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Rules of Probability The additive rule P [ A  B ] = P [ A ] + P [ B ] – P [ A  B ] and if P [ A  B ] = f P [ A  B ] = P [ A ] + P [ B ] The additive rule for more than two events and if A i  A j = f for all i ≠ j. then for any event E The Rule for complements

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the additive rule
The additive rule

P[A  B] = P[A] + P[B] – P[A  B]

and

if P[A  B] = f

P[A  B] = P[A] + P[B]

the additive rule for more than two events
The additive rule for more than two events

and if Ai  Aj = f for all i ≠ j.

then

slide4

for any event E

The Rule for complements

slide7

The multiplicative rule of probability

and

if A and B areindependent.

This is the definition of independent

slide9

Proof

and continuing we obtain

example
Example

What is the probability that a poker hand is a royal flush

i.e.

  • 10 , J , Q , K ,A
  • 10 , J , Q , K ,A
  • 10 , J , Q , K ,A
  • 10 , J , Q , K ,A
solution

10 , J , Q , K ,A

  • 10 , J , Q , K ,A
  • 10 , J , Q , K ,A
  • 10 , J , Q , K ,A
Solution

Let A1 = the event that the first card is a “royal flush” card.

Let Ai = the event that the ith card is a “royal flush” card. i = 2, 3, 4, 5.

slide14

Definition:

The set of k events A1, A2, … , Akare called mutually independent if:

P[Ai1∩ Ai2∩… ∩ Aim] = P[Ai1] P[Ai2] …P[Aim]

For every subset {i1, i2, … , im } of {1, 2, …, k }

i.e.for k = 3 A1, A2, … , Akare mutually independentif:

P[A1∩ A2] = P[A1] P[A2], P[A1∩ A3] = P[A1] P[A3],

P[A2∩ A3] = P[A2] P[A3],

P[A1∩ A2∩ A3] = P[A1] P[A2] P[A3]

slide15

P[A1] = .4, P[A2] = .5 , P[A3] = .6

P[A1∩A2] = (0.4)(0.5) = 0.20

P[A1∩ A3] = (0.4)(0.6) = 0.24

P[A2∩ A3] = (0.5)(0.6) = 0.30

P[A1∩ A2∩ A3] =

(0.4)(0.5)(0.6) = 0.12

A1

A2

0.08

0.08

0.12

0.12

0.12

0.18

0.12

0.18

A3

slide16
Definition:

The set of k events A1, A2, … , Akare called pairwise independent if:

P[Ai∩ Aj] = P[Ai] P[Aj] for all i and j.

i.e.for k = 3 A1, A2, … , Akare pairwise independentif:

P[A1∩ A2] = P[A1] P[A2], P[A1∩ A3] = P[A1] P[A3],

P[A2∩ A3] = P[A2] P[A3],

It is not necessarily true that P[A1∩ A2∩ A3] = P[A1] P[A2] P[A3]

slide17

P[A1] = .4, P[A2] = .5 , P[A3] = .6

P[A1∩A2] = (0.4)(0.5) = 0.20

P[A1∩ A3] = (0.4)(0.6) = 0.24

P[A2∩ A3] = (0.5)(0.6) = 0.30

P[A1∩ A2∩ A3] = 0.14

≠ (0.4)(0.5)(0.6) = 0.12

A1

A2

0.06

0.10

0.14

0.14

0.10

0.16

0.10

0.20

A3

bayes rule
Bayes Rule
  • Due to the reverend T. Bayes
  • Picture found on website: Portraits of Statisticians
  • http://www.york.ac.uk/depts/maths/histstat/people/welcome.htm#h
example20
Example:

We have two urns. Urn 1 contains 14red balls and 12black balls. Urn 2 contains 6red balls and 20black balls.

An Urn is selected at random and a ball is selected from that urn.

Urn 1

Urn 2

If the ball turns out to be red what is the probability that it came from the first urn?

solution21
Solution:

Let A = the event that we select urn 1

= the event that we select urn 2

Let B = the event that we select a red ball

Note: the desired conditional probability is in the reverse direction of the given conditional probabilities. This is the case when Bayes rule should be used

example testing for a disease
Example: Testing for a disease

Suppose that 0.1% of the population have a certain genetic disease.

A test is available the detect the disease.

If a person has the disease, the test concludes that he has the disease 96% of the time. It the person doesn’t have the disease the test states that he has the disease 2% of the time.

Two properties of a medical test

Sensitivity = P[ test is positive | disease] = 0.96

Specificity = P[ test is negative | disease] = 1 – 0.02 = 0.98

A person takes the test and the test is positive, what is the probability that he (or she) has the disease?

solution24
Solution:

Let A = the event that the person has the disease

= the event that the person doesn’t have the disease

Let B = the event that the test is positive.

Note: Again the desired conditional probability is in the reverse direction of the given conditional probabilities.

bayes rule states25
Bayes rule states

Thus if the test turns out to be positive the chance of having the disease is still small (4.58%).Compare this to (.1%), the chance of having the disease without the positive test result.

slide26

An generlization of Bayes Rule

Let A1, A2 , …, Ak denote a set of events such that

for all i and j. Then

slide27
If A1, A2 , …, Ak denote a set of events such that

for all i and j. Then A1, A2 , …, Ak is called a partition of S.

S

A2

Ak

A1

slide28

Proof

for all i and j.

A2

Ak

A1

B

Then

example30
Example:

We have three urns. Urn 1 contains 14red balls and 12black balls. Urn 2 contains 6red balls and 20black balls. Urn 3 contains 3red balls and 23black balls.

An Urn is selected at random and a ball is selected from that urn.

Urn 1

Urn 3

Urn 2

If the ball turns out to be red what is the probability that it came from the first urn? second urn? third Urn?

solution31
Solution:

Let Ai= the event that we select urn i

Let B = the event that we select a red ball

Note: the desired conditional probability is in the reverse direction of the given conditional probabilities. This is the case when Bayes rule should be used

example33
Example:

Suppose that an electronic device is manufactured by a company.

During a period of a week

  • 15% of this product is manufactured on Monday,
  • 23% on Tuesday,
  • 26% on Wednesday ,
  • 24% on Thursday and
  • 12% on Friday.
slide34
Also during a period of a week
  • 5% of the product is manufactured on Monday is defective
  • 3 % of the product is manufactured on Tuesday is defective,
  • 1 % of the product is manufactured on Wednesday is defective ,
  • 2 % of the product is manufactured on Thursday is defective and
  • 6 % of the product is manufactured on Friday is defective.

If the electronic device manufactured by this plant turns out to be defective, what is the probability that is as manufactured on Monday, Tuesday, Wednesday, Thursday or Friday?

slide35

A1 = the event that the product is manufactured on Monday

A2 = the event that the product is manufactured on Tuesday

A3 = the event that the product is manufactured on Wednesday

A4 = the event that the product is manufactured on Thursday

A5 = the event that the product is manufactured on Friday

Solution:

Let

Let B = the event that the product is defective

slide36

Now

P[A1]= 0.15, P[A2]= 0.23, P[A3]= 0.26, P[A4]= 0.24 and P[A5]= 0.12

Also

P[B|A1]= 0.05, P[B|A2]= 0.03, P[B|A3]= 0.01,

P[B|A4]= 0.02 and P[B|A5]= 0.06

We want to find

P[A1|B], P[A2|B], P[A3|B], P[A4|B]and P[A5|B].

We will apply Bayes Rule

the sure thing principle
The sure thing principle

Suppose

Example – to illustrate

Let A = the event that horse A wins the race.

B = the event that horse B wins the race.

C = the event that the track is dry

= the event that the track is muddy

simpson s paradox
Simpson’s Paradox

Does

Example to illustrate

D = death due to lung cancer

S = smoker

C = lives in city, = lives in country

solution42
Solution

similarly

slide43

whether

is greater than

depends also on the values of

slide44

whether

than

and