Conditional Probability
Section 3.6

Public Opinion Polls may categorize respondents by sex, age, race and level of education.  Comparisons are made and trends observed by using conditional probability.

The conditional probability of event A given event B is:

P(A | B) = n(A B)

n(B)

Example 1:

Asked 500 men and women, “Should the driving age be postponed to 18 years?”

(Hypothetical)

 Yes No Don’t Know Total Men 110 85 25 220 Women 185 75 20 280 Total 295 160 45 500

Find:

a)     P( Y )

b)     P( Y | W )

c)      P( Y | M )

d)     P( M | Y )

e)  P( Y M)

a)  P(Y) = n(Y)/n(S)    = 295/500 = .59 = 59%

b)     P(Y | W) = n(Y  W ) =  185 = .66 = 66%
n (W)            280

c)      P( Y | M) = n ( Y   M) = 110  = .50 = 50 %
n(M)            220

d)     P(M | Y) = n ( M   Y) =  110  = .37 = 37%
n(Y)           295

e)     Requires a new formula not yet derived called the Product Rule.

Recall:       P( Y |  M) = n( Y   M)        then       n(M) *P(Y | M) = n( Y   M)
n(M)

n(M) *P(Y | M)  =  n( Y   M)                           P(M)*P(Y | M) = P( Y   M)
n(S)                         n(S)

P( Y M) = 220/500 * 110/220 = 110/500 = .22 = 22%

Example 2:

Find the probability that 2 cards dealt are both aces.

Let

B = 1st card Ace,

let

A = 2nd card Ace.

Find     P (A   B).

P( A   B) = P(A | B) * P(B)  =  3/51 * 4/52

= 3/663 = .0045 =  .45% (less than ˝ %)

TREE DIAGRAM:

ACE                            (4/52)(3/51) = .0045

3/51

ACE

4/52                48/51

NOT ACE                   (4/52)(48/51) = .0723

Start                                                    ACE                (48/52)(4/51) = .0723

48/52                          4/51

NOT ACE

47/51

NOT ACE       (48/52)(47/51) = .8506

a)  What is P(B A )?  Top Branch = .0045

b)     What is P (B’  A)?  Third Branch = .0723

c)      What is P(A)?      = P(B A ) + P(B’ A )

Add Top branch and third branch = .0045 + .0723 = .0768

REVIEW EXAM #3

3.2

Basic Terms of Probability: Experiment, Sample Space, Event, Probability, Odds, Mutually exclusive.  Use in Genetics.

3.3

Basic rules of Probability:  0  P(E)  1; P(S) = 1; P( ) = 0; (6 RULES)

P(E) = n(E)/n(S) = success/total

Odds(E) = n(E): n(E’) = success: failure

3.4

Combinatorics and Probability:  Using Permutations and Combinations to find probability.  5-card hands in poker, Lottery.

3.5

Expected Value: (\$)P(Win) + (-\$)P(Lose) Casino games, Decision Theory.

3.6

Conditional Probability: Conditions on results.  Probabibility of A given B.

P(A|B) = n (A B) / n(B)

P (A B ) = P(A | B )* P(B)

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