Math 105, Topics in Mathematics

 

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Lesson 4: Probability

Introduction

In everyday English the use of the word "probability" is not uncommon. The "probability of occurrence of an event" to a statistician is what "quantified chance of occurrence of that event" is to an ordinary person.

Most people have some intuitive idea about probability:

1. We say that the probability that the face Head will show up in a coin toss is 1/2. In other words, if we toss a coin many times, in approximately half of the times the face Head will show up.
2. We say that, when we roll a die, probability that the face 5 will show up is 1/6. In other words, if we roll the die many times, the face 5 will show up approximately one in six times.
3. We have also heard of loaded dice. For a loaded die, the probability of a certain face to roll over is higher than that of some other face.

Let us look at the coin-toss experiment.

4.1 Random Experiments and Sample Spaces

Tossing a coin or rolling a die are examples of random experiments. Whenever we talk about probability there is a random experiment behind it. We talk about probability in the context of such an experiment. Let us define it more formally.

1. Definition: A random experiment is a procedure that produces exactly one outcome out of many possible outcomes. All the possible outcomes are known. But which outcome will result when you perform the experiment is not known. (A random experiment is also called a statistical experiment.)
2. Definition: We use the word "set" to mean a collection of objects.
3. Definition: Given a random experiment, the set of all possible outcomes is called the sample space. In this class, a sample space is always denoted by "S."

Example 4.1.1. The following are examples of some experiments and their sample spaces.

1. Suppose your experiment is tossing a coin. Then the outcomes are H and T. So the sample space is S={H, T}. (Remark: We will use this brace notation to list the members of the sample space (or a set). Please try to get used to it.)
   
   
2. Suppose your experiment is tossing a coin twice. Then the outcomes are HH, HT, TH, TT. So, the sample space is S={HH, HT, TH, TT}.
   
   
3. Suppose your experiment is rolling a die. Then the outcomes are 1, 2, 3, 4, 5, 6. So, the sample space is S={1, 2, 3, 4, 5, 6}.
   
   
4. Suppose that your experiment is rolling a die twice. Then the possible outcomes are (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3) and so on. The sample space is
   

   		|	(1,1)	(1,2)	(1,3)	(1,4)	(1,5)	(1,6)	|
   		|	(2,1)	(2,2)	(2,3)	(2,4)	(2,5)	(2,6)	|
   S	=	|	(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)	|

5. Suppose your experiment is to record the number of daily road accidents in Lawrence. Then the possible outcomes are 0, 1, 2, 3, … and so on. The sample space is S={0, 1, 2, 3, …}.
   
   
6. Suppose the experiment is to determine the gender of an unborn child (by ultrasonic). Then the possible outcomes are Male and Female. The sample space is S={Male, Female}.
   
   
7. Suppose your experiment is to determine the blood group of a patient, in a lab. The possible outcomes are S={O, A, B, AB}.
   
   
8. Suppose your experiment is to estimate the total wheat production (in tons) in Kansas. Then the possible outcomes are all the positive numbers. S={any positive number} = {x: x is a positive number}.
   
   
9. Suppose your experiment is to give an estimate of the annual rainfall (in inches) in Lawrence. Then the possible outcomes are all the positive numbers. S={any positive numbers} = {x: x is a positive number}.

Events

We are getting ready to talk about probability. Given a sample space, we plan to talk about probability of an outcome. We may also talk about the probability of EVENTS.

What is an EVENT for us? We have the following definitions:

1. Definition: Given a sample space, an event is a collection of outcomes. An event is a subcollection (or subset) from the sample space.
   
   
2. Definition: When we perform a random experiment exactly one outcome results. If E is an event, then we say that E occurred if the outcome is a member of E.
   
   
3. Definitions: There are two special events.
   1. First, there is an event (denoted by ø, called impossible event. The impossible event has no outcome in it. That means it is "empty." The impossible event never occurs.
   2. The whole sample space S is also an event to be called certain event. The certain event occurs whenever you perform the experiment.

Examples 4.1.2. The following are some examples of events with reference to the examples 4.1.1 of sample spaces above.

1. Look at the example of tossing a coin (1). Then {H} is an event, and so is {T}.
2. Refer to the example of tossing a coin twice (2). Let E be the event that there was at least one T, and let F be the event that both the tosses produced the same face. Then E={HT, TH, TT} and F = {HH, TT}.
3. Refer to the example of rolling a die (3). Let E={1,2,3}. Then E is an event. E can be described as the event that the "face value" was less or equal to 3.
4. Refer to the example of rolling a die twice (4). Let E be the event that the first die showed the face 5. Then E={(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}. Let T be the event that the sum of the two "face values" is 5. Then T={(1,4), (2,3), (3,2), (4,1)}.
5. Refer to the example on road accidents (5). Let E be the event that there was no accident in Lawrence on a day. Then E={0}.
6. Refer to the example on annual rainfall in Lawrence. I define a year as a "dry year" if the annual rainfall is less than 5 inches. Let E be the event that a given year will be a dry year. Then E is the set of all positive numbers from 0 to 5.

For now, we will deal with sample spaces that have only a finite number of outcomes. We are still getting ready to talk about probability. In certain cases, computing probability of an event involves counting the number of outcomes in the event and the sample space. We need to learn a little bit about counting techniques.

4.3. Probability

Now we are ready to talk about probability of an outcome or an event. If we toss a coin, then one believes that the probability that the face Head will show up is 1/2. But this is about a "normal" coin. What if we toss a loaded coin? If you have a loaded coin, you may know that the probability that the face Head will show up is 1/5. But what does this mean? How and where did you learn that, for your loaded die, the probability that the face Head will show up is 1/5? When an ordinary person makes such probability statements he/she is, in fact, talking about his/her experience. Regarding your loaded coin, you have tossed your coin many times and have experienced that about once in five tosses the face Head showed up and other times the face Tail showed up. Therefore, "you know" that the probability that the face Head will show up is 1/5.

Similarly, you rolled a die many times. You have experienced that about once in six throws the face 4 shows up. Therefore, "you know" that the probability that the face 4 will show up when you throw the die is 1/6. It is a different story if you are working with a loaded die because your experience would tell you something different.

That was the "real life" idea of probability. In the study of the mathematics of probability, we accept this "experience" as part of the "probability model" and do the mathematics.

The mathematics of probability includes the following:

1. A description of the sample space S (and/or the random experiment).
2. A method or formula to compute the probability of an event. Given an event E, the method or the formula gives us the system of computing the probability P(E) of E. The probabilities P(E) must satisfy certain laws of probability.
3. A probability space is a sample space S with probability assignment as in (2). To describe a probability space, we have to give (1) and (2).
4. Laws of Probability: Let
   S = {o₁, o₂, ..., o_N}
   be a (finite) sample space. (Here o₁, o₂, ..., o_N are the outcomes of the experiments.) Following are the elements of probability spaces:
5. For each outcome o_i, a method or a formula is given so that we can compute the number P(o_i), to be called the probability that the outcome was o_i. The probabilities P(o_i) must satisfy the following two properties:
   1. P(o_i) is a number between 0 and 1;
   2. P(o₁)+ P(o₂)+ ... + P(o_N),=1.
      The sum of probabilities of all the outcomes is 1.

6. Given an event E, the probability P(E) of E is defined as the sum of probabilities of all the outcomes in E.
   
   

   P(E) = the sum of probabilities of all the outcomes in E.

   
   
   
7. The probability of the impossible event is zero, and the probability of the certain event is one.
   

   P(impossible event) = 0

   P(certain event) = P(S) = 1.

4.3. Problems on Probability

Exercise 4.3.l. The following table gives the blood group distribution of a certain population.

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

Suppose you determine the blood group of a randomly selected person from this population.

1. What is the sample space?
2. What are the possible outcomes of the experiment and their probabilities?
3. Write down the event that the selected person has blood group A or B or AB in brace notation.
4. Find the probability that a random sample of blood is of blood group A or B or AB.

Solution

Exercise 4.3.2. A student wants to pick a school based on the past grade distribution of the school. Following is a grade distribution of last year in a school:

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

Suppose you note down the grade of a randomly selected student from this school.

1. What is the sample space?
2. What are the possible outcomes of the experiment and their probabilities?
3. Write down the event that the selected student's grade is at least a B.
4. Find the probability that a randomly picked student has at least a B average.

Solution

Exercise 4.3.3. The following table gives the probability distribution of a loaded die.

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

1. What is the sample space?
2. Write down the event that the die shows 2 or 3 or 6, in brace notation.
3. Find the probability that the face 2 or 3 or 6 shows up when you roll the die.

Solution

4.2 The Multiplication Rule of Counting

A some probability problems involve counting. So, we will spend some time of different methods of counting.

The multiplication rule states that when something takes place in several stages, to find the total number of ways it can occur we multiply the number of ways each individual stage can occur.

It goes as follows: Suppose a certain job (often a selection of an item) is accomplished in r stages.

• The first stage of the job can be accomplished in n₁ ways.
  
• The second stage of the job can be accomplished in n₂ ways.
  
• The third stage of the job can be accomplished in n₃ ways.
  ... ... ...
  
• The final r-th stage of the job can be accomplished in n_r ways.

Then the number of ways the original job can be accomplished is

n₁ n₂ n₃ ... n_r ways.

Example 4.2.1. A household would like to install a storm door. The local store offers two brand names; each brand has four different styles and three colors. Find how many choices he has in the selection. Solution

The job of picking the door is done in three stages. The first stage is to pick the brand, which we can do in two ways. The second stage is to pick the style, which we can do in four ways. Then, the third stage is to pick the color, which we can do in three ways.

Stage	The job	Number of ways
1	Pick the brand	2
2	Pick the style	4
3	Pick the color	3

The whole job of picking the doorr can be done in 2x4x3=24 ways.

Example 4.2.2. Refer to the example (1.4) of rolling a die twice. We want to count the number of outcomes in the sample space S. The whole experiment could be accomplished in two stages. First, the die is rolled, and the number of outcomes for this first stage is six. The second stage is to roll the die again; the second stage also has six outcomes.

Stage	The job	Number of ways
1	Roll the die	6
2	Roll the die again	6

The total number of outcomes in S, by the multiplication principle, is 6x6= 36.

Example 4.2.3. I want to assign the 10 seats on the first row to the 163 students in the class. How many ways we can do it?

Stage	The job	Number of ways
1	Assign the first seat	163
2	Assign the 2nd seat	162
3	Assign the 3rd seat	161
4	Assign the 4th seat	160
5	Assign the 5th seat	159
6	Assign the 6th seat	158
7	Assign the 7th seat	157
8	Assign the 8th seat	156
9	Assign the 9th seat	155
10	Assign the 10th seat	154

So the total number of ways this can be done is =

163 x 162 x 161 x 160 x 159 x 158 x 157 x 156 x 155 x 154

Remark. The multiplication rule of counting has wide applications. You must correctly identify whether your counting problem can be divided into several stages of simple counting problems. Confusion may arise as follows.

Example 4.2.4. Suppose that we want to form a committee of 10 students out of the 163 students in this class. We have just assigned the 10 seats in the first row to the 163 students, which can be done in 163x 162 x . . . x 154 way. Could we say that the number of ways we can form a committee of 10 out of the 163 students in this class is the same? The answer is NO. While we assigned the seats, the different assignments of the seats, in the first row, to the same group of 10 students is considered as distinct. While forming a committee, the group of 10 as a whole is counted as one committee. Without going into details, the number of ways such a committee can be formed is

(163 x 162 x ... x 154) / (1 x 2 x ... x 10).

Ordered and unordered selection

Many counting problems that we consider essentially are like selecting r objects (or people) from a collection of n objects (or people). There are two types of such selections. In example (4.2.3), the assignment of the 10 seats is selection of 10 students where the order in which we selected 10 students did matter. However, in example (4.2.4) order in which we pick 10 to represent in the committee did not count. The selection in (4.2.3) is an ordered-selection of r "objects" from a group n "objects." But the selection in (4.2.4) is an unordered-selection of r "objects" from a group of n "objects."

We have the following definitions, formulas, and notations in this context:

1. Definition 1. A selection of r objects from a collection of n objects where different order of selection counts as distinct is called an ordered-selection. An ordered selection is also called a permutation. An ordered-selection of r objects from a group of n objects is called a permutation of n objects taken r at a time. An assignment of the 10 seats in the first row to 163 students is a permutation of 163 students taken 10 at a time.
2. Definition 2. A selection of r objects from a collection of n objects where different order of selection does not count as distinct is called an unordered-selection. An unordered-selection is also called a combination. An unordered-selection of r objects from a group of n objects is called a combination of n objects taken r at a time. A particular committee of 10 formed out of 163 students is a combination of 163 students taken 10 at a time.
3. Notation: Suppose n is a positive integer. The product of all integers from one through n is called "factorial n" and is denoted by n!.
   

   n! = 1 x 2 x ... x (n-1) x n
   Also 0! = 1.

4. The number of (possible) permutations of n objects taken r at a time is denoted by _nP_r.
   

   _nP_r= n!/(n-r)!   =   n x (n-1) x ... x (n-r+1)

5. The number of (possible) combinations of n objects taken r at a time is denoted by _nC_r.
   
   

   nCr = nPr/r!

   = n!/(r!(n-r)!)

   =(n x (n-1) x ... x (n-r+1))/(1 x 2 x ... x r)

Problems on 4.2 The Multiplication Rule of Counting

Before you attempt any problem, review the diagram.

Exercise 4.2.1.

1. Compute 3!, 6!, 8!.
2. Compute ₅P₂, ₇P₄, ₆P₂, ₄P₃.
3. Compute ₅C₂, ₇C₄, ₆C₂, ₄C₃.

Exercise 4.2.2.

Let me compute ₈P₅ :

1. First method : ₈P₅ = 8 x … x(8-5+1) = 8 x … x 4 = 8 x 7 x 6 x 5 x 4 = 6720.

2. Second Method: ₈P₅ = 8!/(8-5)! = 8!/3! = (1 x 2 x … x 8)/(1 x 2 x 3) = 40320/6 = 6720

Exercise 4.2.3.

Let me also compute ₉C₄. I like to compute as follows: ₉C₄ = ₉P₄ /4!.
We have ₉P₄ = 9 x 8 x 7 x 6 = 3024
and
4! = 1 x 2 x 3 x 4 = 24.
So, ₉C₄ = ₉P₄ /4! = 3024/24 = 126.

Exercise 4.2.4.

How many ways can you deal a hand of 13 cards out of a deck of 52 cards?

Answer = ₅₂C₁₃= 52!/(13! x 39!).

Exercise 4.2.5.

Four financial awards (of different values) will be given to the "best" four students in a class of 163. How many possible ways can these awardees be picked?

Here, the order counts. The answer is ₁₆₃P₄.

Exercise 4.2.5. How many code words of length four you can construct out of the English alphabets? This problem is not like selecting 4 from 26 letters because we can use the same letter more than once. Use the multiplication rule.

Stage	The job	Number of ways
1	Pick the 1st letter	26
2	Pick the 2nd letter	26
3	Pick the 3rd letter	26
4	Pick the 4th letter	26

The total number of such words = 26 x 26 x 26 x 26 =26⁴.

4.4 Probability Spaces with Equally Likely Outcomes

Unlike the above problems, in some probability spaces every outcome has an equal probability. (Such is the case when you toss a "normal" coin or roll a "normal" die.) In such cases, we say that outcomes are equally likely.

When outcomes are equally likely, the probability P(E) can be computed by counting the number of outcomes in E and those of S. If S has N outcomes then we have:

1. P(individual outcome) = P(o_i) = 1/N
   
   
2. For an event E
   

   P(E) = (# of outcomes in E)/(# of outcomes in S)  = (# of outcomes in E)/N

   
   
3. If n(E) = number of outcomes on E, then
   

   P(E) = n(E)/n(S)  = n(E)/N

Remark: Outcomes are equally likely in "normal" situations. Examples are

1. tossing an unbiased coin,
2. throwing a fair dice,
3. picking a card from a shuffled deck of cards and so on.

Example 4.4.1. Suppose we roll a (fair) die three times.

1. Describe the sample space.
2. Count the number of outcomes in the sample space S.
3. What is the probability that the sum of the points on the three faces is 7?
4. What is the probability that the sum of the points on the faces is not equal to 7?
   

The Solution

Answer to 1 is S={(1,1,1), (1,1,2), (1,1,3), (1,1,4), (1,1,5), (1,2,6), (1,2,1), (1,2,2), (1,2,3) ….}. Another way to write the same is S={ (i, j, k): i, j, k=1, 2, 3, 4,5,6 }. Find the total number of outcomes N in S. The job of rolling a die three times can be done in three stages: Stage The job Number of way 1 Throw the die 6 2 Throw the die 2nd time 6 3 Throw the die 3rd time 6 So, the total number of outcomes, by multiplication principle, in S is = n(S) = N= 6 x 6 x 6 = 216. Let E be the event that the sum is 7. E = {(1,1,5), (1,2,4), (1,3,3), (1,4,2), (1,5,1), (2,1,4), (2,2,3), (2,3,2), (2,4,1), (3,1,3), (3,2,2), (3,3,1), (4,1,2), (4,2,1), (5,1,1)} So E has n(E) = 15 outcomes. Therefore, the probability P(E) = n(E)/n(S) = 15/216. Let F be the event that the sum is not equal to 7. So, the n(F) = # of outcomes in F = (# of outcomes in S) -( # of outcomes in E) = 216 - 15 = 201. Therefore P(F) = n(F)/n(S) =201/216.

Example 4.4.2. Suppose we have to form a committee of two from a group of 15 men and 19 women.

1. Describe the sample space and count the number of outcomes in S.
2. What is the probability that both the members of the committee are men?
3. What is the probability that both the members are women?
4. What is the probability that one member is man and the other member is woman?

Hint: This is a problem of unordered-selection.

The Solution

S is the set of all possible pairs selected from this group of 15+19 = 34 people. The number of outcomes in S is n(S) = N =34C2= 34P2/2! =(34 x 33)/1x2 = 561. Let E be the event that both the members of the committee are men. The number of outcomes in E = n(E) = number of ways we can select two from a group of 15 men = 15C2= 15P2/2! = (15x14)/2 = 105. Therefore P(E) = n(E)/n(S) = 105/561. Let F be the event that both members of the committee are women. The number of outcomes in F = n(F) = number of ways we can select two from a group of 19 women = 19C2= 19P2/2! = (19x18)/2 = 171. Therefore P(F) = n(F)/(S) =171/561. Let M be the event that one member is a man and the other is a woman. The job of picking such a committee can be done in two stages: Stage The job Number of ways 1 Pick a male member 15 1 Pick a female member 19 The number of outcomes in M = n(M) = 15x19 = 285. P(M) = n(M)/n(S) = 285/561.

Problems on 4.4: Counting and Probability

Exercise 4.4.1. Find 5!
Solution

Exercise 4.4.2.

Suppose in the World Cup Soccer tournament, group A has eight teams. Now each team of group A has to play with all the other teams in the group. Find how many games will be played among the Group A teams.

Exercise 4.4.3. How many ways you can deal a hand of 13 cards from a deck of 52 cards?

Exercise 4.4.4. How many ways you can deal a hand of 4 Spades, 3 Hearts, 3 Diamonds, and 3 Clubs?
Solution
Solution-variation

Exercise 4.4.5. We have 13 students in a class. How many ways we can assign the four seats in the first row? Solution

Exercise 4.4.6. Programming languages sometimes use 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 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 6-digit hex numbers.