Chapter 3: Conditional Probability and Independence

[Pages:23]Chapter 3: Conditional Probability and Independence

Xiugang Wu

University of Delaware

Spring, 2020

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Example

Roll two dice:

sum of 2 dice

2 3 4 5 6 7 8 9 10 11 12

probability

1

2

3

4

5

6

5

4

3

2

1

36 36 36 36 36 36 36 36 36 36 36

prob. given 1st is 6 0

0

0

0

0

1 6

1 6

1 6

1 6

1 6

1 6

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Outline

? Conditional Probability ? Bayes' Formula ? Independent Events

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Outline

? Conditional Probability ? Bayes' Formula ? Independent Events

4

Conditional Probability

The conditional probability that E occurs given that F has occurred is: P (E|F ) = P (EF ) P (F )

Interpretation: If event F has occurred, then in order for E to occur it is nec-

essary that the actual occurrence is a point in both E and F , i.e. in EF .

Now as F has occurred, F becomes our new (reduced) sample space; hence

P (E|F )

=

P (EF ) P (F )

.

If all outcomes are equally likely, then

P (E|F ) = P (EF ) = number of outcomes in EF /total number of outcomes P (F ) number of outcomes in F /total number of outcomes number of outcomes in EF = number of outcomes in F

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Example

Example: Two dice are rolled. If the first dice is 6, what is the probability that the sum of the dice is 7?

Solution: Let F = {the first dice is 6} and E = {the sum of the dice is 7}. Then we have

P (E|F )

=

P (EF ) P (F )

=

P ({(6, 1)}) P (F )

=

1/36 6/36

=

1 6.

Example: A student is taking a one-hour-limit-exam. Suppose the probability that the student will finish the exam in less than x hour is x/2 for x 2 [0, 1]. Given that the student is still working after 0.75 hour, what is the probability that the full hour is needed?

Solution: Let F = {more than 0.75 hour is needed} and E = {the full hour is needed}.

Then we have P (E) = (1 x/2)|x=1 = 0.5 and P (F ) = (1 x/2)|x=0.75 = 0.625,

and hence

P (E|F ) = P (EF ) = P (E) = 0.5 = 0.8. P (F ) P (F ) 0.625

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Generalization

Conditional on multiple events:

P (E|F1

? ? ? Fn)

=

P (EF1 ? ? ? Fn) . P (F1 ? ? ? Fn)

Multiplication rule:

P (E1E2 ? ? ? En) = P (E1)P (E2|E1) ? ? ? P (En|E1 ? ? ? En 1).

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Example

Example: 52 playing cards are randomly divided into 4 piles of 13 cards each. What is the probability that each pile has exactly one ace?

Solution 1:

4!

48 12,12,12,12

52

13,13,13,13

0.105.

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