slide1 n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Section 5.3: Independence and the Multiplication Rule Section 5.4: Conditional Probability and the General Multiplicat PowerPoint Presentation
Download Presentation
Section 5.3: Independence and the Multiplication Rule Section 5.4: Conditional Probability and the General Multiplicat

Loading in 2 Seconds...

play fullscreen
1 / 12

Section 5.3: Independence and the Multiplication Rule Section 5.4: Conditional Probability and the General Multiplicat - PowerPoint PPT Presentation


  • 175 Views
  • Uploaded on

Section 5.3: Independence and the Multiplication Rule Section 5.4: Conditional Probability and the General Multiplication Rule. Conditional Probability “Probability of event A given event B” Event A happens given that event B will or has occurred. P(A|B) (Not a division sign!)

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Section 5.3: Independence and the Multiplication Rule Section 5.4: Conditional Probability and the General Multiplicat' - alyssa


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
slide1
Section 5.3:

Independence and the Multiplication Rule

Section 5.4:

Conditional Probability and the General Multiplication Rule

slide2

Conditional Probability

  • “Probability of event A given event B”
    • Event A happens given that event B will or has occurred.
    • P(A|B) (Not a division sign!)
    • P(snow|December) is very different from P(snow|July) and P(snow)
    • Roll two dice:
      • P(12)=1/36
      • P(12|first roll 6) = 1/6
      • P(12| first roll 5) = 0
  • P(A|B)=
slide3

Independence

  • Event A and B are independent if the outcome of one event does NOT affect the probability of the other event
  • If A and B are independent then:
    • P(A and B) = P(A) P(B) “multiplication rule for independent events”
    • P( A |B ) = P( A )
    • P( B | A ) = P( B )
    • If one of the above is true, so are the other two.
  • If two events are not independent, they are dependent.
slide4

The multiplication rule generalizes:

If events A, B, C, etc. are independent then

P( A and B and C … ) = P(A) P(B) P(C)…..

slide6

If 5 fair coins are tossed, what is the probability of getting at least one tail?

  • 0
  • (1/2)1+(1/2)2+(1/2)3+(1/2)4+ (1/2)5
  • 1-(1/2)5

D. (1/2)5

slide7

Suppose the probability of a random bank customer defaulting on a loan is 0.2.

Also suppose the probability of a random bank customer who has a FICO score of 700 or higher is 0.03.

Question: Are FICO scores and likelihood of defaulting on a loan dependent or independent?

Why or why not?

slide8

More advanced tree diagram:

Prevalence of disease:

P(disease)=0.02

Sensitivity of test:

P(test positive given person diseased)=0.97

Specificity of test:

P(test negative given person not diseased)=0.95

slide9

So given:

P(disease)=0.02

P(test positive given person diseased)=0.97

P(test negative given person not diseased)=0.95

Determine probability of a person who tested positive actually not having the disease. That is, P(not diseased | person tested positive)=?

p a and b p a p b a
P(A and B) = P(A)∙P(B|A )
  • “General Multiplication Rule”
  • Algebraically equivalent to P(A|B)=
  • Allows us to multiply down branches in tree diagrams
slide12

Suppose Team A and B will play each other in a “Best of 3” series. (First team to win two games wins the series.) Also suppose the probability of the superior Team A winning each game is 0.6. Use a tree diagram to determine the probability of Team B winning the series.

  • P(Team B wins series)=0
  • P(Team B wins series)=0.16
  • P(Team B wins series)=0.352
  • P(Team B wins series)=0.4