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Conditional Probability, Random variables

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• Brief review
• Lecture 15:

|A ∪ B ∪ C|

=|A| + |B| + |C| – |A ∩ B| – |A ∩ C| – |B ∩ C|+ |A ∩ B ∩ C|

• Brief review

Lecture 16: Conditional probability

• If Pr(B) is nonzero, the conditional probability of A given B is

Pr(A|B) = Pr(A∩B)/Pr(B).

Ω A B

• If Pr(B) is nonzero:
• Brief review

Lecture 16: The Bayes’ rule

Question 2:

A room has been burgled twice during the past 20 years. There is a dog in the room, it barks 3 times every week, and the probability that it barks when there are thieves is 0.9.

Then what’s the probability that there are thieves indeed when the dog barks?

Question 2:

A room has been burgled twice during the past 20 years. There is a dog in the room, it barks 3 times every week, and the probability that it barks when there are thieves is 0.9. Then what’s the probability that there are thieves indeed when the dog barks?

Solution:

Suppose that the event dog barks is A, the event there are thieves is B. Then P(A) = 3 / 7, P(B)=2/(20·365)=2/7300, P(A | B) = 0. From Bayes’ rule： P(B|A)=0.9*(2/7300)/(3/7)=0.

• Discrete Random Variables and Distributions:

Important Properties:

• Brief review

Chapter 24.5 of Erwin's Book: RV

Question 3:

The probability distribution of a lift breaking down in a week:

No of breakdown X = xi

0 1 2 3

Probability P(X=xi)= pi

0.10 0.25 0.35 α

( 2 ) Probability of breaking down twice? ( 3 ) Probability of breaking down less than 3 times? ( 4 ) Probability of breaking down at least once?

Example 4: (Chapter 24.5 of Erwin's Book)

In tossing a fair coin, let X=Number of trials until the first head

appears. Then, by independence of events

Then we can check:

Example 4: (Chapter 24.5 of Erwin's Book)

Think about:

what is the expected times we need to waite until the first head appears

Then: