- 760 Views
- Updated On :

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

Related searches for Rules of Probability The additive rule

Download Presentation
## PowerPoint Slideshow about 'Rules of Probability The additive rule' - johana

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

### Conditional Probability,Independence andThe Multiplicative Rue

### Independencefor more than 2 events

### The sure thing principle and Simpson’s paradox

The Rule for complements

Then the conditional probability of A given B is defined to be:

The multiplicative rule of probability

and

if A and B areindependent.

This is the definition of independent

and continuing we obtain

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

- 10 , J , Q , K ,A
- 10 , J , Q , K ,A
- 10 , J , Q , K ,A
- 10 , J , Q , K ,A

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.

Another solution is by counting

The same result

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]

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

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]

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

- 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

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?

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

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?

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 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.

An generlization of Bayes Rule

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

for all i and j. Then

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

…

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?

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

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.

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?

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

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

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

Does

Example to illustrate

D = death due to lung cancer

S = smoker

C = lives in city, = lives in country

Solution

similarly

Download Presentation

Connecting to Server..