Chapter 13 & 14 - Probability PART

Chapter 13 & 14 - Probability

PART IV : PROBABILITY

Dr. Joseph Brennan

Uncertainty

Usually the results of a study, observational or experimental, are

uncertain: if we repeat a study, we will not get exactly the same results. Example 1 (A coin). You toss a coin. Will it landheads or tails?

Example 2 (A die). Aregular die, a D6, is a cube with six faces:

Uncertainty

Example 3 (Box with marbles) A box contains 7marbles: 2 red and5 green. Each marble has an equal chance to be selected. One marble is to be drawn at random from the box. What color will it be?

Example 4 (Box with tickets) A box contains10 tickets labeled 1 through 10. What will be the number of a randomly selected ticket? Example 5 (Height of students) Seven students will be selected at random from the Math 148 class list and their heights will be measured. What will be the average height? Will it change if we choose different seven students?

Probability

What is thechance or probabilitythat we will make the same conclusions every time when we replicate a study? What is the chanceor probability that the histogram will change?

Knowledge of probability theory will help us answer these questions.

NOTE: In this part, the wordschance andprobabilitywill have the same meaning.

Inferential Statistics & Probability Theory

In Parts II and III we used random samples to collect evidence and make inferences about population.

Sample mean, ¯x , estimates unknown population meanµ.

Sample standard deviation, s, estimates unknown population standard deviationσ.

Regression equation is a mathematical model which approximates the true relationship between x and y .

In this part we will think in the opposite direction; we will reason from a known population to randomly selected samples.

Population

Sample Sample

Population

Probability Theory _{Inferential Statistics}

Outcomes and Sample Space

Probability theory deals with studies where the outcomes are not known for sure in advance. Usually, there are many possible outcomes for a study, we just do not know which particular outcome we will observe.

Sample Space:

The set of all possible outcomes of a study. The sample space of a study is denoted byS.

Every repetition of a study, or a trial, produces a single outcome. Usually an outcome is computed from the values of the response variables.

In Example 5 (Height of students) the outcome is the average height,y¯, which is computed from the values of the response variable (y a student’s height).

Sample Spaces

Example 1 (A coin)

When you toss a coin, there are only two possible outcomes: a head or a tail. Then the sample space is

S ={ head, tail}or simply S={H, T}. Example 2 (A die)

The sample space is S = {1, 2, 3, 4, 5, 6}. Example 3 (Box with marbles)

We can draw either a red marble or a green one: S={R,G}.

Sample Spaces

Example 5 (Height of students) Of all our current examples, example 5 is the only continuous quantitative variable. The outcome of this study is y¯, an average height. For any sample of 7 students the average height could be any number between 40 inches and100inches. Then the sample space is the interval:

S = [40, 100]. Example 6 (Children)

For every randomly chosen family the number of boys and girls is recorded. The outcomes in the sample space are pairs of the form

Sample Space

Example 6 (Two dice)

Two dice are thrown. The sample space contains 36 outcomes as shown below:

Mathematically, the sample space can be written as

Probability of Outcomes

Probability:

The theoretical probability of an outcome is the proportion (or percent) of times an outcome occurs in the sample space. Every outcome has a probability.

If an experiment is repeated a large number of times, one would expect the ratio of occurrences of an outcome to number of experiments to be approximately the probability of that outcome.

Notation: We will denote the probability of an outcome as P(outcome),

Example 1 (A coin)

A single coin flip has a simple sample space: S = {H, T }. There are only two potential outcomes: {H}, {T }.

P(H) = number of H in sample space size of sample space =

1 2 P(T ) = number of H in sample space

size of sample space = 1 2

Is it correct to assume that after flipping 10 coins, 5 would have come up heads?

While imprisoned by the Germans during World War II, the South African statistician John Kerrich tossed a coin 10, 000times. Heads was the outcome5,067 times!

We have67 more heads then expected. The difference between the observed percentage and the anticipated result is known as error.

Example 1 (A coin)

The graph shows the percentage of heads minus 50%versus the number of trials in Kerrich’s experiment.

Example 1 (A coin)

From the graph we can see that the error approaches 0 as the number of tosses increases. After 10,000 tosses there are67 extra heads, so the error is:

67

10, 000· 100% = 0.67%.

It is already very close to 0! If we continue Kerrich’s experiment, we can expect the error to become even smaller. We say the error converges to 0, which implies that the observed percentage of heads approaches 50% which suggests that

P(H) = 0.5

has been calculated correctly.

Question Why is the proportion of heads in Kerrich’s experiment not EXACTLY equal to 0.5?

Example 1 (A coin)

Answer The number of heads is approximately half the number of tosses (5000), but it is off by 67 because of the chance error. The error 67 seems to be large in absolute units, but in relative units it is just 0.67%, which is a very small error.

CONCLUSION: Kerrich’s experiment supports the theoretical

P(H) = 0.5. The observed number of heads may be interpreted as

number of heads = half the number of tosses + chance error, where the error may appear to be large in absolute terms, but small relative to the number of tosses.

Example (Blood Distribution)

Data lifted from Wikipedia.

The following Table shows the distribution of Blood Types of people in Australia.

O+ A+ B+ AB+ O- A- B- AB- Total

40% 31% 8% 2% 9% 7% 2% 1% 100%

The above distribution was obtained by computing the proportions of blood types for a huge representative group of Australian people.

Therefore, we can say that the numbers in the table are the probabilities of different blood types for Australians.

Example (Blood Distribution)

1 What is the probability of AB+ blood for Australians? Answer: P(AB+)=0.02

If we randomly sample1000Australians, approximately20of them are expected to have AB+ blood type. However, due to chance

variability, we will not observeexactly20people with an AB+ blood type.

2 In July 2009the population of Australia was 28,395,716people. Approximately, how many people in Australia have an A+ blood type?

Answer: As the probability of having type A+ blood is31%, the population, disregarding chance variability, of is

Example 2 (A die)

An ace is the face of a die with one spot.

Question What is the probability of getting an ace in a single roll?

Answer:

P(ace)= number of faces with 1 number of faces =

1 6

Indeed, a die has 6 faces and every face has the same chance to be observed in a single roll. Then

P(1)=P(2)=P(3)=P(4)=P(5)=P(6)= 1

Equally Likely Outcomes

Assigning correct probabilities to individual outcomes often requires long observation of the random phenomenon. In some circumstances, however, we are willing to assume that individual outcomes are equally likely because of some balance in the phenomenon.

For the equally likely outcomes:

Probability of a single outcome = 1

Equally Likely Outcomes

Ordinary coins have a physical balance that should make heads and tails equally likely. So the outcomes in Example 1 are equally likely:

P(H)=P(T )= 1

2

A fair die in Example 2 is equally likely to show any of its 6 faces in a single roll.

P(1)=P(2)=P(3)=P(4)=P(5)=P(6)= 1

Equally Likely Outcomes

In Example 4, every one of the 10 tickets has the same chance to be drawn:

P(1)=P(2)=P(3)= · · · =P(8)=P(9)=P(10)= 1

10

In Example 6 (two dice), there are 36 equally likely outcomes. The probability of each outcome is ₃₆1.

In Example 3, the outcomes of drawing a red marble and drawing a green marble are not equally likely. There are more green marbles in the box, so the probability of selecting a green marble is greater. What is this probability?

Computing Probabilities of Outcomes

RULE:

In problems involving drawing at random, the probability of getting an object of a particular type in a single draw is equal to proportion of objects of this type in the population.

Example 3 (Box with marbles)

The population is the box with marbles. The proportion of red marbles is 2/7, the proportion of green marbles is 5/7. This implies that in a single draw

P(R)= 2

7 P(G)= 5 7

Notice that P(R) + P(G) = 1. We will explore this phenomenon later in this chapter.

Events

We will give the definition of an event in words and mathematically. Definition of an event in words: An event is some statement about the random phenomenon. Events are usually denoted by capital letters as A, B, etc.

Example of Events:

(a) A= At least one head in two tosses of a coin,

(b) B= Sum of the numbers on the dice is five,

(c) C= A die rolled once shows an even number.

Mathematical definition of an event: An event is a set of outcomes from the sample space S .

Examples of Events

(a) A= At least one head in two tosses of a coin.

The sample space for this experiment isS = {HH, HT , TH, TT }. The event

A = {HH, HT , TH}

(b) B =Sum of the numbers on the dice is five.

The sample space for this experiment contains36 equally likely outcomes which are discussed in Example 2. The event B is

B = { , , , }

(c) C = A die rolled once shows an even number.

The sample space is S = {1, 2, 3, 4, 5, 6}. The eventC is

Occurrence of an Event

We say that an event has occurred if ANYof the outcomes that constitute it occur.

For instance, in Example (c) if we roll a die and it shows 4, we say that the event C has occurred. If the die shows 3, the eventAdoes not occur because 3is an odd number, so the eventC does not contain this outcome.

Let Abe the event of landing at least one heads when tossing a coin 2 times, as in Example (a). If we toss and get {H, T } or {T , H} or {H, H} then we’ll say that Ahas occurred. Only when we have two tails appearing does Anot occur.

Probabilities of Events

The probability of an event Ais denoted byP(A).

In all our examples, except for Example 6, the sample space consisted of finitely many outcomes which we could list. In this case the sample space is called finite. An event is a collection of outcomes which support it.

Theorem 1 (Probability of an event in a finite sample space) The probability of any event is the sum of probabilities of the outcomes making up the event.

In the special case that the outcomes in the sample space are equally likely, the probability of an event E is computed as

P(E )= # of outcomes supporting E

Probabilities of Events

(a) A= At least one head in two tosses of a coin. There are4equally likely outcomes in this experiment. The probability of every outcome is 1 4. Then P(A)= P(HH) + P(HT ) + P(TH)= 1 4 + 1 4 + 1 4= 3 4.

(b) B = Sum of the numbers on the dice is five. There are 36 equally likely outcomes. Out of them,4 outcomes support the event B.

P(B)= 4 36 =

1 9.

(c) C = A die rolled once shows an even number. There are 6equally likely outcomes,3 of them supportC.

Example 3 (Box with Marbles)

A box contains 7 marbles, 2red and5 green. A marble is drawn at

random, its color is recorded, and the marble is put back in the box. Then, the next marble is drawn at random. This example is of sampling with replacement, which we will define later.

Find the probability that 2 red marbles are drawn. Solution: The sample space for this experiment is:

S = {RR,RG,GR,GG}

The outcomes are not equally likely since the red and green marbles have unequal chances to be selected. Intuitively, the RR is the least likely outcome. This outcome can be described:

Example 3 (Box with Marbles)

We can use the following two events to describe eventA: A1 = The first selected marble isred

A2 = The second selected marble is red

Then it follows that A = A1 and A2. Note that if we replace the marble back to the box, a red marble has the same chance to be selected at every draw:

P(A1) = P(A2) =

2 7

We have calculatedP(A1)and P(A2), but how can we use this to calculate P(A1 and A2)?

Special Events

We have the ability to calculate the probability of simple events; those resulting from one repetition of an experiment with equally likely

outcomes. To tackle more complex problems, we must define some special events.

Some special events:

Case 1: Sure-to-happen or certain event. Case 2: Impossible event.

Case 3: Opposite event.

Case 4: Mutually exclusive or disjoint events. Case 5: Independent events.

Certain Event

Certain Event:

An event which isguaranteed to happen at every repetition of the experiment.

The eventA = A head or a tail in a single toss of a coin. In every toss a coin lands up either heads or tails. There are no other possibilities. So Ais a certain event.

Notice thatA = {H, T }which coincides with the sample spaceS for this experiment.

The eventB = rolling a die once will show a number less than 7, is a certain event.

Notice thatB = {1, 2, 3, 4, 5, 6} is the whole sample space. A certain event is equal to the sample space.

Impossible Event

Impossible Event:

An event whichnever can occur.

The eventA = a coin lands up neither heads nor tailsis an impossible event.

In all the coin-related experiment we usually assume that the coin may not lay on the edge. Eliminating this possibility makes eventA impossible.

The eventB = the number that shows up when a die is rolled once is −2is an impossible event.

Opposite Event

Opposite Event:

An event B is opposite to Aif it happens whenever Adoesnot happen.

The event opposite to event Acontains all the outcomes from the sample space S which do not belong to A.

The opposite event for event Ais called not Aevent. An opposite event forA is denoted asA¯.

AandA¯split the sample space into two parts. (Also called a partition) Quite often, events are plotted on a graph as certain areas in the sample space. Such graphs are called the Venn diagrams.

Opposite Event on a Venn diagram

The Venn diagram shows the opposite event to the event Aas the shaded region.

Opposite Events: Examples

(a) EventA = At least one head in two tosses of a coin. The opposite event

¯

A = It is NOT true that at least one H in two coin tosses = No heads in two tosses of a coin

= {TT }

The probability is given by

P( ¯A)= P(TT ) =1

Opposite Events: Examples

(b) EventB = Sum of fivewhen rolling two dice. The opposite event: ¯

B = Sum of the numbers on the dice is NOT equal to five. ¯

B = {(1,1), (1,2), (1,3), (1,5),(1,6), (2,1), (2,2), (2,4), (2,5), (2,6), (3,1), (3,3), (3,4), (3,5), (3,6), (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)}. There are 32 equally likely outcomes inB¯. The sample space for this experiment contains36 equally likely outcomes. This implies

P( ¯B)= 32 36 =

8 9

Opposite Events: Examples

(c) EventC = A die rolled once shows an even number.

EventC can be described via its outcome space as C ={2, 4, 6}. The sample space for this experiment isS ={1, 2, 3, 4, 5, 6}. Then the event C¯ will contain all the outcomes fromS that C doesn’t contain. In particular,

¯

C ={1, 3, 5}

In words,C¯=A die rolled once shows an odd number. The probability P( ¯C )= 1₂.

Rule:

For an eventA,

Example 3 (Box with marbles)

A box contains 7 marbles: 2red and5 green. Two marbles are chosen with replacement. We found the sample space S = {RR,RG,GR,GG}. We reconsider the event

A=two redmarbles in two draws with replacement = {RR}. What is the opposite event?

It follows thatA¯= {RG,GR,GG} can be expressed as: ¯

A = NOT exactly tworedmarbles in two draws = At most onered marble in two draws = Not all the marbles in two draws werered

= At least onegreenmarble in two draws

P( ¯A)= 3 4 = 1 −

1

Mutually Exclusive Events

Mutually Exclusive Events:

Two events Aand B are mutually exclusive if they both cannot happen at the same time.

Examples of mutually exclusive events:

Events A= An odd numbered face showing and B = The 2 face showing are mutual events in a single die role.

Events A= The coin lands up heads andB = The coin lands up tails are mutually exclusive events in a single coin flip experiment.

Events Aand B are mutually exclusive if they do not have common outcomes. We say Aand B do not overlap or do not intersect. For this reason, mutually exclusive events are also called disjoint events.

Mutually Exclusive Events

A Venn diagram depicting two mutually exclusive eventsA andB in the sample space S.

Single outcomes of an experiment which make up the sample space are always mutually exclusive.

An event and its opposite are mutually exclusive: AandA¯are mutually exclusive.

Example (Two dice)

A pair of dice is rolled once. Consider the following events: A = Sum of numbers on the dice is 11. B = Both dice show an even number. C = Both dice show the same number. Are the events A and B mutually exclusive?

Are the events B and C mutually exclusive?

Let’s express the events as sets of outcomes and see if they have common outcomes.

A =

{

;

}

B =

{

;

;

;

;

;

;

;

;

}

Example (Two dice)

A =

{

;

}

B =

{

;

;

;

;

;

;

;

;

}

C =

{

;

;

;

;

;

}

.

Events AandB are mutually exclusive since there are no common outcomes.

Events B and C arenotmutually exclusive since they have three common outcomes.

Intersection of Events

IfA andB are not mutually exclusive events, they have some common outcomes. The set of common outcomes is an event which is called the

intersectionof events A andB. We will denote the intersection of events A andB by

A and B

Example:

B = { ; ; ; ; ; ; ; ; }

C = { ; ; ; ; ; }.

Denote by D the intersection of events B andC. Then

B and C=D = { ; ; }.

Intersection of Events

When we connect two events with AND, we are interested in their intersection.

The intersection of two mutually exclusive events is theimpossible event. If AandB are mutually exclusive, then

Example

Suppose you toss a fair coin twice. You are counting heads, so two events of interest are

A = First toss is a head, B = Second toss is a head. What is the probability of the intersection P(AandB)?

The events AandB are both disjoint, they occur together when both tosses give heads. Indeed, the sample space in this experiment is

S = {HH, HT , TH, TT } All the outcomes are equally likely. The events are

A = {HH, HT } B = {HH, TH} The intersection is A andB={HH}.

Independent Events

Independent Events:

Two events Aand B are independent of each other if knowing that one event occurs does not change the probability that the other event occurs.

Tosses of a fair coin are independent events. Further, it is equally likely that each side show after a toss.

Whether there are two heads in a row or twenty, the chance of getting a head next time is still0.5. It is the memoryless property of the coin.

A fair die also has the memoryless property, i.e., the rolls are independent. For instance, the probability of an ace at every roll is 1₆ no matter how many aces has appeared in the previous rolls.

We will assume that child births are independent events. Child births are independent events.

The actual probability of a boy is slightly higher than 0.5, but we will assume it to be 0.5.

Independent Events

Consider two events arising from rolling two dice: A = Getting a 6 on the first roll.

B =The sum of the numbers seen on the first and second rolls is 11

Are AandB independent events?

Solution 1: The event B consists of the following outcomes:

B =

{

;

}

.

The probability of event B isP(B)= ₃₆2 = ₁₈1.

Suppose eventA has happened. Then the possible outcomes for the second trial are 1, 2, 3, 4, 5, 6. EventB will happen if the second outcome was5. The probability of this is 1₆ which is larger than ₁₈1. Knowing that event Ahas happened changes the probability that eventB

Independent and Disjoint Events

If events AandB are independent, they can notbe mutually exclusive. Independent events always have common outcomes (intersect).

The opposite statement is also true: Mutually exclusive events are not independent (dependent).

Explanation: If we know thatAandB are mutually exclusive events with nonzero probabilitiesP(A)> 0andP(B)> 0, and we know that eventA

has happened, this implies that the probability that eventB also happened is0.

Example: Toss a coin twice. Two eventsA = First outcome is a headandB = Second outcome is a head are independent, but not mutually exclusive.

A = {HH, HT }, B = {TH, HH}

Events AandB are not mutually exclusive since they have a common outcome

Union of Events

Union:

The union of eventsA andB is the event which happens when either event A or eventB or both happen. The union of two events is expressed as

A or B

In mathematics, the word oris not inclusive, rather it is exclusive. Inclusive means either or, but not both; as in a restaurant. Exclusive means either or both.

In other words, the union of events A andB is the event which happens when at least one of events Aor B happens.

Union of Events: Venn diagram

Mathematically, the union of two eventsAandB is found by combining the outcomes fromAandB into a single setA or B.

The following Venn diagram illustrates the union of events AandB.

Example

Find the probability of event

E= at least one 3 in two successive rolls of a die. Solution: The event E consists of the following outcomes:

; ; ; ; ; ; ; ; ; ;

The sample size for the two dice problem has 36 equally likely outcomes. As a consequence:

P(E )= 11

Example

Observe that event E can be represented as the union of two events: A= First outcome is 3= {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)}

B = Second outcome is3= {(1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3)}

Therefore:

E = At least one 3in two successive rolls of a die = First outcome is3 or second outcome is3

= AorB

When combining the outcomes from two events into a single event, do

not repeat the same outcomes twice. Observe that each of the eventsA andB has6 outcomes, but their union (eventE) has11 outcomes. This is because A andB have one common outcome (3, 3)which is written just once in E. NoticeAand B= {(3, 3)}.

Rules of Probability

The following rules simplify many probability computations: Rule 1: The probability P(A) of any eventA satisfies

0 ≤ P(A) ≤ 1.

In other words, the probability of any event is between0 and1. Rule 2: If S is the sample space for an experiment, thenP(S ) = 1. The probability of a certain event is 1.

Rule 3: The probability of an impossible event is 0. Hence,

P(∅) = 0.

Rule 4: If events AandB are independent, then the probability that they both happen is the product of their probabilities:

Rules of Probability

Rule 5: For any events AandB, the probability of their union is equal to the sum of their individual probabilities minus the probability of their intersection:

P(A or B)=P(A)+P(B)−P(Aand B)

Subtracting the probability of the intersection is needed to avoid double counting.

A special case: ifAandB are disjoint events, then the probability of their union is the sum of individual probabilities:

P(A or B)=P(A)+P(B)

Rule 6: For any eventA

A Few Remarks

Some important notes about the rules of probability:

1. The closer the probability of an event is to 1, the more likely this event is to happen.

2. Unlikely events have probabilities close to0.

3. Outcomes partition the sample space:

Every cell in the above picture is an outcome.

4. Probabilities of all the outcomes add up to 1because the set of all outcomes is S andP(S ) = 1.

Example

We reconsider the following events arising from rolling a die twice:

E = At least one 3 in two rolls of a die A = The first outcome is 3

B = The second outcome is 3.

It should be clear that E = A orB, or thatE is the union of both Aand B.

The experiment which consists in throwing 2fair dice has36equally likely outcomes.

P(A)= 6 36 =

1

6 =P(B)

The intersection of events AandB is { } with probability P(A andB)= ₃₆1.

P(E )= P(A or B)=P(A)+P(B)−P(Aand B)= 1 6 + 1 6 − 1 36 = 11 36

Example

Problem 1. A total of 30% of American males smoke cigarettes, 7% smoke cigars, and 5% smoke both cigars and cigarettes. What percentage of males smoke neither cigars nor cigarettes?

Example (Blood Type - 2)

From Moore and McCabe

All human blood can beABO-typed as one ofA,B,O, or AB, but the distribution of the types varies among groups of people. Here is the distribution of blood types for a randomly chosen person in the US:

Blood type O A B AB

U.S. probability 0.45 0.40 0.11 ?

(a) What is the probability of type AB blood in the US? P(AB)= 1 − 0.45 − 0.40 − 0.11 =0.04

(b) Maria has type B blood. She can safely receive blood transfusions from people with blood typesO andB. What is the probability that a randomly chosen American can donate blood to Maria?

Example (Blood Type - 2)

Blood type O A B AB

U.S. probability 0.45 0.40 0.11 0.04

Consider the events:

O = Randomly selected person has O blood type B = Randomly selected person has B blood type.

Events O andB are disjoint since a person can not have typeO and type B of blood at the same time. We are interested in the probability that a randomly selected person has a blood type of either O or B type. Then

Example (Blood Type - 3)

The probability of randomly choosing an individual with blood typeA is 0.43 in the US and0.22 in China. What is the probability of randomly and independently choosing two people, one from the US and the other from China, that both have blood type A?

Solution: Define the following events:

X = An American has A blood type Y = A Chinese has A blood type

Notice that events X and Y are independent. We are interested in the probability of the event X and Y. By the multiplication rule for independent events

Extending the Rules of Probability

1 The multiplication rule for a collection of m mutually independent events.

For any collection of eventsA1, A2, . . . , Am that are mutually independent, the probability of the intersection of all the events is equal to the product of their individual probabilities:

P(A1 and A2and. . . and Am) = P(A1) · P(A2) ·. . . · P(Am). 2 The addition rule for m disjoint events.

If eventsA1, A2, . . . , Am are disjoint in the sense that they do not have common outcomes, then

Example (Universal Donors)

From Moore and McCabe

People with O-negative blood are universal donors. That is, any patient can receive a transfusion of O-negative blood. Only7% of the American population have O-negative blood. If3 people appear at random to give blood, what is the probability that at least one of them is a universal donor?

Solution: Let’s code the outcomes of the experiment in the following way: Ymeans a randomly selected person hasO-negative blood

Nmeans a person does NOT haveO-negative blood. Then the sample space for this experiment is as follows:

Example (Universal Donors)

We are given that P(Y ) = 0.07. Notice thatY andNare opposite events! Therefore

P(N)= 1 − P(Y ) = 1 − 0.07 =0.93. We are interested in the probability of the following event:

A = at least one of the three people is a universal donor = at least oneYin the outcome

= { NNY, NYN, YNN, NYY, YNY, YYN, YYY}

A has all the outcomes fromS except {NNN}, which should be the opposite event.

¯

A = None of the three people is a universal donor = NoYin the outcome

Example (Universal Donors)

S = {NNN, NNY , NYN, YNN, NYY , YNY , YYN, YYY }

A = {NNY, NYN, YNN, NYY , YNY , YYN, YYY } A = {NNN}¯

Recall that each outcome is the result ofthree independent events. Therefore, we may use the multiplication rule of mutually independent events:

P(A¯)= P(N and N and N) = P(N) · P(N) · P(N) = (0.93)3 _=0.804357.

As A¯is opposite Awe have

P(A)= 1 −P(A¯)= 1 − 0.804357 =0.195643. Notice that P(A)can be calculated without finding P( ¯A) by using multiplication and addition formulas:

P(A)= P(NNY or NYN or YNN or NYY or YNY or YYN or YYY ) = P(N)P(N)P(Y ) + P(N)P(Y )P(N) + P(Y )P(N)P(N) +. . .

Two Sampling Schemes

We will consider two important sampling schemes: Sampling with Replacement

Sampling without Replacement

We explore these sampling techniques through the following problem:

Problem. A box contains r red andg green marbles. Define the following events:

Ri = The marble selected on the i th draw is red. For example,R1is the event for which the first drawn marble isred. We are interested in the probabilities of events R1, R2, R3, . . ..

Since every marble has the same chance to be picked, we see that P(R1) =

r r + g

Two Sampling Schemes

What are the probabilities of events R2,R3, . . .? That depends on the

sampling scheme:

Sampling with replacement:

Suppose we return the marble to the box after each draw. In this case the conditions of experiment do not change since each subsequent draw is made from the same population.

1 _{Subsequent draws areindependent.}

2 _{The probability to select aredmarble is the same at every draw:}

P(Ri) = r

Two Sampling Schemes

Sampling without replacement:

After each draw the drawn marble is notreturned to the box.

Subsequently, conditions of the experiment change from draw to draw because the population of marbles in the box changes after each draw.

1 _{The probability of selecting aredmarblechangeswith each draw!} 2 _{Subsequent draws are NOT independent.}

Example (Box with Marbles)

A box contains 7 marbles -2 red and5green. We have considered the event A:

A= The first two chosen marbles are red. We used the following two events to describe event A:

A1 = The first selected marble isred,

A2 = The second selected marble is red. We found that A = A1 and A2

We know A1 but we are not sure how to calculateA2; is the sampling done with or without replacement?

Example (Box with Marbles)

Sampling with replacement: Subsequent draws are independent.

P(A1)=P(A2)=

2 7

⇒P(A)= P(A1) · P(A2) =

2 7 · 2 7 = 4 49

Sampling without replacement:

The probability of A1 is still 2₇. The probability of A2 depends on the outcome of the first draw:

The probability to select aredmarble on the second draw, given that the first selected marble wasred is 1₆.

The probability to select aredmarble given that the first selected marble wasgreenis 2

Conditional Probability

Conditional Probability:

Let AandB be events. If the events are not independent, then the occurrence of B alters the probability that A will occur. The conditional probability of eventA given that eventB has happened is denoted P(A|B).

Example: (Single roll of a fair die) Consider two events:

A= The die shows 3 B = The die shows odd. The ordinary, unconditional, probabilities of events AandB are

P(A)= 1

6 P(B)= 1 2

Suppose we know that event B happened. Then the conditional probability of Agiven B is

P(A|B)=1

3

Conditional Probability

The conditional probability of an event Agiven an event B is P(A|B) =P(A and B)

P(B)

A B

S

Conditional Probability: Examples

(a) Two machines, I and II, produce bolts. Five percent of those from I and ten percentof those from II are defective.

This can be written as

P(defective | machine I)=0.05 P(defective | machine II)=0.10.

Notice that we do not know P(defective), nor can we calculate it with the information supplied.

(b) Suppose a mortality table shows that the probability of dying within one year for a 25-year-old male is0.00193.

This can be stated as

Conditional Probability

Note 1: Knowledge thatB has occurred effectively reduces the sample space fromS to B. This is sometimes called the reduced sample space. Therefore, when interpreting the area of an event on the Venn diagram as its probability, P(A|B)is the proportion of the area ofB occupied byA.

Note 2: If events AandB are mutually exclusive, thenP(A|B) = 0 as P(A and B) = 0.

Note 3. Recall the formula:

P(A|B) = P(A and B) P(B) With algebraic manipulation we find:

The General Multiplication Rule

As andis commutative,

P(A|B) · P(B)=P(A and B)=P(A and B)=P(B|A) · P(A) The order of conditioning may be changed if needed.

Extension of the General Multiplication Rule:

The events A1, A2, . . . , An are not necessarily independent.

P(A1 and A2 and A3 and A4 and . . .)=

Intersection of 3 Events on a Venn Diagram

The intersection of three events A,B, and C has probability

P(A and B and C )=P(A) · P(B|A) · P(C |A and B)

A

B

S

C

Example (Athletes)

(From Moore and McCabe)

Only 5%of male high school basketball, baseball, and football players go on to play at the college level. Of these, only 1.7%enter major league professional sports. About40%of the athletes who compete in college and then reach the pros have a career of more than3years. Define these events

A = Competes in college, B = Competes professionally,

C = Professional career longer than 3 years.

What is the probability that a high school male athlete competes in college, reaches a professional level and then goes on to have a pro career of more than 3 years?

Example (Athletes)

We are given that

P(A)=0.05

P(B|A)=0.017

P(C |A and B)=0.4. The probability we want is, therefore,

P(A and B and C ) = P(A) · P(B|A) · P(C |A and B) = 0.05 · 0.017 · 0.4

= 0.00034

Interpretation: Only about 3of every 10,000 high school male athletes can expect to compete in college, reach a professional level and have a professional career for more than 3years.

Conditional Probability and Independence

When events AandB are independent, knowing that eventB has

occurred gives no additional information about the occurrence of event A. This can be expressed as

P(A|B) = P(A)

To check whether two events Aand B are independent, you should check either equality: P(A|B) = P(A)or P(A and B) = P(A) · P(B).

Example (Degrees)

(From Moore and McCabe)

The counts (in thousands) of earned degrees in the US in the 2005-2006 academic year, classified by level and by the sex of the degree recipient:

Bachelor’s Master’s Professional Doctorate Total

Female 784 276 39 20 1119

Male 559 197 44 25 825

Total 1343 473 83 45 1944

Example (Degrees)

(a) If you choose a degree recipient at random, what is the probability that the person that you choose is a woman?

P(W ) = 1119 thousands

1944 thousands ≈0.5756.

(b) What is the probability of choosing a woman, given that the person chosen received a professional degree?

P(W |P) = 39

83 ≈0.4699.

(c) Are the events choose a woman and choose a professional degree recipient independent? Why or why not?

We found in (a) that P(W ) = 0.5756. In (b) we found

P(W |P) = 0.4699. SinceP(W ) 6= P(W |P), the events arenot

Example (Degrees)

(d) A randomly chosen person is a man. What is the probability that he received a bachelor’s degree?

P(B|M) = 559

825 ≈0.6776.

(e) Use the multiplication rule to find the probability of choosing a male bachelor’s recipient. Check your result by finding this probability directly from the table of counts.

P(M and B) = P(B|M) · P(M) = 559 825· 825 1944 = 559 1944,

Example (Box with Marbles)

Recall that there are 2 red and5green marbles in a box. We defined the following events:

A= Two randomly chosen marbles are red A1 = First selected marble is red

A2 = Second selected marble is red. We found A = A1 and A2, and

P(A1)=P(A2)=

2 7.

We want to compute the probability of Aunder sampling without replacement:

P(A) = P(A1 and A2)=P(A2|A1)P(A1)=

1 6 · 2 7 = 1 21.

We previously calculated that the probability of event Aunder sampling with replacement is ₄₉4. Since ₄₉4 > 1

21, event Ais more likely to occur

Examples

Example: Two birds are selected at random without replacement from a cage containing five male and two female finches. What is the probability that both are males?

P(M and M) = P(M) · P(M|M) = 5 7 · 4 6 = 10 21

Example: A large basket of fruit contains3 oranges,2 apples and 5 bananas. If two fruit are chosen at random without replacement, what is the probability that one of the selected fruits is an apple and the other one is an orange?

P(A and O) = P(A) · P(O|A) = 2 10·

3 9 =

1 15

Examples

Example: A box of10 items has 2 defective items. Three items are selected by the researcher without replacement.

(a) What is the probability that the researcher will obtain no defective items?

P( ¯D and ¯D and ¯D) = P( ¯D) · P( ¯D| ¯D) · P( ¯D| ¯D and ¯D) = 8 10 · 7 9· 6 8 = 7 15

(b) Given that the researcher finds the first item selected as defective, what is the probability that the researcher will also have the other defective item? P( ( ¯D and D) |D) = P( ¯D|D) · P(D| ¯D and D) = 8 9· 1 8 = 1 9

Example (Roulette)

Example 20 (Roulette), from Moore and McCabe

A roulette wheel has 38 slots, numbered0,00, and1 through 36. The slots0and00 are coloredgreen,18 of the others arered, and18 are black. The dealer spins the wheel and at the same time rolls a small ball along the wheel in the opposite direction. The wheel is carefully balanced so that the ball is equally likely to land in any slot when the wheel slows. Gamblers can bet on various combinations of numbers and colors. (a) What is the probability that the ball will land at any one slot?

P(any slot)= 1

Example (Roulette)

(b) If you bet on red, you will win if the ball lands on a red slot. What is the probability of winning?

P(red)= # of red slots

38 = 18 38 = 9 19 < 1 2

Example (Roulette)

(c) The slot numbers are laid on a board on which gamblers place their bets. One column of numbers on the board contains all multiples of3, i.e.,

3,6,9, ..., 36. You place acolumn betthat wins if any of these numbers comes up. What is you probability of winning?

P(Slot # is a multiple of 3)= 12

38 =

6 19 <13.

In fact, every game in a casino offers options to the gambler that has less than a 50%chance of winning. In later chapters we will discuss probability and payout.

Example (Lottery)

(From Moore and McCabe.)

A state lottery’s Pick 3 game asks players to choose a three-digit number,

000to 999. The state chooses the winning three-digit number at random. You win if the winning number contains the digits in you number, in any order.

(a) Your number is123. What is your probability of winning? (b) Your number is 112. What is your probability of winning?

Solution: First of all, note that the sample space S for the experiment, which consists in choosing a 3-digit number at random, contains1000 equally likely outcomes:

Example (Lottery)

(a) Consider the eventA: a chosen number has digits 1, 2, and 3. A = {123, 132, 213, 231, 312, 321}

Then P(A)= ₁₀₀₀6 =0.006, quite a small chance!

(b) Define the event B: a chosen number has two 1’s and one 2. B = {112, 121, 211}

Then P(B)= ₁₀₀₀3 =0.003.

Concluding Remarks

Observation 1: and

andmeans that we are looking at the intersection of events. We need to compute the probability that ALL events in the statement occur simultaneously. Probability computations in this case usually involve using one of the multiplication rules.

Observation 2: or

ormeans that we are looking at the union of events. We need to compute the probability thatat least oneevent in the union occurs. Probability computations in this case usually involve either one of the addition rules or the rule for opposite events.

Observation 3: at least,at most,not exactly,not equal

Most of the problems of this type reduce to computing probabilities of the unions of events. It is often easier to compute the probability of the opposite event first.