1 probability part 1 – definitions * event * probability * union * intersection * complement part...

1

Probability

ProbabilityProbability

Part 1 – Definitions* Event* Probability* Union* Intersection* Complement

Part 2 – Rules

Part 1 – Definitions* Event* Probability* Union* Intersection* Complement

Part 2 – Rules

2

Probability

Definitions - EventDefinitions - Event

* An event is a specific collection of possible outcomes.* For example, the event ‘A’ could be the event that one Head occurs in two tosses of a coin.

* An event is a specific collection of possible outcomes.* For example, the event ‘A’ could be the event that one Head occurs in two tosses of a coin.

HH HT TH TT

A

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Probability

Definitions - ProbabilityDefinitions - Probability

The probability of an event is the ratio of the number of outcomes matching the event description to the number of possible outcomes.

P(1 head in 2 tosses) = 2/4 = .5

Probabilities are ALWAYS between 0 and 1.

The probability of an event is the ratio of the number of outcomes matching the event description to the number of possible outcomes.

P(1 head in 2 tosses) = 2/4 = .5

Probabilities are ALWAYS between 0 and 1.

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Probability

Definitions - UnionDefinitions - Union

The union of two events A and B is the event that occurs if either A or B or both occur on a single measurement.

* For example, suppose:* A = a car has two doors* B = a car is red

* Then, if we randomly select a car from a large parking lot, P(A U B) is the probability that the car is either a two-door model or red or both two-door and red.

The union of two events A and B is the event that occurs if either A or B or both occur on a single measurement.

* For example, suppose:* A = a car has two doors* B = a car is red

* Then, if we randomly select a car from a large parking lot, P(A U B) is the probability that the car is either a two-door model or red or both two-door and red.

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Probability

Definitions – UnionDefinitions – Union

Define:

C = being female

D = being nineteen years old

What is P(C U D) for a randomly selected person from among the set of people currently in this room?

Define:

C = being female

D = being nineteen years old

What is P(C U D) for a randomly selected person from among the set of people currently in this room?

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Probability

Definitions – UnionDefinitions – Union

# of women = ________

plus

# of 19 year olds = ________

minus

# of 19 year old women = ________

equals

# either 19 or a woman or both = ________

# of women = ________

plus

# of 19 year olds = ________

minus

# of 19 year old women = ________

equals

# either 19 or a woman or both = ________

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Probability

Definition – UnionDefinition – Union

Total # of people in this room = _________

Therefore, P(C U D) = __________ =

Total # of people in this room = _________

Therefore, P(C U D) = __________ =

8

Probability

Definitions – IntersectionDefinitions – Intersection

The intersection of two events A and B is the event that occurs if and only if both A and B occur on a single measurement.

* Suppose we randomly select a person from the set of people currently in this room. With A and B defined as before, the probability that a randomly-selected person is a 19 year old woman is:

P(A ∩ B) =

The intersection of two events A and B is the event that occurs if and only if both A and B occur on a single measurement.

* Suppose we randomly select a person from the set of people currently in this room. With A and B defined as before, the probability that a randomly-selected person is a 19 year old woman is:

P(A ∩ B) =

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Probability

Definitions – ComplementDefinitions – Complement

The complement of an event A is the event that A does not occur.The complement of an event A is the event that A does not occur.

HH TT

HT TH

A’

A

10

Probability

Definitions – ComplementDefinitions – Complement

P(A’) = 1 – P(A)

P(A) + P(A’) = 1

The second equation says, “Either A happens or A doesn’t happen. There are no other possibilities.” That may seem obvious, but keep it in mind on exams.

P(A’) = 1 – P(A)

P(A) + P(A’) = 1

The second equation says, “Either A happens or A doesn’t happen. There are no other possibilities.” That may seem obvious, but keep it in mind on exams.

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Probability

Part 2 – RulesPart 2 – Rules

* Additive Rule* Mutually Exclusive Events* Conditional Probability* Multiplicative Rule* Independence

* Additive Rule* Mutually Exclusive Events* Conditional Probability* Multiplicative Rule* Independence

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Probability

Rules – Additive RuleRules – Additive Rule

P(A U B) = P(A) + P(B) – P(A ∩ B)

* Recall 19 year old women example.

* When you add # of 19 year olds to # of women, you count the 19 year old women in both groups – so you count them twice. Subtract that number once as a correction.

P(A U B) = P(A) + P(B) – P(A ∩ B)

* Recall 19 year old women example.

* When you add # of 19 year olds to # of women, you count the 19 year old women in both groups – so you count them twice. Subtract that number once as a correction.

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Probability

Rules – Mutually Exclusive EventsRules – Mutually Exclusive Events

Two events are mutually exclusive when

P(A ∩ B) = 0

and

P(A U B) = P(A) + P(B)

Be careful! Note that the first equation uses ∩ while the second one uses U.

Two events are mutually exclusive when

P(A ∩ B) = 0

and

P(A U B) = P(A) + P(B)

Be careful! Note that the first equation uses ∩ while the second one uses U.

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Probability

Rules – Conditional ProbabilityRules – Conditional Probability

When you have information that reduces the set of possible outcomes, you work with new probabilities that are conditional on that new information.

*Remember that the # of possible outcomes is the denominator of the ratio that gives probability of an event.*The probability changes on the basis of new information because the numerator stays the same but the denominator decreases.

When you have information that reduces the set of possible outcomes, you work with new probabilities that are conditional on that new information.

*Remember that the # of possible outcomes is the denominator of the ratio that gives probability of an event.*The probability changes on the basis of new information because the numerator stays the same but the denominator decreases.

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Probability


Example: Suppose before class I picked a card at random from a standard deck of cards. What is P(E) if:

E = Card I picked is a Club?

* Note that this is a question about the ordinary (non-conditional) probability.

Example: Suppose before class I picked a card at random from a standard deck of cards. What is P(E) if:

E = Card I picked is a Club?

* Note that this is a question about the ordinary (non-conditional) probability.

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Probability


P(E) = 13/52 = .25 (Do you see why?)

Now, suppose I tell you that the card I picked is black. What is the conditional probability that the card is a club given that it is black?

P(E │black) = 13/26 = .5

* Only 26 cards in a normal deck are black.

P(E) = 13/52 = .25 (Do you see why?)

Now, suppose I tell you that the card I picked is black. What is the conditional probability that the card is a club given that it is black?

P(E │black) = 13/26 = .5

* Only 26 cards in a normal deck are black.

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Probability


We write:

P(A │B) = P(A ∩ B)

P(B)

We write:

P(A │B) = P(A ∩ B)

P(B)

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Probability

A

B

A ∩ B

P(A │B) = P(A ∩ B) P(B)

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Probability

Rules – Multiplicative ProbabilityRules – Multiplicative Probability

P(A ∩ B) = P(A) * P(B│A)

P(A ∩ B) = P(B) * P(A│B)

P(A ∩ B) = P(A) * P(B│A)

P(A ∩ B) = P(B) * P(A│B)

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Probability

Rules – Multiplicative ProbabilityRules – Multiplicative Probability

To see why, begin with the conditional probability formula, and multiply both sides by either P(A) or P(B):

P(A │B) = P(A ∩ B)

P(B)

P(B │A) = P(A ∩ B)

P(A)

To see why, begin with the conditional probability formula, and multiply both sides by either P(A) or P(B):

P(A │B) = P(A ∩ B)

P(B)

P(B │A) = P(A ∩ B)

P(A)

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Probability

Rules - IndependenceRules - Independence

Events A and B are independent if the occurrence of one does not alter the probability of the other.

P(A│B) = P(A)

P(B│A) = P(B)

Events A and B are independent if the occurrence of one does not alter the probability of the other.

P(A│B) = P(A)

P(B│A) = P(B)

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Probability

Rules – IndependenceRules – Independence

We can now re-write the multiplicative rule for the special case of independent events:

P(A ∩ B) = P(A) * P(B)

* This is because P(B│A) = P(B) for independent events.

We can now re-write the multiplicative rule for the special case of independent events:

P(A ∩ B) = P(A) * P(B)

* This is because P(B│A) = P(B) for independent events.

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Probability

Probability – ExamplesProbability – Examples

60% of Western students are female. 60% of female students have a B average or better. 80% of male students have less than a B average.

a. What is the probability that a randomly selected student will have less than a B average?

60% of Western students are female. 60% of female students have a B average or better. 80% of male students have less than a B average.


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Probability

A = Being femaleB = Having a B average or better

A

A’

.6

.4

B

B’

B

B’.8

.2

.6

.4

.6 x .6 = .36

.6 x .4 = .24

.4 x .2 = .08

.4 x .8 = .32

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Probability



P(B’) = P(B’ ∩ A) + P(B’ ∩ A’)

* Either we get (B’ and A) or we get (B’ and A’).* That is, either our randomly selected student who has less than a B average is a female or he is a male.


P(B’) = P(B’ ∩ A) + P(B’ ∩ A’)

* Either we get (B’ and A) or we get (B’ and A’).* That is, either our randomly selected student who has less than a B average is a female or he is a male.

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Probability


P(B’) = P(B’ ∩ A) + P(B’ ∩ A’)

By Multiplicative Rule:

P(B’) = P(B’│A)*P(A) + P(B’│A’)*P(A’)

= (.4*.6) + (.8*.4)

= .56

P(B’) = P(B’ ∩ A) + P(B’ ∩ A’)

By Multiplicative Rule:

P(B’) = P(B’│A)*P(A) + P(B’│A’)*P(A’)

= (.4*.6) + (.8*.4)

= .56

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Probability

Reminder – Multiplicative ProbabilityReminder – Multiplicative Probability

P(A ∩ B) = P(A) * P(B│A)

P(A ∩ B) = P(B) * P(A│B)

P(A ∩ B) = P(A) * P(B│A)

P(A ∩ B) = P(B) * P(A│B)

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Probability


b. If we randomly select a student at Western and note that this student has a B or better average, what is the probability that the student is male?

b. If we randomly select a student at Western and note that this student has a B or better average, what is the probability that the student is male?

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Probability

A = Being femaleB = Having a B average or better

A

A’

.6

.4

B

B’

B

B’.8

.2

.6

.4

.6 x .6 = .36

.6 x .4 = .24

.4 x .2 = .08

.4 x .8 = .32

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Probability


We know that overall, 40% of students (A’) are male. But among the students who get a B or better average, what proportion are male?

The probability of getting a B or better average is .44 (from .36 for women and .08 for men). Thus,

P(A’│B) = P(A’ ∩ B) = .08 = .1818

P(B) .44

We know that overall, 40% of students (A’) are male. But among the students who get a B or better average, what proportion are male?

The probability of getting a B or better average is .44 (from .36 for women and .08 for men). Thus,

P(A’│B) = P(A’ ∩ B) = .08 = .1818

P(B) .44

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Probability


You’re on a game show. You’re given a choice of 3 doors you can open. Behind one door is a car. Behind each of the other two doors is a goat. You win what is behind the door you open. You pick a door, but don’t get to open it yet. The host opens another door, behind which is a goat. The host then says to you, “Do you want to stick with the door you chose or switch to the other remaining door?”

Is it to your advantage to switch?

You’re on a game show. You’re given a choice of 3 doors you can open. Behind one door is a car. Behind each of the other two doors is a goat. You win what is behind the door you open. You pick a door, but don’t get to open it yet. The host opens another door, behind which is a goat. The host then says to you, “Do you want to stick with the door you chose or switch to the other remaining door?”

Is it to your advantage to switch?

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Probability


Yes – you should switch to the other door. The door you originally chose has a 1/3rd chance of winning the car. The other remaining door has a 2/3rd chance of winning the car.

Yes – you should switch to the other door. The door you originally chose has a 1/3rd chance of winning the car. The other remaining door has a 2/3rd chance of winning the car.

33

Probability


Suppose you originally pick Door #1. The door the host opens is indicated by the boldface type below. What you win if you switch is underlined.

Door #1 Door #2 Door #3

Goat Goat Car

Goat Car Goat

Car Goat Goat

Suppose you originally pick Door #1. The door the host opens is indicated by the boldface type below. What you win if you switch is underlined.

Door #1 Door #2 Door #3

Goat Goat Car

Goat Car Goat

Car Goat Goat

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Probability

Probability - ExamplesProbability - Examples

When you picked Door #1, there was a 2/3rd probability that the car was behind one of the other doors.* That is still true after the host opens one of those other doors.* But since the host knows where the car is, he opens a door that has a goat behind it. So now the 2/3rd probability is all associated with the one remaining door.* The critical point is that the host has information – so we are dealing with the conditional probability of the car being behind Door #3 GIVEN THAT the host opened Door #2 (after you picked Door #1).

When you picked Door #1, there was a 2/3rd probability that the car was behind one of the other doors.* That is still true after the host opens one of those other doors.* But since the host knows where the car is, he opens a door that has a goat behind it. So now the 2/3rd probability is all associated with the one remaining door.* The critical point is that the host has information – so we are dealing with the conditional probability of the car being behind Door #3 GIVEN THAT the host opened Door #2 (after you picked Door #1).

1 probability part 1 – definitions * event * probability * union * intersection * complement part...

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