experiment is to determine the number of road accidents in Lawrence on a particular day. So, the sample space is S = {0,1,2,3 ... }.


6. Suppose the experiment is to determine the sex of an unborn chlid. Then the sample space is S = {Female, Male}.


7. Suppose your experiment is to determine the blood group of a patient in a lab. Then the sample space is S = {O,A,B,AB}.


8. Suppose your experiment is to observe the annual wheat production in Kansas. Then the sample space is S={x : x is a nonnegative Number} = {x ∈ R : x ≥ 0} =[0, ∞).

Definition. The sample space S is called a finite sample space if S has only a finite number of outcomes. If S has infinite elements, it is called an infinite sample space. Note that examples 1, 2, 3, 4, 6, and 7 above have finite sample spaces, and 5 and 8 have infinite sample space.

Events

Definitions. Given an experiment and its sample space S, the following are important definitions.

  1. A subset of the sample space S is called an event. So, an event E consists of outcomes, and we have

    E ⊆ S.

  2. An event ∅ that has no outcome and is called the empty event or impossible event. The impossible event consists of no outcome; if you perform the experiment, the impossible event will never occur.

  3. Since S is also a subset of S, S is an event. This event S is called the sure event. If you perform the experiment, this event is sure to occur.

  4. A simple event consists of a single outcome.

Remark. Often, we will describe events in "English," and we may have to identify them as a subset of the sample space and also conversely.


Examples. The following are some examples of events.

  1. Look at example 2 above—the experiment on the coin toss. Let E be the event that at least one of the tosses gave T, and let F be the event that both tosses gave the same face. Then

    E = {HT, TH, TT} and F = {HH,TT}.

  2. Look at example 4 above—the experiment on rolling a die. Let E5 be the event that first die showed 5. Then

    E5 = {(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}.

    Let T5 be the event that the sum of the two "rolls" is 5. Then

    T5 = {(1,4), (2,3), (3,2), (4,1)}.

    Let T1 be the event that the sum of the two rolls is 1. Because T1 has no outcome, it is an impossible event. Let T13 be the event that the sum of the two rolls is 13. Then T13 is also an impossible event.

  3. Look at the example 5 above—the experiment on road accidents. Let E be the event that there is no accident on that day. Then

    E = {0}.

  4. Look at example 8 above—the experiment on annual wheat production. Let E be the event that there will be more than 1000 units of wheat production in 1998. Then

    E = (1000, ∞).

The Theory of Probability

Given a sample space S, in the MATHEMATICS of probability we have rules for how to compute the probability of an event E. Although the MATHEMATICS of probability was inspired by the empirical concept of probability, we do not derive anything from our intuitive ideas. We are guided by the precise rules and laws that we set up.

For now we will be dealing with finite sample spaces.


Definition. Let

S = { e1, e2, ... ,en }.

be a finite sample space. The probability of a simple event {e} is a number (possibly given) denoted by P({e}) which has the following properties:

  1. 0 ≤ P({e}) ≤ 1.

  2. The sum of the probabilities of all the simple events is 1:

    P({e1}) + P({e2}) + ... + P({en}) = 1.

  3. If E is an event, then the probability E, P(E), is defined as the sum of the probabilities of all the sample events in E:

    P(E)= ∑ P({e})
      e ∈ E  
  4. So, we also have

    P(impossible Event)=P (∅)=0

    P(Sure Event)=P(S)=1


Remark. If we know the probabilities P({e}) of all the simple events {e}, we will be able to compute the probability of any event E using 3. The probabilities of the simple events will

  1. either be given
  2. or we will be given a rule how to compute it.

Probability with Equally Likely Outcomes

One of the most frequently used models to compute probabilities of simple events is called EQUALLY LIKELY OUTCOMES.

Definition. Let S = {e1, ... , eN} be a finite sample space. We say that all the outcomes are equally likely if all the outcomes have the same probability. So, in this case, we have

P({e1}) = P({e2}) = … = P({eN}) = 1/N.

Also, in this case, for an event E

P(E) = ∑ P({e}) = ∑ 1/N
  e ∈ E   e ∈ E  

=(Number of Outcomes in E)/(Number of Outcomes in S)

If n(E) denotes the number of outcomes in E then

P(E) = n(E)
n(S)
 .

Problems on 3.2

Probability of Simple Events Given in a Table

Exercise 3.2.1. The following table gives the blood group distribution of a certain population.

Blood Group Distribution
Blood Group O A B AB
Percentage of
Population 		47 42 8 3

Find the probability that a random sample of blood will be of Blood Group A or B or AB. (Here S={O, A, B, AB} and we want to compute the probability P(E) of the event E={A, B, AB}.
Solution

Exercise 3.2.2. A student wants to pick a school based on its grade distribution. Following is the most recent grade distribution in a school:

Grade Distribution
Unreal Data
Grades A B C D F
Percentage of
Students 		19 33 31 14 3

Find the probability that a randomly picked student will have at least a B average.
Solution

Exercise 3.2.3. The following table gives the probability distribution of a loaded die.

Probability Distribution for a Die
Face 1 2 3 4 5 6
Probability 0.20 0.15 0.15 0.10 0.05 0.35

Find the probability that the face 2 or 3 or 6 will show up when you roll the die.
Solution

Find the Probability with Equally Likely Outcomes

Exercise 3.2.4. An urn contains 7 apples and 3 oranges and 5 pears. One piece of fruit is picked at random. Find the probability that

  1. the fruit is an apple,
  2. the fruit is either an apple or a pear, and
  3. the fruit is an orange.

Solution

Exercise 3.2.5. A die is rolled twice. Find the probability that

  1. the sum is 8,
  2. only 2 or 3 showed up in both the rolls, and
  3. the first roll produced a bigger number.

Solution

Exercise 3.2.6. A letter is chosen at random from the letters of the English alphabet. Find the probability that

  1. the letter is either I or U,
  2. the letter is in the word ALWAYS, and
  3. the letter is not in the word NEVER.

Solution

3.3 Laws of Probabilityback to top

Notations from Set Theory

Following are a few notations from the set theory, which we will be using in the context of sample spaces and events.

Notations. Let S be a set and E, F be two subsets of S.

  1. The union E ∪ F, of E and F is the set defined as follows:

    E ∪ F = {x ∈ S : x ∈ E  or x ∈ F}.

    So, if you put together the elements of E and F in a single collection, you get the union E ∪ F.

  2. The intersection E ∩ F, of E and F is defined as follows:

    E ∩ F = {x ∈ S : both x ∈ E and x ∈ F}.


    So, if you take all the elements common to both E and F, you get the intersection of E and F.

  3. The complement Ec, of E is defined as follows:

    Ec = {x ∈ S : x ∉ E}.

    So, the complement Ec of E is the collection of all the elements in S that are not in E.


Remark. If we can understand and interpret the above definitions in our context of sample spaces and events, that is adequate. For us, S will be a fixed sample space and E,F will be events.

  1. E ∪ F is the event that consists of all outcomes that are either in E or in F (or both). So the occurrence of either E or F is the same as the occurrence of E ∪ F. That is why some textbooks use the notation (E or F) for E ∪ F. So, notationally, as in some textbooks,

    E ∪ F = E or F.


  2. E ∩ F is the event that consists of all the outcomes that are both in E and F. So the simultaneous occurrence of E and F is the same as the occurrence of E ∩ F. That is why E ∩ F is denoted by (E and F) in some texts. Notationally, as in some textbooks,

    E ∩ F = E and  F.


  3. Similarly, Ec is the event that consists of all the outcomes in S that are not in E. So, the occurrence of Ec is the same as the nonoccurrence of E. Notationally, as in some textbooks,

    E c = (not E)

Laws of Probability

Following are some of the laws of probability.

First, probability behaves like area and the laws of probability are like that of area.

Some formulas and definitions: Let S be sample space and let E and F be two events.

  1. We have

    P(E ∪ F) = P( E or F) = P(E)+P(F)- P(E ∩ F)

    = P(E) + P(F) -P(E and F)

    We subtract P(E ∩ F) because we counted it twice: once in P(E) and once in P(F).



  2. Definition. We say E and F are mutually exclusive if E ∩ F = ∅, i.e., E and F have no outcome in common. Since P(∅) = 0, it follows from 1 that

    if E and F are mutually exclusive then

    P(E ∪ F) = P(E) + P(F)

  3. We also have

    P(Ec) = 1 - P(E).

  4. Definition. Let E be an event. We say that the odds of an event E occuring are a to b if

    P(E) = a/(a+b)

Remark: This concept of ODDS is used often in gambling. When the odds in favor of a horse are 2 to 3, essentially this means that the probability the horse will win is 2/5. We say "essentially" because in actual betting, the probability is actually slightly less than 2/3, so that in the long run the gambling establishment makes more money than it gives. (This instructor is not particularly experienced in such betting or horse races.)

Problems on 3.3: Laws of Probability

Exercise 3.3.1. Let E, F, G be three events. It is given


P(E)=0.3         P(F)=0.7          P(G)=0.6  
P( E ∩ F) = 0.2      P( E ∪ G) = 0.7                      

Find the probability that
  1. E or F occur,
  2. both E and G occur, and
  3. E does not occur.

Solution

Exercise 3.3.2. Let E, F, G be events .

  1. If the odds in favor of E are 3 to 5, find the probability that E occurs.
  2. If the odds against F are 3 to 4, find P(F).
  3. If P(G) = 7/10, what are the odds in favor of G?


Exercise 3.3.3. The probability that a Christmas tree is taller than 6 feet is .30; the probability that a Christmas tree weighs more than sixty pounds is 0.25; and the probability that a Christmas tree is either taller than 6 feet or more than sixty pounds is .4.

  1. Find the probability that a Christmas tree is both taller than 6 feet and weighs more than sixty pounds.
  2. Find the probability that a Christmas tree is not taller than 6 feet.
  3. Find the probability that a Christmas tree is either less than 6 feet tall or less than sixty pounds in weight.
  4. Find the probability that a Christmas tree is neither taller than 6 feet nor heavier than sixty pounds.

Solution

Exercise 3.3.4. The probability that a student majors in liberal arts is .44; the probability that a student majors in business is .33; and the probability that a student majors in either liberal arts or business is .65. Find the probabilities

  1. that a student majors in both liberal arts and business.
  2. that a student majors in neither liberal arts nor business.
Solution

3.4 Counting Techniques and Probabilityback to top

Counting techniques are important and useful to learn. You might like to know, for example,

  1. the number of English words (formal) of 5 letters, (A formal word is any sequence of letters from the English alphabet. For example, eezq is a formal word.)
  2. the number of ways you can deal a hand of 13 cards from a deck of 52 cards, or
  3. the number of ways you can assign the first row of 11 seats to 231 guests.
Before we go further into counting, let us recall the factorial notation.

Notations. Let n be a positive integer. Then the n! (read as factorial n) is defined as

n!= 1 . 2 . … (n-2) . (n-1) n

0!=1.

Factorial n is the product of all integers from 1 up to n.

One of the main tools for such counting is the following principle:

The Basic Counting Principle. Suppose we have an experiment that is a combination of r sub-experiments, performed one after the other, such that

  1. the first sub-experiment has n1 outcomes;
  2. corresponding to each outcome of the first sub-experiment, the second sub-experiment has n2 outcomes;
  3. corresponding to each outcome of the first and the second sub-experiments, the third sub-experiment has n3 outcomes;


    r. corresponding to each outcome of each of the previous r-1 sub-experiments, the rth sub-experiment has nr outcomes.

Then our original experiment will have n1n2 ... nr outcomes.


Remark. Here we have used the word "experiment" in a slightly different sense than the statistical experiments. The basic counting principle will be used to count the number of outcomes in sample spaces and events.

Examples.

3.4.1. Count the number of words of length four that you can construct from the English alphabet. Answer: 26x 26x26x26

We use the counting principle by splitting this experiment into four sub-experiments:

Stage Job to do Number of Ways
1. Pick the first letter 26
2. Pick the 2nd letter 26
3. Pick the 3rd letter 26
4. Pick the 4th letter 26
  Answer = Product = 456976


3.4.2. Count the number of ways you can assign the 11 seats in the first row in a concert hall to 231 guests.

Stage Job to do Number of Ways
1. Assign seat 1 231
2. Assign seat 2 230
3. Assign seat 3 229
4. Assign seat 4 228
5. Assign seat 5 227
6. Assign seat 6 226
7. Assign seat 7 225
8. Assign seat 8 224
9. Assign seat 9 223
10. Assign seat 10 222
11. Assign seat 11 221
  Answer = Product = 221*222*...*230*231


3.4.3. Contrast: How many ways can you form a committee of 11 members from a group of 231 people? Unlike assigning seats, here the order of selection of the members will be ignored. The 11 members, when permuted around, will have different seat assignments but in the same committee. Forming the committee is a "combination" problem that comes below.


Remark. The difference between assigning 11 seats in a row and forming a committee of 11 is that in the first case the order of assignment is important. Assigning the first row to the same 11 guests in two different ways will count as two different outcomes. When we form a committee, the order in which we pick 11 members does not make any difference.


Definition. Suppose we have n objects. We pick r of them one by one (without ever puttting them back) and arrange them in a row. Such an ordered arrangement will be called a permutation of n objects taken r at a time. The number of permutations of n objects taken r at a time is denoted by nPr. It follows from the basic counting principle that


nPr = n (n-1) (n-2) ... (n-r+1) = n!/(n- r)!

Number of permutations nPr = product of r integers starting from n downward.

In contrast, we can pick r objects from a collection of n objects one by one but place the object back in the collection before the next pick, and arrange all of them in a row. Such selection and arrangement is called picking with replacment. Constructing a formal word of length 4 is an experiment of picking with replacement.

Remark: Example 3.1 is a problem on picking with replacement because a letter can be selected more that once. Example 3.2 is a permutation problem.


Definition. Suppose we have n objects in a container. We pick r of them all at a time. In this case the order of selection does not come into consideration. Such a selection is called a combination of n objects taken r at a time. The number of combinations of n objects taken r at a time is denoted by nCr and is given by

nCr =       n!      
(r! (n-r)!)


Examples. 1. Count the number of ways you can form a committee of 11 from a group of 231 people. Answer: 231C11
2. Count the number of ways you can deal a hand of 13 cards from a deck of 52 cards. Answer: 52C13.

Problems on 3.4: Counting Techniques and Probability

Exercise 3.4.1. Find 5!
Solution

Exercise 3.4.2. A homeowner would like to install a new storm door. The local store offers 2 brand names; each brand has 4 different styles and 3 colors. How many choices does the homeowner have?
Solution

Exercise 3.4.3. Suppose in the World Cup soccer tournament, group A has 8 teams. Each team of group A has to play all the other teams in the group. How many games will be played among the group A teams. Answer: 8C2

Exercise 3.4.4. How many ways can you deal a hand of 13 cards from a deck of 52 cards? Answer: 52C13

Exercise 3.4.5. How many ways can you deal a hand of 4 spades, 3 hearts, 3 diamonds, and 3 clubs?
Solution
Solution-variation

Exercise 3.4.6. We have 13 students in a class. How many ways can we assign the 4 seats in the first row? Solution

Exercise 3.4.7. Programming languages sometimes use a hexadecimal system (also called "hex") of numbers. In this system, 16 digits are used and denoted by 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F. Suppose you form a 6-digit number in a hexadecimal system.

  1. What is the probability that the number will start with a letter digit?
  2. What is the probability that the number is divisible by 16 (i.e., ends with 0)?

Solution
Here the sample space is the collection of all the 5-digit hex numbers.

  • Using the counting principle, the number of hex = n(S) = 166.
  • Let E be the event that the number starts with a letter digit.
  • Again, by the counting principle, the number of hex in E = n(E) = 6*165.
  • So, P(E) = n(E)/n(S) = 6/16.
  • Let F be the event that the number is divisible by 16. Since a number is divisible by 16 means, in hex, the first digit is 0.
  • So, the number of hex in F = n(F) = 165*1 = 165.
  • So, P(F) = 165/166 = 1/16.

Exercise 3.4.8. You are playing Bridge and you are dealt a hand of 13 cards.

  1. What is the probability that you will get a hand of 4 spades, 3 hearts, 3 diamonds and 3 clubs?
  2. What is the probability that you will get all 4 aces?
  3. What is the probability that you will get all 13 spades?

Solution

Exercise 3.4.9. A committee of 9 is selected at random from a group of 11 students, 17 mothers and 13 fathers.

  1. What is the probability that the committee has 3 students, 3 mothers, and 3 fathers, i. e., is a balanced committee?
  2. What is the probability that the committee has 4 mothers and 5 fathers?
  3. What is the probability that the committee has all students?

Solution

Exercise 3.4.10. Three scholarships of unequal value will be awarded from a group of 35 applicants. How many ways can such a selection be made?
Solution

3.5 Conditional Probability and Independent Eventsback to top

Sometimes when new information becomes available, the probability of an event may have to be reevaluated in light of this new information. Suppose we have a sample space S and an event E. Now suppose we have new information that an event C has occurred. We will have to reevaluate the conditional probability of E given that C has occurred. The conditional probability of E given that C has occurred is denoted by P(E|C). Clearly, P(E|C) may be different from P(E). In fact, now that C has occurred, our old sample space is no longer relevant. And C assumes the role of the new sample space.

Example. Suppose we pick a KU student at random and let E be the event that the student is taller than 6 feet. Then we have the following observations.

  1. The sample space S is the whole KU student population.

  2. Since all the outcomes are equally likely, we have
    P(E) = number of KU students who are taller than 6 feet
    Total number of KU students
    		 = n(E)
    n(S)
    		 .

  3. Now suppose we know that the student selected is a male. Let us denote the event that the student is a male by C. The probability that the student is taller than 6 feet, given that the student is a male, is higher than "simple" P(E). In fact, our new sample space is C, which is the whole KU male student population, not S, which is the whole KU student population.

  4. We now have the probability that the student is taller than 6 feet in height given that the student is a male
    = P(E|C) = number of MALE students who are taller than 6 feet
    Total number of male KU- students

    = n(E∩C)
    n(C)
    		 .
    Simple computations show that
    P(E|C) = n(E∩C)/n(S)
    n(C)/n(S)
    		 = P(E∩C)
    P(C)
    		 .

Based on the above example, we give the following definition and formula.

Definition. Let S be a sample space and E, C be two events.

  1. The conditional probability of E given that C has occurred is
    P(E|C) = P(E∩C)
    P(C)

    if P(C) ≠ 0.

  2. We get the following formula
    P(E∩C) = P(E|C)P(C).

Independent Events

If the conditional probability P(E|F) = P(E) the "simple" probability, then we say that E and F are independent. In this case,

P(E∩F) = P(E)P(F).

Definition. We say that two events E and F are independent if

P(E∩F) = P(E)P(F).

If two events are not independent, then they are said to be dependent.

Remark. Let us also describe what we mean by independence of 3 or more events. For events E1,E2, … , En, we say they are independent if the "multiplication rule" applies. For example E,F,G,H are independent if all of the following holds:

2 events

P(E∩F) = P(E)P(F),   P(E∩G) = P(E)P(G),

P(E∩H) = P(E)P(H),   P(F∩G) = P(F)P(G),

P(F∩H) = P(F)P(H),   P(G∩H) = P(G)P(H)

3 events

P(E∩F∩G) = P(E)P(F)P(G),   P(E∩F∩H) = P(E)P(F)P(H),  

P(E∩G∩H) = P(E)P(G)P(H),   P(F∩G∩H) = P(F)P(G)P(H)  

4 events

P(E∩F∩G∩H) = P(E)P(F)P(G)P(H)

Problems on 3.5: Conditional Probability and Independent Events

Exercise 3.5.1. Let A, B be two events. Given that

P(A) = .66   P(A ∩ B) = .11

Find P(B|A).
Solution

Exercise 3.5.2. Given

P(A|B) = .8 P(B) = .1

Find P(A∩B).
Solution

Exercise 3.5.3. In a certain county, the probability that a person took a flu shot is .45 and the probability that a person will get flu, given that he/she took a flu shot is .06. What is the probability that a randomly selected person took a flu shot and will get flu?
Solution

Exercise 3.5.4. Consider the following two circuit diagrams:

circuit diagram 1
Circuit 1		    circuit diagram 2
Circuit 2


For each of the two circuits do the following:
As you can see, current flows through two switches A and B to the radio and back to the battery. It is given that the probability that the switch A is closed is 0.91 and the probability that the switch B is closed is 0.83. Assume that the two switches function independently. Find the probability that the radio is playing.
Solution

Exercise 3.5.5. An airplane has two engines. The probability that engine 1 fails is 0.023 and the probability that engine 2 fails is 0.06. Assume that the engines function independently.

  1. What is the probability that both engines fail?
  2. What is the probability that both will not fail?
  3. What is the probability that neither will fail?

Solution

Exercise 3.5.6. Following are data from a hospital emergency room:

  1. The probability that a patient in the emergency room will have health isurance is 0.75.
  2. The probability that a patient in the emergency room will survive the treatment 0.85.
  3. The probability that a patient in the emergency room will have health insurance and will also survive is 0.7.

What is the conditional probability that a patient in the emergency room will survive, given that he/she has health insurance.
Solution

Exercise 3.5.7. The probability that you will receive a wrong number call this week is 0.3; the probability that you will receive a sales call this week is 0.8; and that the probability that you will receive a survey call this week is 0.5. What is the probability that you will receive one of each this week? (Assume that all these calls are independent.)
Solution

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