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STAT 1012: Statistics for Life Science

2013/14 Term 1

Solutions to Practice Problems (Chapter 2) Problem 1:

(a) Possible Outcomes ={HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, where H and T denotes a

head or a tail appearing in each of the three flips.

(b) Event of one head and 2 tails = {HTT, THT, TTH}. Hence Pr(one head and 2 tails) = 3/8

Problem 2:

(a) Since there are two possible outcomes (True or false) for each of the 5 questions, the

number of possible outcomes for answering 5 questions should be 25=32.

(b) Out of the 32 possible outcomes, there is only one outcome which gives all the 5

answers correct. Hence Pr(at least one question wrong) = 1- Pr(all answers

correct)=1-1/32=31/32 = 0.96875

Problem 3:

(a) False, because the cards are draw without replacement. The correct probability is

(26/52)x(25/51)x(24/50)=2/17=0.1176

(b) A and B are NOT independent, because (1) if the first card is red (i.e. A occurs),

Pr(B)=25/51, (2) if the first card is black (i.e. A not occur), Pr(B)=26/51.

(c) True in (a) that Probability is 0.5x0.5x0.5 = 0.125. In (b), A and B are independent,

because Pr(B)=1/2 regardless of whether A occurred or not.

Problem 4: (a) Probability = 1/9 (b) (i) Based on Benford’s Law, Pr(5 or 6 as leading digit) =

0.08+0.07 = 0.15. (ii) Random selection: Pr(5 or 6 as leading digit) = 1/9+1/9= 2/9 = 0.2222.

Problem 5: False. Equally likely with probability (0.5)10 each

Problem 6: False. The chance of obtaining 1 head is twice as that for obtaining 0 head or 2

heads. Hence Pr(1 head) = 2/4 =1/2

Problem 7:

(a) Pr(Ac∩Bc∩Cc)=Pr(Ac)Pr(Bc)Pr(Cc) (because Ac, Bc and Cc are independent)

=(1-1/4)x(1-1/3)x(1-1/2)=1/4

(b) Pr({One event will occur})

=Pr(A∩Bc∩Cc)+ Pr(Ac∩B∩Cc)+ Pr(Ac∩Bc∩C)

=Pr(A)Pr(Bc)Pr(Cc)+ Pr(Ac)Pr(B)Pr(Cc)+ Pr(Ac)Pr(Bc)Pr(C) (because of the independence)

=(1/4)(2/3)(1/2)+(3/4)(1/3)(1/2)+(3/4)(2/3)(1/2)=11/24

Problem 8:

(a) Let R be the event of recessive gene ‘a’, M be the event that the maternal gene is of type

‘a’, and P be the event that the paternal gene is of type ‘a’. Then

Pr(R)=Pr(M ∩ P) = Pr(M)Pr(P) (because of independence) = (0.25)(0.25)=0.0625

(b) Let AA be the event that both parents are from population A, AB be the event that

parents are from different population, and BB be the event that both parents are from

population B.

Now, given that Pr(R|A)=40%, Pr(R|B)=10%, then

Pr(R|AA)=Pr(R|A)2=0.16, Pr(R|BB)=Pr(R|B)2=0.01 and Pr(R|AB)=2Pr(R|A)Pr(R|B)=0.08.

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Also Pr(AA)=0.25, Pr(AB)=0.10 and Pr(BB)=0.65.

Note that events AA, AB and BB are mutually exclusive and exhaustive. Based on the

total probability rule, Pr(R)=Pr(R|AA)Pr(AA)+Pr(R|AB)Pr(AB)+Pr(R|BB)Pr(BB)

=0.16x0.25+0.08x0.10+0.01x0.65=0.0545

(c) Events AA, AB and BB are mutually exclusive and exhaustive. Based on the Bayes’ rule,

1468.00545.0

)10.0(08.0

)Pr()|Pr()Pr()|Pr()Pr()|Pr(

)Pr()|Pr()|Pr(

1193.00545.0

)65.0(01.0

)Pr()|Pr()Pr()|Pr()Pr()|Pr(

)Pr()|Pr()|Pr(

7339.00545.0

)25.0(16.0

)Pr()|Pr()Pr()|Pr()Pr()|Pr(

)Pr()|Pr()|Pr(

BBBBRABABRAAAAR

BBBBRRB

BBBBRABABRAAAAR

ABABRRAB

BBBBRABABRAAAAR

AAAARRAA

Problem 9:

(a) Since an Ace cannot be a face card, hence A∩B=ϕ and therefore A and B are mutually

exclusive.

However, B∩C={♣J, ♣Q, ♣K} ǂ ϕ and A∩C={♣A} ǂ ϕ, meaning that (1) B and C are NOT

mutually exclusive, and (2) A and C are NOT mutually exclusive.

(b) Pr(B∩C)=Pr({♣J, ♣Q, ♣K})=3/52.

Problem 10: (d) Based on the De Morgan’s Law, Ac U Bc= (A∩B)c.

Hence Pr(Ac U Bc)=1-Pr(A∩B) = 1-Pr(A)Pr(B)= 1-0.5x0.4=0.8.

Problem 11: Want to compute Pr(AUB | (Ac∩Bc∩Cc)c). Now,

Pr(Ac∩Bc∩Cc) =Pr(Ac)Pr(Bc∩Cc) (because of independence)

=(1-1/4)x[1-Pr(B U C)]

=(3/4)x[1-(Pr(B)+Pr(C)-Pr(B∩C))] (using the addition law of probability)

=(3/4)x[1-1/3-1/2+1/8]=7/32

2/1)3/1)(4/1(3/14/1

ce)independen of (because )Pr()Pr()Pr()Pr()Pr()Pr()Pr()Pr( Also,

BABABABABA

Hence 25

16

32/71

2/1

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)Pr(

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cccccccc

cccccccc

CBA

BA

CBA

CBABACBABA

Problem 12:

(a) Let S be the event that the mother actually smoke, and D be the event that a positive

report (on mother’s smoking) was made by her daughter.

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)Pr()Pr(

)Pr(

)Pr(

)Pr()|Pr(

DSDS

DS

D

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c

(b) 9500.0232271222

23227

)Pr()Pr(

)Pr(

)Pr(

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CcC

CC

C

CCCC

DSDS

DS

D

DSDSPV

Problem 13: (b) 1 and 3 only. (3) is true because given AUB = Ω, we have Pr(AUB)=1,

which gives Pr(A)+Pr(B)=1+Pr(A∩B)≥1.

(1) is true because Pr(A)+Pr(B)≥1 => Pr(B) ≥1-Pr(A)=Pr(Ac).

(2) is false because Pr(C|A)Pr(A)+Pr(C|B)Pr(B)=Pr(C∩A)+ Pr(C∩B) = [Pr(C)- Pr(C∩Ac)]+

Pr(C∩B) = Pr(C)+ [Pr(C∩B)- Pr(C∩Ac)] ≥ Pr(C)

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Problem 14: (b) 3 only

1. is not true because Pr(A|Bc)≠ Pr(Ac|B) in general.

2. is not true because Pr(A|B)=Pr(B|A) Pr(A∩B)/Pr(B)= Pr(A∩B)/Pr(A) Pr(A)=Pr(B).

3. is true because Pr(B)= Pr(A∩B)+ Pr(Ac∩B) ≥Pr(A∩B).

Problem 15: Let C be the event that A is in class, and D is the event that B is in class,

Pr({at least one student in class})=Pr(C υ D)

=Pr(C)+Pr(D)-Pr(C ∩ D) by the addition law of probability

=0.8+0.6-0.48=0.92

Therefore probability that A is in class given at least one students is in class is given by

Pr(C|C υ D)=Pr(C ∩ (C υ D))/Pr(C υ D)= Pr(C)/Pr(C υ D)=0.8/0.92=0.8696

Problem 16:

(a) Let D be the event of having disease, test+ be the event of test positive and test- be the

event of test negative.

Pr(D)=0.02, Pr(test+|D)=0.99, Pr(test+|ND)=0.005. Using the Bayes’ rule,

Pr(D|test+) = Pr(D)xPr(test+|D)/[ Pr(D)xPr(test+|D)+ Pr(ND)xPr(test+|ND)]

= (0.02x0.99)/(0.02x0.99+0.98x0.005)=0.0198/0.0247=0.8016

(b) Pr(D|test-) = Pr(D)xPr(test-|D)/[ Pr(D)xPr(test-|D)+ Pr(ND)xPr(test-|ND)]

= (0.02x0.01)/(0.02x0.01+0.98x0.995)=0.0002/0.9753=0.000205

Therefore RR= Pr(D|test+)/Pr(D|test-)=0.8016/0.000205=3909

Since RR >>1, the test and the disease are higher dependent with each other. Therefore,

the test is useful in indicating the disease.

Problem 17:

(a) Let HIV be the event of HIV positive from the high risk group, test+ be the event of test

positive and test- be the event of test negative. Then Pr(HIV)=0.5, Pr(test+|HIV)=0.999

and Pr(test-|no HIV)=0.9999. Based on the Baye’s Rule,

Pr (HIV|test+)=Pr(test+|HIV)Pr(HIV)/[Pr(test+|HIV)Pr(HIV)+Pr(test+|no HIV)Pr(no HIV)]

=(0.999)(0.5)/[0.999(0.5)+(1-0.9999)(0.5)]=9990/9991

(b) If Pr(HIV)=0.1, then

Pr (HIV|test+)=Pr(test+|HIV)Pr(HIV)/[Pr(test+|HIV)Pr(HIV)+Pr(test+|no HIV)Pr(no HIV)]

=(0.999)(0.1)/[0.999(0.1)+(1-0.9999)(0.9)]=9990/9999

If Pr(HIV)=0.01, then

Pr (HIV|test+)=Pr(test+|HIV)Pr(HIV)/[Pr(test+|HIV)Pr(HIV)+Pr(test+|no HIV)Pr(no HIV)]

=(0.999)(0.01)/[0.999(0.01)+(1-0.9999)(0.99)]=9990/10089

In conclusion, when the prevalence rate (i.e. Pr(HIV)) decreases from 0.50 to 0.10 and

0.01, Pr(HIV|test+) decreases as well, suggesting that the a positive test result becomes

less accurate/reliable.