probability calculations

8/10/2019 Probability Calculations

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ProbabilityMirza Amin ul Haq


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ProbabilityThe calculated likelihood that a given eventwill occur


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Methods of Determining ProbabilityEmpirical

Experimental observationExample Process control

TheoreticalUses known elementsExample Coin toss, die rolling

Subjective AssumptionsExample I think that . . .


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Probability Components

Experiment An activity with observable results

Sample Space A set of all possible outcomes

Event A subset of a sample space

Outcome / Sample Point The result of an experiment


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ProbabilityWhat is the probability of a tossed coin

landing heads up?

Probability Tree

Experiment

Sample Space

Event

Outcome


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Relative FrequencyThe number of times an event will occur

divided by the number of opportunities= Relative frequency of outcome x

= Number of events with outcome x

= Total number of events

x

x

n f N

Expressed as a number between 0 and 1fraction, percent, decimal, odds

Total frequency of all possible events totals 1

x f

xn

N


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Probability

x x

a

F P F

What is the probability of a tossed coin

landing heads up?

How many possibleoutcomes? 2

How many desirableoutcomes? 1

12

P .5 50%

Probability Tree

What is the probability of the coinlanding tails up?


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Probability

x x

a

F P

F

How many possibleoutcomes?


14

P

What is the probability of tossing a coin

twice and it landing heads up both times?

4

HH

HT

TH

TT

.25 25%


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Probability

x x

a

F P

F

How many possibleoutcomes?


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P

What is the probability of tossing

a coin three times and it landingheads up exactly two times?

8

1 st

2nd

3 rd HHH

HHT

HTH

HTT

THH

THT

TTH

TTT

.375 37.5%


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Law of Large Numbers

Trial 1: Toss a single coin 5 timesH,T,H,H,TP = .600 = 60%

Trial 2: Toss a single coin 500 timesH,H,H,T,T,H,T,T,T P = .502 = 50.2%

Theoretical Probability = .5 = 50%

The more trials that are conducted, the closer

the results become to the theoretical probability


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Probability

Independent events occurring simultaneouslyProduct of individual probabilities

If events A and B are independent, then theprobability of A and B occurring is:

P = P(A) P(B)

AND (Multiplication)


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Probability AND (Multiplication)What is the probability of rolling a 4 on a single die?

How many possible outcomes?How many desirable outcomes? 1

6 416

P

What is the probability of rolling a 1 on a single die?


6 116

P

What is the probability of rolling a 4 and then a1 using two dice?

4 1P = (P )(P ) 1 1=

6 6

1.0278

36

2.78%


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Probability

Independent events occurring individuallySum of individual probabilities

If events A and B are mutually exclusive, thenthe probability of A or B occurring is:

P = P(A) + P(B)

OR (Addition)


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ProbabilityOR (Addition)What is the probability of rolling a 4 on a single die?


6 416

P

What is the probability of rolling a 1 on a single die?


6 116

P

What is the probability of rolling a 4 or a 1 on asingle die?

4 1( ) ( ) P P P 1 1

6 6

2 .3333 33 3 %6

. 3


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Probability

Independent event not occurring1 minus the probability of occurrence

P = 1 - P(A)

NOT

What is the probability of not rolling a 1 on a die?

11 P P 11 6 5 .8333 83 3 %

6. 3


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How many tens are in a deck?

ProbabilityTwo cards are dealt from a shuffled deck.

What is the probability that the first card is anace and the second card is a face card or aten?

How many cards are in a deck? 52

4

12

4

How many aces are in a deck?

How many face cards are in deck?


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Probability

What is the probability that the first card is an ace?4 1

.0769 7.69%52 13

12 4.2353 23.53%

51 17

Since the first card was NOT a face, what is theprobability that the second card is a face card?

Since the first card was NOT a ten, what is theprobability that the second card is a ten?

4.0784 7.84%

51


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ProbabilityTwo cards are dealt from a shuffled deck.

What is the probability that the first card is anace and the second card is a face card or aten?

1 4 4= +

13 17 51

A F 10P = P (P + P )

1 12 4= +

13 51 51

1 16=

13 51

.0241 2.41% If the first card is an ace, what is theprobability that the second card is a

face card or a ten? 31.37%


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Contingency Table

A contingency table is a particular way to view a samplespace. Say we talk to 100 customers who just made apurchase in a store and not only do we note the gender ofthe customer we ask if they paid cash or used a credit card.

The responses are summarized on the next screen


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Contingency Table

Payment Method

Gender Cash Credit Card Total

Female 3 12 15

Male 17 68 85

Total 20 80 100

So, each of the 100 people observed had to be put in agender category and had to be given a payment method.If you divide each number in the table by the grand total(here 100 and this represents the total number of peopleobserved) the table is then called a joint probability table.

Lets do this and see what results on the next screen.


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Marginal Probabilities

The total column and the total row are called marginalprobabilities because the are written in the margins of thtable. But, note that adding across the Female row gives atotal of .15. This is the P(F), or the probability that you wouldselect a female when talking to someone involved in thestudy.

The other marginal probabilities are

P(M) = .85P(Cash) = .2 and P(Credit Card) = .8


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Joint Probability

Payment MethodGender Cash (B) CC (B) TotalFemale (A) P(A and B) P(A and B) P(A)Male (A) P(A and B) P(A and B) P(A)

Total P(B) P(B) 1.0From the example I have labeled the gender A and A andthe payment method B and B and then I have filled thetable out in definitional form.


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Joint Probability Table

Payment Method


Female .03 .12 .15

Male .17 .68 .85

Total .20 .80 1


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Marginal Probability

Note in the joint probability table on the previous screen(which is a contingency table that has been modified bydividing all numbers by the grand total!) that

1) In any row the marginal probability is the sum of the jointprobabilities in that row, and

2) In any column the marginal probability is the sum of the joint probabilities in that column.

(Also note that the sum of the probability of complementsequals 1 > for example P(A) + P(A c) = 1


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n on o ven s e eneraAddition Rule

Sometimes we want to ask a question about the probability of Aor B, written P(A or B) = P(A B).

By the general addition rule

P(A or B) = P(A) + P(B) P(A and B).In our example we have P(A or B) = .15 + 0.20 0.03 = 0.32.Lets think about this some more. How manyare Female? 3 +12 = 15! How many paid cash? 3 + 17 = 20! But 3 of thesewere in both A and B. So, when we ask a question about A orB we want to include all that are A or B, but we only want toinclude them once. If they are both we subtract out theintersection because it was included in both the row andcolumn total.


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Venn Diagram

A Venn diagram, named in honor of Mr. Venn, is anotherway to present the sample space for two variables.

A

B

The rectangle here

represents the samplespace. On one variable wehave event A and that takesup the space represented

by circle A. Ignoring circleB, all the rest of therectangle is A c (thecomplement of A). A similarinterpretation holds for B.This area represents the intersection

of A and B.


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Conditional Probability


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Conditional Probability

As we have seen, P(A) refers to the probability that event A will occur. A newidea is that P(A|B) refers to the probability that A will occur but with the

understanding that B has already occurred and we know it. So, we say theprobability of A given B. The given B part means that it is known that B hasoccurred.

By definition

P(A|B) = P(A and B)/P(B).

Similarly

P(B|A) = P(A and B)/P(A).

Note P(A and B) = P(B and A)


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Now we have by definitionP(A|B) = P(A and B)/P(B).

In this definition, B has already occurred. The P(B) is the denominator ofP(A|B) and is thus the base of the conditional probability. The intersection of Aand B is in the numerator. Since B has occurred, the only way A can haveoccurred is if there is an overlap of A and B. So we have the ratio

probability of overlap/probability of known event.

Lets turn tothe example from above. The joint probability table is repeated onthe next slide.


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Payment Method


Female .03 .12 .15

Male .17 .68 .85

Total .20 .80 1

Lets say you saw someone pay cash for a purchase, butyour view was blocked as to the gender of the person. TheP(femalecash) = .03/.20 = .15


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Payment Method


Female .03 .12 .15

Male .17 .68 .85

Total .20 .80 1

Lets say you saw someone a female leave the store havingmade a purchase, but your view was blocked as to the typeof payment. The P( cashfemale) = .03/.15 = .20


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Independent Events

Events A and B are said to be independent ifP(A|B) = P(A) or P(B|A) = P(B).

In the example we have been using P(Female) = .15, and P(Female|Cash) = .15.

What is going on here? Well, in this example it turns out that the proportion offemales in the study is .15 and when you look at those who paid cash theproportion of females is still .15. So, knowing that the person paid cash doesnchange your view of the likelihood that the person is female.

But, in some case (not here) having information about B gives a different viewabout A. When P(A|B) P(A) we say events A and B are dependent events.Similarly, when P(B|A) P(B) events A and B are dependent.


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Does a coin have a memory? In other words, does a coin remember how manytimes it has come up heads and will thus come up tails if it came up heads a lotlately? Say A is heads on the third flip, B is heads on the first two flips. Is headson the third flip influenced by the first two heads. No, coins have no memory!

Thus A and B are independent. (Note I am not concerned here about theprobability of getting three heads!)

Have you ever heard the saying, Pink sky in the morning, sailors take warningpink sky at night sailors delight. I just heard about it recently. Apparently it irule of thumb about rain. Pink sky in the morning would serve as a warning for

rain that day. If A is rain in the day and B is pink sky in the morning, then it seemsthat the P(A|B) P(A) and thus the probability of rain is influenced by morning skycolor (color is really just an indicator of conditions).


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Lets think about one more example. If you watched a football team all yearyou could use the empirical approach to find the probability that it will throw apass on a given play. Say P(pass)=0.4. This means the probability it willpass on a given play is 0.4.

But, if there are 5 minutes left in the game and the team is down 14 points theteam will want to pass more. SO, P(pass|down 14 with 5 minutes left) = 0.75,for example. This means the probability of a pass depends on the score andtime remaining!


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I have used some examples to give you a feel about when events areindependent and when they are dependent.

By simple equation manipulation we change the conditional probability

definition to the rule called the multiplication law or rule for the intersectionof events:

P(A and B) = P(B)P(A B) or P(A and B) = P(A)P(B A) .

Now this rule simplifies if A and B are independent . The conditional

probabilities revert to regular probabilities. We would then haveP(A and B) = P(B)P(A) = P(A)P(B).

Does this hold in our running example? Sure it does!

Note the given part shows up in the other term.


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Say, as a new example, we have A and B with P(A)=.5, P(B)=.6 and P(A and B)=.4

Then

a. P(A B) = .4/.6 = .667

b. P(B A) = .4/.5 = .8

c. A and B are not independent because we do NOT have P(A B) = P(A), orP(B A) = P(B).

Say, as another example, we have A and B with P(A)=.3 and P(B)=.4 and here wewill say A and B are mutually exclusive. This means P(A and B) = 0 (in aVenn Diagram A and B have no overlap), then

a. P(A B) = 0/.4 = 0 Here A and B are not independent.


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Y

Y1 Y2 Totals

X X1 P(X1 and Y1) P(X1 and Y2) P(X1)

X2 P(X2 and Y1) P(X2 and Y2) P(X2)

Totals P(Y1) P(Y2) 1.00

Here I put the joint probability table again in general terms. Question X hasmutually exclusive and collectively exhaustive events X1 and X2. For Y we have

a similar set-up. Note here each has only two responses, but what we will seebelow would apply if there are more than 2 responses.

Lets review some of the probability rules we just went through and then we wadd one more rule.


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Inside the joint probability table we find joint probabilities (like P(X1 and Y1)and in the margins we find the marginal probabilities (like P(X1)).

Marginal Probability RuleP(X1) = P(X1 and Y1) + P(X1 and Y2)

General Addition RuleP(X1 or Y1) = P(X1) + P(Y1) P(X1 and Y1)

Conditional ProbabilityP(X1|Y1) = P(X1 and Y1) /P(Y1)

Multiplication RuleP(X1 and Y1) = P(X1|Y1)P(Y1)


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The new part is to view the marginal probability rule as taking each part anduse the multiplication rule. So,

Marginal Probability RuleP(X1) = P(X1 and Y1) + P(X1 and Y2)

=P(X1|Y1)P(Y1) + P(X1|Y2)P(Y2)Where Y1 and Y2 are mutually exclusive and collectively exhaustive.

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