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[PDF] 27 Conditional Probability on the Independence of Events

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Section 7. Conditional Probability on the Independence of Events (ATTENDANCE 3)39

2.7 Conditional Probability on the Independence

of Events Often we know that a particular event, eventBsay, is "known to have occurred". We can use this information to upgrade our knowledge about the probability of some other event, eventA, say. More specifically, we calculate the probability of eventA, conditional on the occurrence of eventB. This probability is denotedP(A|B). For any two eventsAandBwithP(B)>0, theconditionalprobability ofAgiven that

Bhas occurred is defined by

P(A|B) =P(A∩B)P(B).

Furthermore, two events areindependentif any one of the following is true:

P(A∩B) =P(A)P(B),

P(A|B) =P(A),

P(B|A) =P(B).

Exercise 2.7 (Conditional Probability on the Independenceof Events)

1.Conditional probability versus unconditional probability: Interviews.Seven can-

didates, three of which are females, are being interviewed for a job. The three female"s names are Kathy, Susan and Jamie; the four male"s names are Tom, Tim, Tyler and Toothy. Assume the candidates are chosen at random for a job interview. (a) The chance that Tom is chosengiven thata male is chosen,14, is an example of (i)conditional probability(ii)unconditional probability. (b) The chance Tom is chosengiven thata female is chosen,03, is an example of (i)conditional probability(ii)unconditional probability. (c) The chance Tom is chosen, 1

7, is an example of

(i)conditional probability(ii)unconditional probability. (d) The chance Tom isnotchosen,67, is an example of (i)conditional probability(ii)unconditional probability. (e) Conditional probability essentially involves taking a subset, defined by the conditional event, of the original sample space and then calculatingthe probability within this subset. (i)True(ii)False

2.Using conditional probability formula: coins.Consider the box of coins in Fig-

ure 2.12. A coin is chosen at random.

40Chapter 2. Probability (ATTENDANCE 3)

S (b) P(74 | n) = 1/3 = = P(74 and n)

P(n) 1/10

3/10

74c 78c 78c 76c 80c

78d 81d

74n 78n 80n

n74 n event conditional on event occuring

P(c and 78)

P(78) S(a) P(c | 78) = 2/4 = = 2/10

4/10 78
78 c
event conditional on event occuring

74c 78c 78c 76c 80c

78d 81d

74n 78n 80n

Figure 2.12: Conditional probability: coins

(a)CalculatingP(C|78). The probability of picking a 1978 out of the box is

P(78) = (circle one) (i)3

10(ii)410(iii)610(iv)710.

The probability of picking out a coin which is both a cent and a 1978, is

P(C∩78) = (circle one) (i)2

10(ii)410(iii)610(iv)710.

The probability of picking out a centgiven thatit is a 1978 coin, is

P(C|78) =P(C∩78)

P(78)=2/104/10= (circle one) (i)24(ii)25(iii)610(iv)710. (b)CalculatingP(74|N).

P(N) = (circle one) (i)3

10(ii)410(iii)610(iv)710.

P(74∩N) = (circle one) (i)1

10(ii)210(iii)510(iv)710.

P(74|N) =P(74∩N)

P(N)=1/103/10= (circle one) (i)12(ii)13(iii)24(iv)34. (c)P(78|N) =P(78∩N) P(N)=1/103/10= (circle one) (i)03(ii)13(iii)23(iv)33. (d)P(C|¯80) =P(C∩¯80) P(¯80)=4/108/10= (circle one) (i)15(ii)25(iii)48(iv)710.

3.Independence (sampling with replacement)versus dependence (sampling without replacement): Box of Tickets.

Two things areindependentif the chances for the second given the first are the same, no matter how the first turns out; otherwise, the two things aredependent.

Consider the following box with six tickets.

1a2a1b3b2c3c

(a) Two tickets are drawn at random from the box. The first ticketdrawn is notreplaced in the box; that is, there are only five tickets remaining in the box when the second ticket is drawn. The chance the second ticketis a "2"given thatthe first ticket is a "2" is (circle one) (i)1

5(ii)25(iii)35.

(b) Two tickets are sampled withoutreplacement at random from the box. That is, there are only five tickets remaining in the box when the second Section 7. Conditional Probability on the Independence of Events (ATTENDANCE 3)41 ticket is drawn. The chance the second ticket is a "2"given thatthe first ticket is a "1" is (circle one) (i) 1

5(ii)25(iii)35.

(c) When sampling at random withoutreplacement, the chance the second ticket of two drawn from the box is any given number willdependon what number was drawn on the first ticket. (i)True(ii)False (d) Two tickets are sampledwithreplacement at random from the box. That is, allsixtickets remain in the box when the second ticket is drawn. The chance the second ticket is a "2"given thatthe first ticket is a "1" is (circle one) (i) 1

6(ii)26(iii)36.

(e) When sampling at random with replacement, the chance the second ticket of two drawn from the box is "2", no matter what the first ticket is,will alwaysbe (circle one) (i)1

6(ii)26(iii)36.

(f) When sampling at random with replacement, the draws are independent of one another; without replacement, the draws are dependent. (i)True(ii)False

4.Independence versus dependence: fathers, sons and college.The data from a

sample of 80 families in a midwestern city gives the record of college attendance by fathers (F) and their oldest sons (S). A family is chosen at random. son attended son did not college attend college father attended college18 725 father did not attend college22 3355

40 4080

(a)One way to demonstrate independence or dependence.The probability a son attended college, isP(S) = (circle one)

(i) 18

25(ii)1840(iii)2580(iv)4080.

The probability a father attended college, isP(F) = (circle one) (i) 18

25(ii)1840(iii)2580(iv)4080.

The probability a sonanda father both attended college isP(S∩F) = (i) 18

25(ii)1880(iii)2580(iv)4080.

SinceP(S∩F) =18

80?=P(S)×P(F) =?4080?×?2580?, the events "son attends

college" and "father attends college" are (choose one) (i)independent(ii)dependent. (b) EventsAandBareindependent if and only if

P(A and B) =P(A)·P(B).

EventsAandBare dependent if and only if

P(A and B)?=P(A)·P(B).

(i)True(ii)False

42Chapter 2. Probability (ATTENDANCE 3)

(c)Another way to demonstrate independence or dependence.P(S) = (circle one) (i)18

25(ii)1840(iii)2580(iv)4080,

P(S|F) = (circle one) (i)18

25(ii)1840(iii)2580(iv)4080.

SinceP(S|F) =18

25?=P(S) =4080, the events "son attends college" and

"father attends college" are (choose one) (i)independent(ii)dependent. (d)A third way to demonstrate independence or dependence.P(F) = (circle one) (i)18

25(ii)1840(iii)2580(iv)4080,

P(F|S) = (circle one) (i)18

25(ii)1840(iii)2580(iv)4080.

SinceP(F|S) =18

40?=P(F) =2580, the events "son attends college" and

"father attends college" are (choose one) (i)independent(ii)dependent.

2.8 Two Laws of Probability

Themultiplicative law of probabilityfor two events is given by

P(A∩B) =P(B)P(A|B),

and, more generally, fornevents,

P(A1∩A2∩ ··· ∩An) =P(A1)P(A2|A1)P(A3|A1∩A2)···P(An|A1∩ ··· ∩An-1),

and if eventsA1,...,Anare independent, then, for every subsetAi1,...,Airof them, P(Ai1∩Ai2∩ ··· ∩Air) =P(Ai1)P(Ai2)···P(Air). Theadditive law of probabilityfor two events is given by

P(A?B) =P(A) +P(B)-P(A∩B),

and, for three events, it is P(A?B?C) =P(A)+P(B)+P(C)-P(A∩B)-P(A∩C)-P(B∩C)+P(A∩B∩C).

Finally,

P(A) = 1-P(¯A).

Exercise 2.8 (Two Laws of Probability)

1.Multiplicative law: cards.Three cards are taken out of the deck at random. Let

A irepresent the event theith card is taken from the deck. (a) The chance the first card dealt is an ace is

P(A1) = (circle one) (i)1

51(ii)451(iii)352(iv)452.

Section 8. Two Laws of Probability (ATTENDANCE 3)43 (b) The chance thesecondcard dealt is a jack, given the first card dealt is an ace, isP(A2|A1) = (circle one) (i) 1

51(ii)451(iii)352(iv)452.

(c) The probability the first card is an aceandthe second card is a jack is11

P(A1∩A2) =P(A1)P(A2|A1) = (circle one)

(i) 1 (d) The probability that thethirdcard dealt is a jack, conditional on the first two cards dealt are a jack and an ace, isP(A3|A1∩A2) = (circle one) (i) 3

50(ii)450(iii)451(iv)452.

(e) The probability that the first card is an aceandthe second card is a jack andthe third card is another jack isquotesdbs_dbs20.pdfusesText_26