02 Introduction to Probability

“Don’t worry about failures, worry about the chances you miss when you don’t even try.”

Chapter 2

INTRODUCTION TO PROBABILITY

Motivation

We already introduced the field of probability theory during the Part 1 of the course. It was seen that it really helps in studying the uncertainty or error. More than just a mere tool for gambling gambling well, this field transcends transcends to developi developing ng the area of Inferential Statistics. This is the reason why we need to study this first before the other one as it will be a tool for our complete understanding. understanding. Statistics was an art of deducing estimates for quantities that people cannot directly observe. And it is still a science of analyzing systematic systematic patterns on dice and cards to adjust their bets to the “odds” of success. The word 'chance' then expresses only our ignorance of the causes of the phenomena that we observe to occur and to succeed one another in no apparent order. Probability is relative in part to this ignorance, and in part to our knowledge.

Chapter 2


Motivation

We already introduced the field of probability theory during the Part 1 of the course. It was seen that it really helps in studying the uncertainty or error. More than just a mere tool for gambling gambling well, this field transcends transcends to developi developing ng the area of Inferential Statistics. This is the reason why we need to study this first before the other one as it will be a tool for our complete understanding. understanding. Statistics was an art of deducing estimates for quantities that people cannot directly observe. And it is still a science of analyzing systematic systematic patterns on dice and cards to adjust their bets to the “odds” of success. The word 'chance' then expresses only our ignorance of the causes of the phenomena that we observe to occur and to succeed one another in no apparent order. Probability is relative in part to this ignorance, and in part to our knowledge.

Chapter 2


Learning Objectives By the end of this chapter, chapter, each student should be able to:  To

explain the basic concepts of probability  To count correctly  To compute for the probability of an event  To apply probability theorems and properties  To turn impossibilities into probabilities

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MODEL BUILDING Math and/or Stat is a continuous search for patterns. Naturally, we would like to describe this long-term pattern, or trend, in mathematical terms. It is possible to do this through the process called . It is not surprising that the major objective of many types of researches is actually the construction of such models. This is because these models will provide us with a better understanding of the phenomenon under study.

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ABSTRACT MODEL An is a description of the essential properties of a phenomenon that is formulated in mathematical terms. The use of abstract models is actually not new to us. We apply the mathematical formula provided by an abstract model to come up with an approximation of reality. In fact, most of us have already been exposed to these models in real life.

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DETERMINISTIC MODEL A is a type of abstract model that describes a phenomenon through known relationships among the states and events, in which a given input will always produce the same output. A deterministic model will always produce the same output if the given input is the same. The model does not leave any room for random variation. For this reason, we cannot describe the outcomes in a game of chance using this type of model.

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EXAMPLE The states that the pressure and volume of a gas may both vary simultaneously when the temperature of the gas is changed. This verbal model can be translated into a mathematical equation by writing:

PV = RT where P is the pressure of the gas, V is its volume, T is its temperature, and R is a constant called the gas constant.

Once we know the values assumed by two of the three variables P, V, or T, we can calculate the value of the third via this mathematical model. It is deterministic in the sense that it allows us to determine an exact value for the variable of interest under specified conditions.

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PROBABILISTIC MODEL A or is a type of abstract model that describes a phenomenon by assigning a likelihood of occurrence to the different possible outcomes of the process. The outcomes now depend on some random component. This random component is an additional input that is not present in any deterministic model which takes care of the variability in the output, even after fixing the values of all other inputs.

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EXAMPLE The Perfect Gas Law does describe some real gases at moderate temperatures and pressures. Unfortunately, many real gases cannot be described by this or any other deterministic model especially at extreme temperatures and pressures! Under these circumstances, we must find another way to predict the behavior of the gas with some degree of certainty. This can be done with the aid of statistical methods, namely, experimental designs and such.

V = T(X) + P(X) + T(X)P(X) + error

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RANDOM EXPERIMENT A is a process that can be repeated under similar conditions but whose outcome cannot be predicted with certainty beforehand. There are many examples of random experiments. Tossing a pair of dice, tossing a coin, selecting 5 cards from a well-shuffled deck of cards, and selecting a sample of size n from a population of N using a probability sampling method are some of them. All of these processes generate outcomes that may vary each time we repeat the process, even under exactly the same conditions. Regardless of the number of times we repeat the process, it is still not possible to determine in advance what the next outcome will be.

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SAMPLE SPACE The , denoted by Ω, is the collection of all possible outcomes of a random experiment. An element of the sample space is called a . The sample space is a set because it is a collection of elements. In set theory, this set is referred to as the universal set since it contains all elements under consideration. This is why it is only appropriate to denote the sample space by the last letter in the Greek alphabet omega.

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SPECIFYING A SET

Ω List down all the elements belonging in the set then enclosing them in braces.

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State a rule that the elements must satisfy in order to belong in the set then enclosing this rule in braces.


TREE DIAGRAM We can construct the tree diagram to help up . Step 1: It begins with as many nodes as there are possible outcomes on the first stage of the experiment. Step 2: We label each node by the notation used to represent each outcome. Step 3: Branches will emanate from each one of these nodes. The number of branches that will emanate from each node is equal to the number of possible outcomes on the second stage of the experiment. Step 4: Once again, we label each new node by the notation used to represent each outcome. The different paths along the branches of the tree represent the outcomes for an experiment with two stages.

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ILLUSTRATION Consider the random experiment of tossing a coin once. Using H to denote a head and T to denote a tail, we can specify the sample space by roster method as follows:

Ω

= {H, T}

We can also use the rule method to specify the same sample space as follows:

Ω

= {x | x ϵ {H,T}}

Now, to visualize the possible outcomes, we have the following tree diagram. • H

•

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T INTRODUCTION TO PROBABILITY

ILLUSTRATION Now, consider the random experiment of tossing a coin twice. Using H to denote a head and T to denote a tail, we can specify the sample space by roster method as follows:

Ω

= {HH, HT, TH, TT}

We can also use the rule method to specify the same sample space as follows:

Ω

= {(x,y) | x ϵ {H,T}, y ϵ {H,T}}

Now, to visualize the possible outcomes, we have the following tree diagram. H H T H T T

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ILLUSTRATION Recall that we defined our sample space previously as follows:

Ω

= {HH, HT, TH, TT}

But take note that, in the same experiment, instead of recording what comes up on the first and second tosses, what we can also do is to count the number of heads that will come up in the two tosses. In this case, the sample space is

Ω

= {0, 1, 2}

Or, sometimes the “pilosopo” will say that the sample space is

Ω

= {head, tail, edge} Chapter 2


SAMPLE SPACE  The

previous illustration shows that the description of the sample space is .

 There

are in which we can specify the collection of all possible outcomes of the experiment.



Which representation must we choose to use? Well, naturally, the choice depends on the characteristic of interest and .

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EXERCISES Specify a sample space for the following random experiments: a. rolling a die three times b. tossing a coin until “heads” appears c. selecting a person at random and observing his favorite color in the rainbow d. selecting a student at random and observing his grade in the exam (measured in percentage) e. counting the number of car accidents in a randomly selected day along the North Luzon Expressway

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D E S C R I P T I V E

S T A T I S T I C S

EVENT An is a subset of the sample space whose probability is defined. We say that an if the outcome of the experiment is one of the sample points belonging in the event. Otherwise, the event did not occur. We will use any capital Latin letter to denote an event of interest.

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ILLUSTRATION Consider the experiment of rolling a die. The sample space is given to be: Ω

= {1, 2, 3, 4, 5, 6}

Let A = event of observing odd number of dots in a roll of a die = {1, 3, 5} B = event of observing even number of dots in a roll of a die = {2, 4, 6} C = event of observing less than 3 dots in a roll of a die = {1, 2}

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IMPOSSIBLE AND SURE EVENT The The

∅

is the empty set . is the sample space Ω.

Two subsets of the sample space that will always be events are the empty set and the sample space. Their probabilities are always defined. Remember that an event occurs if the outcome of the experiment belongs in it. But is the empty set so it does not contain any sample point and thus it is impossible for this event to happen. On the other hand, Ω is the sample space so it contains all possible outcomes of the experiment and thus we are sure that it will always occur.

∅

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IMPOSSIBLE AND SURE EVENT

In a toss of a coin, it is impossible to observe a Jack of Hearts.

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In a roll of a die, it is a sure event that the number of dots is less than 7. INTRODUCTION TO PROBABILITY

OTHER EVENTS Aside from the impossible event and sure event, the other subsets of the sample space are also required to be events.

C A

A complement the collection of sample points in the sample space that do not belong in A

A

∪B

A union B

the collection of sample points that belong in at least one of A and B

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A∩B A intersection B

the collection of sample points that belong in both A and B.


OTHER EVENTS

∪ ∪∪

A1∩A2∩…∩An

the (finite) union of n events

the (finite) intersection of n events

A1 A2 … An the collection of sample points that belong in at least one of A1 , A2 , … , An

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the collection of sample points that belong in each one of A1 , A2 , … , An


ILLUSTRATION Consider the experiment of tossing a pair of colored dice, one is green and the other is red. For each sample point, the first coordinate represents the number of dots that comes up on the green die while the second coordinate represents the number of dots on the red die. Let Ω = {(x,y) | x ϵ {1,2,3,4,5,6} and y ϵ {1,2,3,4,5,6}}. This sample space contains 36 sample points.

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ILLUSTRATION The following following are considere considered d events: events: A = event of having the same number of dots on both dice = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)} B = event of 3 dots on the green die = {(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)}

∅

C = event of 7 dots on the green die = { } = .

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ILLUSTRATION The other events are: AC = {(1,2),(1,3),(1,4),(1, {(1,2),(1,3),(1,4),(1,5),(1,6),( 5),(1,6),(2,1),(2,3), 2,1),(2,3),(2,4),(2,5),(2 (2,4),(2,5),(2,6),(3,1), ,6),(3,1),(3,2),(3,4), (3,2),(3,4),(3,5),(3,6) (3,5),(3,6),, (4,1),(4,2),(4,3),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,6),(6,1),(6,2),(6,3),(6,4), (6,5)} A

∪ B = {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(3,1),(3,2),(3,4),(3,5),(3,6)}

A ∩ B = {(3,3)}

∪ B ∪ C = {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(3,1),(3,2),(3,4),(3,5),(3,6)} A B C=∅ A

∩

∩

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MUTUAL MU TUALL LY EX EXCLUS CLUSIV IVE E EVE EVENT NTS S Two events A and B are if and only if A ∩ B = , that is, A and B have no elements in common.

∅

The concept of mutually exclusive events can be extended to more than two events. Accordingly, Accordingly, any collection of events is said to be mutually exclusive if the collection is pairwise disjoint, which means that when one event in the collection occurs then any one of the other events in the collection cannot occur.

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EXAMPLE The following are examples of mutually exclusive events:

A∩B and

A and

C A

A and

C B ∩A

A and

C A∩B

∅

Their intersections are all equal to the null set.

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EXAMPLE In a toss of a coin, tails and heads can never happen or occur simultaneously.

In a roll of a die, observing an odd number of dots can never happen at the same time with observing an even number of dots.

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SET THEORY & PROBABILITY THEORY the sure event

the universal set, Ω Ω

the set A

A

Event A will occur Ω

A complement, AC

A

Event A will not occur Ω

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SET THEORY & PROBABILITY THEORY A union B, A

∪B

A

Event A or B will occur

B Ω

A union B union C, A B C

A

A intersection B, A ∩ B

A

∪ ∪

B C

Ω

B

At least one of A, B, and C will occur

Events A and B will occur Ω

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SET THEORY & PROBABILITY THEORY A intersection B intersection C, A ∩ B ∩ C

A

B C

A intersection B complement, A ∩ BC

A

A and B are disjoint

A

Ω

B Ω

B Ω

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All events A, B, and C will occur

Only event A will occur but not event B

Events A and B are mutually exclusive


EXERCISES A Japanese industry is considering on setting up plants in the Philippines. We define the following events: L = event of putting up a plant in Luzon V = event of putting up a plant in Visayas M = event of putting up a plant in Mindanao Express the following events as a composition of the events L,V, M. a. b. c. d. e.

event of putting up a plant both in Luzon and in Visayas event of putting up a plant in Luzon but not in Visayas event of not putting a plant in Luzon event of putting up a plant in at least one of the three subdivisions, Luzon, Visayas, and Mindanao. event of putting up a plant in all 3 geographic subdivisions

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S T A T I S T I C S

PROBABILITY The , denoted by P(A), is a function that assigns a measure of chance that event A will occur and must satisfy the following properties: (a) Nonnegativity . 0 ≤ P(A) ≤ 1 for any event A (b) Norming Axiom. P(Ω) = 1 (c) Countable Additivity . If A1, A2, A3, … is a sequence of mutually exclusive events, that is, A = A1 A2 A3 … , then

∪ ∪ ∪

P(A) = P(A1) + P(A2) + P(A3) + … OR

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INTERPRETATION A probability measure that is means that the event has a . On the other hand, if the probability measure is , then the event has a . A probability of , the midpoint of the interval [0,1], means that the event has a 50-50 chance of occurrence, that is, the chance that the In fact, if you are assigned a must always be

that an event is going to happen, then it must be . Similarly, the probability of the event .

Impossible

Hmmm…

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Certain


FINITE V.S. COUNTABLE

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EXERCISES The definition of probability function is useful not only in computing for probabilities but also in determining if our assignment of probabilities is valid or not. Find the errors in each of the following assignments of probabilities:

a. The probabilities that a couple will have 0, 1, 2, 3, or 4 or more children are, 0.42, 0.36, 0.25, 0.12, and -0.15 respectively. b. A person tosses a biased die three times. The probability that the sum of the number of dots in all three tosses is 2 is 0.0625. c. The probability that a selected student will pass the Physics exam is 0.35 and the probability that this student will fail the same exam is 0.62. d. The probabilities that a salesperson will sell exactly 0, 1, 2, or 3 items on any given day are, 0.23, 0.42, 0.25, and 0.20 respectively.

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S T A T I S T I C S

EXERCISES A die is loaded so that all the numbers have the same chances of occurrence except for a 6 whose chance of coming up is three times the chance of any other number coming up. Find the probabilities of the following events: a. event that a 6 comes up in a single toss b. event of observing an even number in a single toss c. event of observing a number less than 5 in a single toss

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S T A T I S T I C S

APPROACHES TO ASSIGNING PROBABILITIES

A priori or

Subjective

Classical

A posteriori or

Relative Frequency

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CLASSICAL PROBABILITY The method of using or assigns probabilities to events before the experiment is performed using the following rule: If an experiment can result in any one of the N different equally likely outcomes, and if exactly n of these outcomes belong to event A, then

no. of elements in A n P A = = no. of elements in Ω N Chapter 2


REMARKS •

•

•

•

A priori probability is also referred to as the “ ” because it was the first formula that provided a theoretical computation of probability. Its use is restricted to experiments whose sample space contains , and consequently, the sample space must have only a . of such experiments are the following: (i) die-throwing experiments where the die used is fair (ii) coin-tossing experiments where the coin used is balanced (iii) selecting n cards at random from a well-shuffled deck of cards (iv) selecting a sample of size n from a population of size N using simple random sampling P A =

no. of elements in A proportion of elements possessing = no. of elements in the population the characteristic of interest

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STEPS IN ASSIGNING PROBABILITY We assign probabilities to events using a priori probability by following these steps:

Step 1: Specify the sample space. Make sure that the outcomes are equiprobable. Step 2: Specify event A whose probability you are interested in. Step 3: Count the number of sample points in Ω and denote this by N. Count the number of sample points in event A and denote this by n.

Step 4: Compute for the probability of event A using the formula, n P A = . N

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EXAMPLE A fair coin is tossed 4 times. Find the probability of the following events. a. A = event of observing heads in the first 2 tosses b. B = event of observing exactly 2 heads c. C = event of observing at most 2 heads

{

Ω

=

}

H H H H , H H H T, H H T H , H H T T, H T H H , H T H T, H T T H , H T T T, T H H H , T H H T, T H T H , T H T T, T T H H , T T H T, T T T H , T T T T

Note that this contains N=16 equiprobable sample points since the coin is fair.

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EXAMPLE a. A = event event of observ observing ing heads in the the first 2 tosses tosses b. B = event event of observing observing exactly exactly 2 heads c. C = event event of observing observing at most 2 heads A = {HHHH, HHHT, HHHT, HHTH, HHTT} Since A contains n = 4 sample points then P(A) = 4/16 = 0.25. B = {HHTT, HTHT, HTTH, THHT, THTH, TTHH} Since B contains n = 6 sample points then P(B) = 6/16 = 0.375. C = {HHTT,HTHT,HTTH,HTTT,THHT,THTH,THTT,TTHH,TTHT,TTTH,TTTT} Since C contains contains n = 11 sample sample points points then P(C) P(C) = 11/16 11/16 = 0.6875 0.6875

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EXERCISES Consid Cons ider er the the expe experi rim men entt of dr draw awin ing g a card card from from a well well-shuffled standard deck of playing cards. Specify the following events using the roster method and compute for their a priori probabilities: a. A = event event of drawin drawing g an ace b. B = event event of drawing drawing a red card c. C = event event of drawin drawing g a heart heart d. D = event event of drawing drawing a king or a queen e. E = event event of drawin drawing g a Queen Queen of Hearts Hearts

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S T A T I S T I C S

RELATIV RELA TIVE E FRE FREQUE QUENCY NCY The method of using or assigns probabilities to events by repeating the experiment a large number of times and using the following rule: If a random experiment is repeated many times under uniform conditions, use the empirical probability of an event A to assign its probability as follows: no. of times event A occured empi empiri rica call P A = no. of times experiment was repeated

The definition of the probability of event A is the limiting value of its empirical probability if we repeat the process endlessly.

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REMARKS •

•

•

•

The empirical probability of event A will provide us with a for the probability of event A if the random experiment were repeated many times under uniform conditions. This empirical probability will be a good approximate of the actual probability if we perform the process a large number of times and under uniform conditions. In fact, for any event A, the a posteriori approach defines the . The advantage of using a posteriori probabilities instead of a priori probabilities is that its use is random experiments that generate a sample space containing . The advantage, on the other hand, of using a priori probabilities instead of a posteriori probabilities is that its use and can be determined prior it.

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SUBJECTIVE PROBABILITY assigns probabilities to events by using intuition, personal beliefs, and other indirect information. This method is more personal in its approach to assigning probabilities. The assignments may vary from one person to another, depending on their personal assessment of the available information on the situation at hand. But whatever the assignments may be, these measures must still conform to Kolmogorov’s definition of a probability. Ideally, we want to use objective methods in assigning probabilities. However, it is sometimes impractical or not at all possible to use a priori or a posteriori probabilities.

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RULES OF COUNTING

Combinatorial Analysis

• the branch of mathematics that is primarily concerned with determining the number of elements in a welldefined collection

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GENERALIZED PRINCIPLE OF COUNTING Suppose an experiment can be performed in k stages.

… 1st stage

2nd stage

3rd stage

kth stage

n1 distinct possible outcomes



nk distinct possible outcomes

Then there are of the experiment.

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EXAMPLE How many even 3-digit numbers can be formed from the digits 1, 2, 5, 6, and 9 if each digit can be used only once?

1 9

2 6

5

2 or 6

1st stage

2nd stage

3rd stage

selecting the ones digit

selecting the tens digit

selecting the hundreds digit

2

x

4

x

3

There are 24 possible outcomes.

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FACTORIAL The is a compact representation for the product of the first n consecutive positive integers. It is denoted by n!, read as “n factorial”, and

where n is a positive integer. We also define 0! = 1.

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PERMUTATION V.S. COMBINATION An of set Z is an ordered arrangement of r distinct elements selected from set Z. It can be represented by an ordered r-tuple with distinct coordinates. If set Z contains n distinct elements, then the number of r-permutations of set Z is denoted by P(n,r) or nPr read as “permutation n taken r”. An of set Z is a subset of set Z that contains r distinct elements. If set Z contains n distinct elements, then the number of r-combinations n of set Z is denoted by C(n,r) or read as “n taken r”. r

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EXAMPLE Suppose Z = {A, B, C, D}. List down all the possible 3-permutations of Z and all the possible 3-combinations of Z. The possible 3-permutations of Z are:

(A,B,C) (A,C,B) (B,A,C) (B,C,A) (C,A,B) (C,B,A)

(A,B,D) (A,D,B) (B,A,D) (B,D,A) (D,A,B) (D,B,A)

(A,C,D) (A,D,C) (C,A,D) (C,D,A) (D,A,C) (D,C,A)

(B,C,D) (B,D,C) (C,B,D) (C,D,B) (D,B,C) (D,C,B)

{A,C,D}

{B,C,D}

The possible 3-combinations of Z are:

{A,B,C}

{A,B,D}

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PERMUTATION V.S. COMBINATION The number of distinct r-permutations that we can form from the n distinct elements of set Z is

The number of distinct r-combinations that can be formed from the n distinct elements of set Z is

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EXAMPLE Suppose we toss a balanced die 4 times. a. How many equiprobable outcomes are there in the sample space? b. Let A be the event where no two tosses show the same number of dots. How many sample points are in A? c. What is P(A)? a. The experiment can be performed in 4 stages. In each stage, we observe the number of dots. By the Generalized Basic Principle of Counting, we have the following: 6 x 6 x 6 x 6 = 1,296 possible outcomes. b. Event A contains 4-permutations of the set Z = {1,2,3,4,5,6} because no two tosses show the same number of dots (i.e. the elements are distinct from each other.) Thus, P(6,4) = 6 x 5 x 4 x 3 = 360. c. We compute for P(A) using the classical definition of probability. no. of elements in A 360 P A = = 0.2778 no. of elements in Ω 1296

=

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EXAMPLE A nongovernment organization is awarding 5 scholarships to children of poor families. Fifty children are qualified for the scholarship. Among these 50 children, only 10 are boys while the other 40 are girls. a. How many ways can the nongovernment agency choose the 5 children who will be awarded the scholarship? b. Let A = event that the agency selects 4 boys and only 1 girl. How many sample points are in A? c. Assuming that the organization selected the children at random, find the P(A). a. We ignore the order in which the 5 children are selected. Thus, C(50,5) = 2,118,760. b. The experiment can be divided into two stages: (i) selection of boys and (ii) selection of girls. The first stage has C(10,4) = 210 possible outcomes while the second stage has C(40,1) = 40 outcomes. Thus, the number of sample points in A is (210)(40) = 8400. c. P A =

no. of elements in A 8,400 = no. of elements in Ω 2,118,760

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= 0.00396 I N T R O D U C T IO N T O P R O B A B I L I T Y

RULES OF COUNTING

Getting x number of objects out of n objects…

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USEFUL THEOREMS The number of distinct ways of arranging n objects of which n 1 are of one kind, n2 are of a second kind, …, nk are of kth kind is

k

where

 ni=n

i=1

The number of distinct ways of grouping n distinct objects into k groups such that n1 objects belong in the first group, n 2 objects belong in the second group, …, nk objects belong in the kth group is

k

where

 ni=n

i=1

Chapter 2


EXAMPLE Consider our favorite word, “STATISTICS” How many distinct ways can we arrange the letters contained in this word? 10! =50,400 3! 3! 1! 2! 1! How many ways can we assign twenty new applicants into the 5 committees of an organization so that each committee will get 4 new applicants each? 20! = 305,540,235,000 4! 4! 4! 4! 4!

Chapter 2


EXERCISES A man has four shirts. One is red, the others are yellow, white, and green. He has three pairs of pants. One is red, the others are white and blue. How many ways can he match his shirts with his pants? How many ways can we form a committee consisting of four members if we were to select them from the 15 officers of an organization? In how many different ways can 2 red, 3 yellow, and 4 blue bulbs be arranged in a string of Christmas tree lights with 9 sockets? In how many ways can the 5 starting positions on a basketball team be filled with 8 men who can play any position?

Chapter 1


S T A T I S T I C S

EXERCISES If a multiple choice test consists of 5 questions each with 4 possible answers of which only 1 is correct, a. How many different ways can a student answer the 5 questions? b. How many ways can a student answer all the 5 questions incorrectly? c. If the student is choosing the answers at random, what is the probability of getting a score of 0?

Chapter 1


S T A T I S T I C S

PROPERTIES OF PROBABILITY FUNCTION

1 2 3 4 5

If A is an event, then P(AC) = 1 – P(A). If A and B are events, then P(A BC) = P(A) – P(AB). If A and B are events, then P(A B) = P(A) + P(B) - P(AB). If A and B are mutually exclusive, then P(A B) = P(A) + P(B).

If A and B are events, then P(A B)c = P(AcBc). P(AB)c = P(Ac  Bc).

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A

A

A

–

P(A)

B

Ω

P(ABC) = P(A) P(AB) P(AB) = P(A) + P(B) - P(AB) –

A

A

B Ω

Ω

P(AB) = P(A) + P(B)

B

Ω

Ω

P(AC) = 1

A

B

P(AB)c = P(AcBc)

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B Ω

P(AB)c = P(Ac  Bc)


EXAMPLE Suppose A and B are events for which it is known that P(A) = 0.6, P(B) = 0.7 and P(A B) = 0.4. Find the following: a. P(AB) = P(A) + P(B) - P(AB) = 0.6 + 0.7 – 0.4 = 0.9 b. P(ABC) = P(A) - P(AB) = 0.6 – 0.4 = 0.2 c. P(BAC) = P(B) - P(AB) = 0.7 – 0.4 = 0.3 d. P((AB)C) = 1 - P(AB) = 1 – 0.4 = 0.6 e. P((A  B)C) = 1 – P(A  B) = 1 – 0.9 = 0.1

Chapter 2



6

Inclusion-Exclusion Formula

7

Monotonicity Property

If A, B, and C are events, then P(A BC) = P(A) + P(B) + P(C) - P(AB) – P(AC) – P(BC) + P(ABC).

If A and B are events, and B

Chapter 2

⊂ A, then P(B) ≤ P(A).


EVENT COMPOSITION METHOD Here, the probabilities are computed by expressing the event of interest as a composition of other events. Step 1: Define the basic events. The basic events are those events in the problem that cannot be expressed as a composition of other events. Step 2: List the known probabilities of events as stated in the problem. Step 3: Express the event of interest as a composition of the basic events using the set operations. Step 4: Use theorems or formulas for the computation of the probabilities.

Chapter 2


EXAMPLE The probability that a randomly selected student passes Stat 101 is 0.60 and the probability that he passes Math 17 is 0.85. If the probability that he passes at least one of the two courses is 0.95, a. b.

What is the probability that the selected student passes both courses? What is the probability that he will fail both Stat 101 and Math 17?

Let S = event that the selected student passes Stat 101 M = event that the selected student passes Math 17 Given:

P(S) = 0.60

P(M) = 0.85

P(S M) = 0.95

a.) P(SM) = P(S) + P(M) - P(S M) = 0.60 + 0.85 – 0.95 = 0.5 b.) P(ScMc ) = P(SM)c = 1 - P(S M) = 1 – 0.95 = 0.05

Chapter 2


EXERCISES Reniel, a health worker, is studying the prevalence of certain diseases in a particular community. Based on her previous studies, she came up with the following figures: 10% of the people in the community will contract disease A sometime during their lifetime; 25% will contract disease B; and 5% will contract both diseases. Find the probability that a randomly selected person from this community will contract: a. at least one of the 2 diseases b. disease B but not disease A c. exactly one of the 2 diseases

Chapter 1


S T A T I S T I C S

EXERCISES A smoke detector system uses two devices, A and B. If smoke is present, the probability that it will be detected by device A is 0.95; by device B, 0.90; and by both devices, 0.88. If smoke is present, find the probability of the following events: a. event that the smoke will be detected by at least one of A and B b. event that the smoke will not be detected c. event that the smoke will be detected by device A alone

Chapter 1


S T A T I S T I C S

CONDITIONAL PROBABILITY Let A and B be two events where P(B)>0. The of event A given the occurrence of event B, denoted by P(A|B), read as “probability of A given B”, is

P(A∩B) PAB = P(B) The definition of P(A|B) requires that P(B) must be positive. This means that the concept of conditional probability is left undefined when P(B) = 0 because division by 0 is undefined. We do not condition on an impossible event!!

Chapter 2


EXAMPLE In a previous example, the probability that a randomly selected student passes Stat 101 is 0.60, the probability that he passes Math 17 is 0.85, and the probability that he passes at least one of the two courses is 0.95. a. b.

If the selected student passed Stat 101, what is the probability that he passes Math 17? If the selected student received a failing grade in Stat 101, what is the probability that he passes Math 17?

a.) Find P(M|S). By definition, this is P M S = b.) Find P(M|Sc). By definition, this is C P(M∩S ) P M − P(M∩S) 0.85−0.5 C 1−0.60 1 − P(S) P(S )

=

=

Chapter 2

P(M∩S) 0.5 = = 0.8333. 0.60 P(S)

= 0.875.


PROPERTIES OF CONDITIONAL PROBABILITY The conditional probability is a probability function so it enjoys all of the properties of a probability function.

∅



P( |B) = 0



If A1, A2, …, An are mutually exclusive events, then P(A1 A2 ... An |B) = P(A1 |B) + P(A2 |B) + +P (An |B



If A is an event, then P(AC|B) = 1 – P(A|B).



If A and B are events, then P(ABC|C) = P(A|C) – P(AB|C).

∪ ∪∪

…



 If

A and B are events, then P(A B|C) = P(A|C) + P(B|C) P(AB|C).

Chapter 2


THEOREM OF TOTAL PROBABILITY If {B1, B2, …, Bn} is a collection of mutually exclusive events wherein each event has a nonzero probability and Ω = B1  B2  …  B n, then for any event A,

That is, P(A) Let’s take the simplest case!

P(A) = P(A|B)P(B) + Chapter 2

C C P(A|B )P(B )


BAYES’ THEOREM If {B1, B2, …, Bn} is a collection of mutually exclusive events wherein each event has a nonzero probability and Ω = B1  B2  …  Bn, then for any event A for which P(A)>0,

Let’s take the simplest case!

Chapter 2


B1

B4

B2

B3

B5 Chapter 2

B6


B1

B2 Chapter 2


EXAMPLE Statistics is very useful in analyzing biological or medical data. This area of Statistics is called Biostatistics. One special interest in biostatistics is to determine the effectiveness of a medical test in detecting a particular disease. The sensitivity of a given test is the probability of correctly diagnosing a person who has the disease while the specificity of a test is the probability of correctly diagnosing a person who does not have the disease. Naturally, we would want both the sensitivity and specificity of a test to be very close to 1. Suppose that it is known that only 0.005 of the people in a town have diabetes. A diabetes test is available. The sensitivity of this test is 0.999 and its specificity is 0.995. Suppose a person from this town is selected at random and the diabetes test was performed. a. What is the probability that the test will indicate that he has diabetes? b. If the test shows that the person has diabetes, what is the probability that the test is correct, that is, he really does have diabetes? c. If the test shows that the person does not have diabetes, what is the probability that the test is correct, that is, he really does not have diabetes?

Chapter 2


MULTIPLICATION RULE Let A1, A2, …, An be events for which P(A 1A2…An-1) > 0, then

Let’s take the simplest case!

P(AB) = P(A) P(B|A) or P(B) P(A|B)

Chapter 2


EXAMPLE There are 3 boxes, each containing 3 balls. A box is selected at random and a ball is selected from that box. What is the probability that Box 2 will be selected and the ball selected is white? Find P(AB). In the 1st stage, we randomly select a box from the 3 boxes so that P(A)=1/3. P(B|A) = 2/3 because if event A occurred then Box 2 was selected and we know that Box 2 contains 2 white balls and 1 black ball. Since we randomly select a ball from this box, then the probability of selecting a white ball given that Box 2 was selected is 2/3. Thus, P(AB) = P(B|A)P(A) = (2/3)(1/3) = 2/9

Chapter 2


INDEPENDENT EVENTS Two events A and B are said to be if and only if any one of the following conditions is satisfied: (i) P(A|B) = P(A) if P(B)>0; or (ii) P(B|A) = P(B) if P(A)>0; or (iii) P(AB) = P(A)P(B) Otherwise, the events are said to be

Chapter 2


MUTUALLY EXCLUSIVE V.S. INDEPENDENT

P(AB) = 0

mutually exclusive

P(AB) = P(A)P(B)

Chapter 2

independent


EXAMPLE The probability that Jerico will correctly answer the toughest question in an exam is ¼. The probability that Joshua will correctly answer the same question is 4/5. Find the probability of the following, assuming that they do not copy from each other: a. event that both Jerico and Joshua will answer the question correctly b. event that only Joshua will answer the question correctly Let A = event that Jerico will correctly answer the toughest question B = event that only Joshua will correctly answer the toughest question Given: P(A) = ¼

P(B) = 4/5

a.) P(AB) = P(A) P(B) = (1/4) (4/5) = 1/5 b.) P(BAC) = P(B) P(AC) = P(B) (1-P(A)) = (4/5) (1 – 1/4) = (4/5)(3/4) = 3/5

Chapter 2


AY DEPENDENT??

Chapter 2


AY DEPENDENT??

Chapter 2


REMARKS •

•

•

If A and B are independent events, then:  A and BC are independent.  AC and B are independent.  AC and BC are independent. If A1, A2, …, An are independent events, then the complements A1C, A2C, …, AnC are independent. Further, A1, A2C, …, AnC are independent, and so on. If there are more than 2 events, pairwise independence does not imply independence of all events. That is, if A and B are independent, A and C are independent, and B and C are independent, we cannot say that A, B, and C are independent.

Chapter 2


EXERCISES A movie critic feels that the probabilities that a certain movie will get an award for best actress is 0.18, for best actor is 0.33, and at least one of these two awards is 0.40. Suppose it was just announced that the movie won the best actor award, what is the probability that it will win the best actress award?

Chapter 1


S T A T I S T I C S

EXERCISES A random sample of 200 adults is classified below according to sex and the level of education attained. If a person is picked at random from this group, find the probability that the person a. is a male, given that the person has a secondary education b. does not have a college degree, given that the person is a female. Male Female Elementary 38 45 Secondary 28 50 College 22 17

Chapter 1


S T A T I S T I C S

EXERCISES The probability that a Japanese industry will locate in Cebu is 0.7. The probability that it will locate in Bataan is 0.3, and the probability that it will locate in at least one of the two provinces is 0.79. Define the following events: A = event that a Japanese industry will locate in Cebu B = event that a Japanese industry will locate in Bataan

Are A and B independent events?

Chapter 1


S T A T I S T I C S

EXERCISES Inquiries to an online computer system arrive on five communication lines. The percentage of messages received from lines 1, 2, 3, 4, 5, are 20, 30, 10, 15, and 25, respectively. The corresponding probabilities that the length of an inquiry will exceed 140 characters are 0.4, 0.6, 0.2, 0.8, and 0.9. What is the probability that a randomly selected inquiry will be longer than 140 characters?

Chapter 1


S T A T I S T I C S

EXERCISES Methods A, B and C are available for teaching a certain skill. Only one of these methods is used in teaching a particular worker. If used to teach a certain skill, Method A fails 30% of the time, Method B fails 20% of the time, while Method C fails 10% of the time. Also, Method B is used 20% of the time while Method C is used 40% of the time. Suppose a worker was selected at random and failed to learn the skill correctly, what is the probability that the worker was taught using Method B?

Chapter 1


S T A T I S T I C S

EXERCISES A computer program is tested by 3 independent tests. When there is an error, these tests will discover it with probabilities 0.2, 0.3, and 0.5, respectively. Suppose that the program contains an error. What is the probability that it will be found by at least one test?

Chapter 1


S T A T I S T I C S

ΩMG ACTIVITY!!

Summation! Top 3 Learning Points 3

 ith learning point

i=1

Chapter 2


02 Introduction to Probability

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