Probability: Chapter15: Satyagopal Mandal

Satyagopal Mandal

Department of Mathematics

University of Kansas

Office: 624 Snow Hall Phone: 785-864-5180

e-mail: mandal@math.ukans.edu

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Chapter 15: Probability

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

We have some intuitive idea about probability.

One says that probability that the face HEAD will show up in a coin toss is one-half. What do we understand by this? It means, if we toss a coin for a large number of times, approximately half the times the face HEAD will show up.

One says that, when we roll a die, that the face 5 will show up is 1/6. By this we mean that if we roll the die for a large number of times, approximately, 1/6^th times we will the face 5 will show up.

We also heard of loaded dice. For them the probability of certain face to roll over is higher than that of some other face.

Random Experiments and Sample Spaces

The 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 me define it more formally.

Definition 1: 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.)

Remark: We use the word "set" to mean a collection of objects.

Definition: Given a random experiment, the set of all possible outcomes is called the sample space. In this class, a sample space will always be denoted by "S".

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

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 be using this brace notation to list the members of the sample space (or a set). Try to get used to it.)

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}.

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}.

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), Е and so on. So the sample space is S={(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), Е}.

Suppose your experiment is to record the number of daily road accidents in Lawrence. Then the possible outcomes are 0, 1, 2, 3, Е and so on. So, the sample space is S={0, 1, 2, 3, Е}.

Suppose the experiment is to determine the sex of an unborn child (by ultrasonic). Then the possible outcomes are Male and Female. So, the sample space is S={Male, Female}.

Suppose your experiment is to determine the blood group of a patient, in a lab. So, the possible outcomes are S={O, A, B, AB}.

Suppose your experiment is to estimate the total wheat production (in tons) in Kansas in 1999 then the possible outcomes are all the positive numbers. So, S={any positive number} = {x: x is a positive number}.

Suppose your experiment is to give an estimate of the annual rainfall (in inches) in Lawrence in 1999. Then the possible outcomes are all the positive numbers. So, 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 like to talk about the probability of EVENTS.

What is an EVENT for us?

Definition: Given a sample space, an event is a collection of outcomes. So, an event is a subset of the sample space.

Remark. When we perform a random experiment exactly one outcome will result. If E is an event, we say that E occurred if the outcome is a member of E.

Definitions: There are two special events. First, there is an event (denoted by f) called impossible event. The impossible event has no outcome in it. That means it is "empty". Then, the whole sample space S is also an event, to be called certain event.

When we perform a random experiment, there will be an outcome. So, the impossible event will never occur and the certain event will always occur.

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

2.1) Look at example (1.1) of tossing a coin. The {H} is an event and so is {T}.

2.2) Refer to the example (1.2) of tossing a coin twice. 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}.

2.3) Refer to the example (1.3) of rolling a die. Let E={1,2,3} then E is an event. So, E is the even that the "face value" was less or equal to 3.

2.4) Refer to the example (2.4) of rolling a die twice. 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)}.

2.5) Refer to the example (1.5) on road accidents. Let E be the event that there was no accident in Lawrence on a day. Then E={0}.

2.9) Refer to the example (1.9) on annual rainfall in Lawrence. I define a year is be a "dry year" if the annual rainfall is less than 5 inches. Let E be the event that 1999 will be a dry year. Then E is the set of all positive numbers from 0 to 5.

For now, we will be dealing with sample spaces, that has only finitely many outcomes. We are still getting ready to talk about probability. In certain cases, computing probability of an event will involve counting the number of outcomes in the event and the sample space. So, we need to learn a little bit about counting techniques.

The Multiplication rule 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 certain job (often a selection of an item) is accomplished in r stages.

The first stage of the job can be accomplished in n₁ways.

The 2nd stage of the job is can be accomplished in n₂ways.

The 3rd stage of the job is can be accomplished in n₃ways.

ЕЕЕ.

ЕЕЕЕЕ

r) The final r-th stage of the job can be accomplished in n_rways.

Then the number of ways the original job can be accomplished is n= n₁ n₂ n₃ Еn_r ways.

Example 3.1: Suppose you are out to buy a car. You have a choice of 3 makes, 5 body styles and 6 colors. How many choices of cars you have?

The job of picking the car is done in 3 stages. First stage is to pick the make, which we can do in 3 ways. The second stage is to pick the body style, which we can do in 5 ways. Then, the third stage is to pick the color that we can do in 6 ways.

Stage	The job	Number of way
1	Pick the Make	3
2	Pick the Body Style	5
3	Pick the Color	6

So the whole job of picking the car can be done in (Answer =) 3x5x6=90 ways.

Example 3.2: 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 is could be accomplished in two stages. First, the die is rolled and the number of outcomes for this first stage is 6. The second stage is to roll the die again. So, the second stage also has 6 outcomes.

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

So, the total number of outcomes in S is 6x6= 36.

Example 3.3: 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	Job	To	Number of ways
1	Assign the first seat	163	163
2	Assign the 2^nd seat	162	162
3	Assign the 3^rd seat	161	161
4	Assign the 4^th seat	160	160
5	Assign the 5^th seat	159	159
6	Assign the 6^th seat	158	158
7	Assign the 7^th seat	157	157
8	Assign the 8^th seat	156	156
9	Assign the 9^th seat	155	155
10	Assign the 10^th seat	154	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

Suggested Problems: Before you proceed, please have a look at Examples 1-13 (pp. 486-).

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

Example 3.4: Suppose that we want to form a committee of 10 students out of the 163 students in this class. We have just (Example 3.3 above) assigned the 10 seats in the first row to the 163 students. Which can be done in 163x 162 x Е 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 will be considered as distinct. While forming a committee the group of 10 as a whole would be counted as one committee. Without going into detail, the number of ways such a committee can be formed is

(163 x 162 x Е x 154) / (1 x 2 x Е 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 (3.3), the assignment of the 10 seats is selection of 10 students where the order in which we selected 10 students did matter. On the other hand, in example (3.4) order in which we pick 10 to represent in the committee did not count. So, the selection in (3.3) is an ordered-selection of r "objects" from a group n "objects". The selection I (3.4) is an unordered-selection of r "objects" from a group of n "objects".

Definition 2.1: A selection of r objects from a collection of n objects where different order of selection count as distinct will be 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. So, an assignment of the 10 seats in the first row to 163 students is a permutation of 163 students taken 10 at a time.

Definition 2.2: A selection of r objects from a collection of n objects where different order of selection does not count as distinct will be 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. So, a particular committee of 10 formed out of 163 students is a combination of 163 students taken 10 at a time.

Notations and Formulas: 1) Suppose n is a positive integer. The product of all integers from 1 to n is called "factorial n" and is denoted by "n!". So,

n! = 1 x 2 x Е x (n-1) x n

Also 0! = 1.

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

_{nP_r= n!/(n-r)!}

= n x (n-1) x Е x (n-r+1)

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

_{nC_r = _nP_r/r!}

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

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

Suggested Problems: Examples 14-17 (pp. 491-); Ex. 1a-c, 2a.2b, 2d, Ex.5, 6, 8 (pp. 504-).

Example 4.1:

Compute 3!, 6!, 8! .

Compute ₅P₂, ₇P₄, ₆P₂, ₄P₃

Compute ₅C₂, ₇C₄, ₆C₂, ₄C₃

Example 4.2: Let me compute ₈P₅: First method : ₈P₅= 8 xЕx(8-5+1) = 8 x Е x 4 = 8 x 7 x 6 x 5 x 4 = 6720.

Second Method:

_{8P₅= 8!/(8-5)! = 8!/3!
= (1 x 2 x Е x 8)/(1 x 2 x 3)
= 40320/6 = 6720}

Example 4.3: 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.

Example 4.4: How many ways you can deal a hand of 13 cards out of a deck of 52 cards?
Answer = ₅₂C₁₃= 52!/(13!x39!).

Example 4.5: Four financial awards (of different values) will be given to the "best" four students in a class of 163. How many possible ways these awardees can be picked?

The order counts. So, the answer is ₁₆₃P_4.}

Example 4.6: 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 way
1	Pick the 1^st letter	26
2	Pick the 2^nd letter	26
3	Pick the 3^rd letter	26
4	Pick the 4^th letter	26

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