math 3 tutorial 1-3 solutions.pdf

7/30/2019 Math 3 Tutorial 1-3 Solutions.pdf

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MP2006-Math 3 PROBABILITY THEORY Tutorial Solutions

1. (a) (i) 2, 3, 4, 5, 12 = 11 outcomes (ii) (1,1), (1,2), (1,3), (6,6) = 36 outcomes (iii)1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 30, 36 = 18 outcomes

(b) only (ii) has equally likely outcomes and allows to use the first theorem for the

definition of probability

(c) S: (1,1), (1,2), (1,3), (2,1), (2,2) (3,1), (6,6) (note the presence of (1,2) and(2,1), etc) ; A: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) ; B: (4,6), (6,4), (5,5) (5,6), (6,5),(6,6) ; C: (1,6), (6,1), (2,5), (5,2), (3,4), (4,3) ; D: (1,1), (1,3), (3,1), (1,5), (5,1), (2,2),(2,4), (4,2), (2,6), (6,2), (3,3), (3,5), (5,3), (4,4), (4,6), (6,4), (5,5), (6,6)

(d) E and F are not mutually exclusive because E={(1,2), (2,1), (1,5), (5,1), (2,4), (4,2),(3,3), (3,6), (6,3), (4,5), (5,4), (6,6)} and F={(1,3), (3,1), (2,2), (2,6), (6,2), (3,5), (5,3),(4,4), (6,6)} have one element in common )}6,6{(=FE .

2. The sample space has 5x5x5=125 outcomes that are equally likely:S={(1,1,1), (1,1,2), (1,1,3),. (5,5,5)}The compliment events Ec the sum of the three chips is smaller or equal to 4 has only4 outcomes: Ec = {(1,1,1), (1,1,2), (1,2,1), (2,1,1)}We are in a case (i.e. finite number of equally likely outcomes) where we can use thefirst definition of probability, thus P(Ec) = 4/125 and from the compliment theorem weget : P(E) = 1 - P(E

c) = 1 - 4/125

3. Let A ii ( , , , )= 1 2 3 4 be the event that the ith IC fails, then

P A ii( ) . ( , , , )= =0 03 1 2 3 4 Let B be the event that the apparatus fails, then

B A A A Ac c c c c= 1 2 3 4I I I

= =P B P A A A A P A P A P A P Ac c c c c c c c c( ) ( ) ( ) ( ) ( ) ( )1 2 3 4 1 2 3 4I I I

since these events are independent.

P Bc( ) ( . ) .= =1 0 03 0 97 4 4

= = P B P Bc( ) ( ) .1 1 0 97 4 .

4. LetA be the event of drawing a defective fuse the first time,B be the event of drawing a defective fuse the second time, then

P A P B( ) . , ( ) .= =0 02 0 02

(a) Let Cbe the event of getting no defectives, then

C A Bc c= I P C P A B P A P Bc c c c( ) ( ) ( ) ( ) ( . ) .= = = =I 1 0 02 0 982 2


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(b) LetD be the event of getting one defectives, then

D A B A B P D P A B P A Bc c c c= = +( ) ( ) ( ) ( ) ( )I U I I I

P D P A P B P A P Bc c( ) ( ) ( ) ( ) ( ) . .= + = 2 0 98 0 02

(c) LetEbe the event of getting two defectives, then

E A B P E P A P B= = = =I ( ) ( ) ( ) . . .0 02 0 02 0 022

The sum of these probabilities is 1.

5(a) The event C: two rods of the desired length.

The sample space Shas 99001

99

1

100=

outcomes.

The event Chas 24501

49

1

50=

outcomes.

= =P C( ) / . %2450 9900 24 75 .

Alternative method:

LetA: the first rod drawn is of desired length,B: the second rod drawn is of desired length.

Then event C: two rods of the desired length becomes C A B= I .

Q P A( ) /= 50 100 , P B A( | ) / = 49 99 ,

= = = =P C P A B P B A P A( ) ( ) ( | ) ( ) / / . %I 49 99 50 100 24 75 .

5(b) The event C: one of the desired length.


99

1

100=

outcomes.

The event Chas

5000)(1

50

1

50)(

1

50

1

50=

+

defectivesecondthedefectivefirstthe outcomes.

= =P C( ) / . %5000 9900 50 5 .

5(c) The event C: none of the desired length.


99

1

100=

outcomes.


49

1

50=

outcomes.

= =P C( ) / . %2450 9900 24 75 .


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5(d) The event C: two undersized rods.


99

1

100=

outcomes.


125 =

outcomes.

= =P C( ) / . %600 9900 6 06 .

6. Every time we throw two dices we have 35/36 to not have a pair of 6.Thus after n

throws (independent events) the probability that we did not get a pair of 6 is (35/36)n.

Thus the probability to get a pair of six aftern throws is P(n) = 1 (35/36)n.The Knight want to know for which n we have P(n)>0.5, thus:

6.24)36/35ln(/5.0ln

5.0ln)36/35ln(5.0)36/35(5.0)36/35(15.0)(

>>

nn

nnP nn

(note ln(35/36)


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9. We have a finite number of equally likely outcomes and may use the first definition ofprobability, P = nb of favorable cases/total nb of cases.

The total number of hands of 5 cards than we can get from a set of 52 cards is 552C .

(i) the number of hands of 5 cards that we can get with 4 aces is 48 : the fifth card can

be any of the 48 remaining cards. Thus the probability is: %0018.0/485

52 == CP

(ii) the number of hands of 5 cards having at least one king will be obtain by looking atthe compliment event, hands having no king. The number of hands having no king

is 548C , thus the number of hands having at least one king is:5

48

5

52 CC . Thus the

probability of the event is: %3415

52

5

48

5

52

5

48

5

52 ==

=C

C

C

CCP

(note: we may also look at the compliment event probability which gives the sameresult)

10. No. of ways to select 3 out of 13 letters = 286P(getting 2 Bs) =11/286; P(getting only 2 I s) =310/286, P(getting 3 I s) =1/286

the probability of choosing at least two letters which are the same:

11. (i) 11 outcomes possible:H = 0 1 2 3 4 5 6 7 8 9 10T = 10 9 8 7 6 5 4 3 2 1 0H-T = 10 -8 -6 -4 -2 0 2 4 6 8 10(ii) 1000, -998, -996, .. 998, 1000

12. (a) the outcomes are: -2, -1, 0, 1, 2, 4

(b) 3077.0/)2( 2142

8 === CCXP : both balls are white

1758.0/)1( 2141

8

1

2 === CCCXP : 1 white 1 orange

01099.0/)0( 2142

2 === CCXP : both balls are orange

3516.0/)1( 2141

4

1

8 === CCCXP : 1 white 1 black

0879.0/)2( 2141

2

1

4 === CCCXP : 1 black 1 orange

0659.0/)4( 2142

4 === CCXP : both balls are black(note that the sum of the probability is 1)

(c) The mean value of X is:

040659.020879.013516.000199.011758.023077.0)( =+++++=== iiiX xpXE

13. Let X be the amount of dollars a person wins, then X has the possible values of:X 100 200 300 400 100nwith the corresponding probabilities (independent events) of

p 1/8 1/16 (1/2)nHence the mean winning is:

11 3 10 1 21

286 286 286 143

+ + =


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=

=

=====11

2002100100)2/1()(n

n

n

n

i

iiX nnxpXE dollars

One would be willing to play this game for 150$ since on the long run, he will bewinning in the average 50$ per time he plays.

14(i) By definition, +

= 1)( dxxf .

313

13

11

0

1

0

32 ==== kk

xk

dxkx .

14(ii)

464.01.01.0)(

3)()(

1

3

11

0

3

10

32

1

1 11

===

====

cccXP

cxdxxdxxfcXPc cc

965.09.09.0)(

3)()(

2

3

21

0

3

20

32

2

2 22

===

====

cccXP

cxdxxdxxfcXPc cc

15.

55

)5/(

5

1

5)(

0

5/

0

5/

0

5/

0

5/

0

5/

0

5/

0

=

==

===

+

+ + +

+ +

+

=

=

x

xxx

xx

e

dxedexxdex

dxexdxe

xdxxfx

+

+

==0

5/2

0

22

5)5()()( dx

exdxxfx

x

Let u x= /5, then dudxux 5,5 == ,


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( )

2522125

2)1(2125

)1(2125

)1(2)1(25

)1(25)1(255

)5(

0

00

0

00

2

0 0 0

225/

22

=

++=

+=

+=

=

===

+

+

+

+ + +

u

uu

u

uu

uux

e

dueeu

deu

dueueu

edudueudxe

x

%17.98115

1

)20()535()3(

45/2020

0

20

0

5/5/ ==+===

=+=+

eeedxe

XPXPXP

xx

16. Binomial distribution with 10,1.0 == np ,

LetA: hit the target at least once, then

A

c

becomes none of the 10 shots hits the target.

3487.09.0)1()1(0

10)( 1010100 ===

= pppAP c

%13.656513.03487.01)(1)( ==== cAPAP .

LetB: hit the target at least twice,

%39.269.01.0109.01

)1(10

1)1(10

)(

910

101

0

1010

2

==

=

=

=

= kkk

kk

k

ppk

ppk

BP

17. Binomial distribution with p n q= = =0 95 100 0 05. , , . .


7/7

%29.9695.005.095.01001

100

100

99

1001

1001

100)98(

10099

01009910099

100100

99

10098

1

==

=

=

=

=

=

qpqp

qpx

qpx

XP xx

x

xx

x

18. Binomial distribution with 97.0,50,03.0 === qnp .

xxqpx

xfxXP

=== 50

50)()(

When n becomes large, binomial distribution approaches Poissonsdistribution with = = =np 50 0 03 1 5. . .

5.150

!

5.1

!

50)()( =

=== e

xe

xqp

xxfxXP

xxxx .

19. We have a binomial distribution with p=4 10-6

and n=500,000. Since n is very large

(and p small) we may approximate it with a Poissons distribution with mean =np=500,000 4 10-6 =2.

2

!

2)( == e

kkXP

k

Thus:

%7.947!4

2

!3

2

!2

2

!1

2

!0

2

!

2)4( 2

4321022

4

0

==

++++==

=eee

kXP

k

k

This problem can also be approximated by normal distribution.

20. (a) %01.0

6.351)5(1)5(1)5(

===> FXPXP

(b) (i) the distribution is symmetric around the mean, thus c==3.6.(ii) 9.0)(9.0)(1.0)(11.0)( ====> cFcXPcXPcXP

728.328.11.0

6.39.0

1.0

6.3

=

c

cc

(iii) )10()10()6.3()6.3()6.36.3( cccFcFcXcP =+=+

math 3 tutorial 1-3 solutions.pdf

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