m1210

===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003.

Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631052/34===================================================================================================

HINTS/SOLUTIONS for M1210(Probability and Linear Programming)

Classroom Discussion Exercise

1. (b) P(B) = 1 – P(B1) =21

P(AB) = P(A) + P(B) – P(AB)

31

21)A(P

65

P(A) =32

P(A) P(B) = )AB(P31

21

32

A and B are independent.

2. (c) S =

6,65,64,63,62,61,66,55,54,53,52,51,56,45,44,43,42,41,46,35,34,33,32,31,36,25,24,23,22,21,26,15,14,13,12,11,1

A =

5,61,66,52,53,41,44,32,35,23,21,26,14,12,11,1

Probability P(A) =125

3615

3. (c) Bag I Bag II4 white and 3 white and2 black 5 black

P (ball from I is black and from II is white)

=81

83

62

P (ball from I is white and from II is black)

=125

85

64

Required probability =.24

13125

81

4. (b) Probability =157

C10C4

C10C6

2

2

2

2

5. (c) Probability

=43

31

21

41

32

21

41

31

21

=41

243

242

241

6. (d) In a leap year, 52 weeks and 2 extra days. Theycan be distributed as {(Sunday, Monday),(Monday, Tuesday), (Tuesday, Wednesday),(Wednesday, Thursday), (Thursday, Friday),(Friday, Saturday), (Saturday, Sunday)}For 53 Sundays we select Sunday from thesample

required probability =72

7. (c) P(Bag A is chosen) = P(A) =31

P(Bag B is chosen) = P(B) =32

R : event that red ball is chosen.

P(R/A) =53 P(R/B) =

52

P(A/R) =)B/R(P)B(P)A/R(P)A(P

)A/R(P)A(P

=73

52

32

53

31

53

31

8. (a) P(M1) =61

6000010000

P(M2) =31

P(M3) =21 , P (D/M1) = 0.02,

P(D/M2) = 0.03, P(D/M3) = 0.04

P(M1/D) = MDP)M(P

)M/D(P)M(P

1

11

=04.0

2103.

3102.

61

02.61

= 1.0101

20.002.0

12.006.002.02.0

9. (c) Two wheelers : W1 , Cars W2, Buses W3

Trucks W4

P(W1) =115 , P(W2) =

113 ,

P(W3) =112 , P(W4) =

111

Accident: A.

P(A/W1) =167 P(A/W2) =

165 ,

P(A/W3) =163 ,P(A/W4) =

161 .

P(W1/A) =

4

1iii

11

)W/A(P)W(P

W/AP).W(P =5735

10. (c) 1)xX(P

2K2 + K 1 = 0, K =21 or 1

But k = 1 is not admissible K =21 .

11. (c) n = 6. 9 422

6244

6 qpCqpC

9 p2 = q2 = (1 – p)2 = 1 – 2p + p2

i.e., 8 p2 + 2p – 1 = 0

(2p + 1) (4p – 1) = 0 p =41 .



12. (b) Mean = np = 4

variance = npq =38 q =

32

41

38

p = 1 q =31 and n = 34 = 12.

13. (c) There are four vertices

14. (d) unbounded, obviously.

15. (c) conceptual

16. (a) conceptual

17. (a) Maximum value of 7x y is 35at (5, 0). Hence maximum of Z is 15500.

18. (d) Minimum value is 150, at (15, 20)

19. (d)

Feasible region is ABCDE with vertices A (2, 0),

B (3, 0), C (0, 3), D (0, 2), E

32,

32

Maximum value is 6 at (3, 0)

20. (a)

The feasible region is ABCDE with verticesA(2, 0), B(3.5, 0), C(2, 2), D(0, 3.5), E(0, 2) Minimum is 2 at E

21. (d) p : 2 is a factor of 3~p : 2 is not a factor of 3q : 8 is a multiple of 4~q : 8 is not a multiple of 4Given statement is p q ~ (p q) = ~ p ~q

22. (d) ~(p ~q) ~p q p q

23. (d)

From column (8) of above truth table it is clear that~ (p q) ~p ~ q is neither a tautology nor acontradiction.

24. (d) p q is true if either both p and q are true or bothare false. Therefore options (a), (b), (c) are correct

25. (c) Given that ~ q r is true and r is false ~ q is true q is false q r is falseAlso given that ~(p q) (q r) is false~ (p q) is true i.e; p q is false p is truep q is true and q p is true

Regular Homework Exercise

1. (b) P(A) =85 , P(B) =

117

P (Problem is solved) = 1 – P(A1) P(B1)

= 1 –2219

114

83

2. (d) n (M) = 70, n (P) = 50, n (MP) = 30n (MP) = 70 + 50 – 30 = 90

Probability =43

12090

3. (b) P(A) = 0.35 P(B) = 0.5P(AB) = 0.2P(AB) = 0.65P(A1B1) = P[(AB)1] = 0.35.

4. (c) P(ace or spade) =521

524

5213

=134

5. (d) P(A) =32 , P(B) =

43

Since both of them agree same, statement is eithertrue or false

Probability that the statement is true

=)falseortruespeaktheythatobability(Pr

truthspeakboththatobabilityPr

=76

41

31

43

32

43

32

1 2 3 4 5 6 7 8p q ~p ~q p q ~(p q) ~p ~ q (6) (7)

TTFF

TFTF

FFTT

FTFF

TFTT

FTFF

FTTT

TTFF

E

C

D

BA

E CD

BA



6. (b) TW – represents transferred ball is whiteDR represents drawn balls is red

P(TW) = 103TP;

105

R P(TG) =102

115

TDP;

114

TDP

RR

WR

;

GR

TDP =

114

Required probability =

RR

DTP

=

WR

W

RR

R

TDPTP

TDPTP

=

114.

102

115.

103

114.

105

115.

103

=

4315

8152015

7. (a) A denotes B.A; B denotes B.Com and C denotesB.Sc classes. S represent selected studentshaving scholarship

P(A) =94 ; P(B) =

93 P(C) =

92

P 100

5C

SP;100

3B

SP;100

2A

S

Required probability =

A

SP).A(PC

SP.CPS

CP

=

1005.

92

1003.

93

1002.

94

1005.

92

=2710

109810

8. (c) Total probability = 1

3p2 + 6p = 1 p =6

486

p 0 p =6

486 .

9. (c) np = 8, and npq = 6

p =41

, q =43 n = 32.

10. (b) np = 2npq p = 2q

But p +q = 1 3q = 131q , p

32

11. (c) (1 – p)4 =8116 or p =

31 ;

P(X =1) =8132

32

314

3

.

12. (c)

Feasible region is OABC number of vertices is 4

13. (d)

14. (b)

The feasible region is unbounded with vertices

A(0, 3), B

23,

21 , C

21,

23 , D(3, 0)

15. (b) Vertices are (3, 0), (4, 0), (0, 4), (0, 3) (1, 1)Minimum value = 5

16. (d) The feasible region have vertices A(1, 0),

B

0,23 , C(1, 1), D(0, 3) with maximum 9 at

B

0,23

17. (c) The vertices are (2, 0),

0,27 ,

(2, 2),

27,0 , (0, 2)

Maximum value =2

35

18. (d) p : The coefficients are real numbersq : The roots are complex number Given proposition is p qbut p q ~ p q

~ q ~ p both (a) and (c) are logically equivalents withgiven proposition

19. (c) ~ (p (~p q) ~ p (~(~ p q)) ~ p (~(~p) ~ q) ~ p (p ~ q)

20. (d)p q ~p ~p q p ~ p qTTFF

TFTF

FFTT

TFTT

TFTT

From the above truth table it is clear thatp ~ p q is neither a tautology nor acontradiction

O A

BC

A

B

C

D



Assignment Exercise

1. (a) Scores : 10, 11, 12 – Event A.Same number appears on each=Event BA = { (4, 6) (5, 5) (5, 6) (6, 4) (6, 5) (6, 6)}B = { (1,1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)}A B = {(5, 5), (6, 6) }

P(A) =366 , P(B) =

366 , P(AB) =

362

P(AB) = P(A) + P(B) – P(A B)

=362

366

366

=185

3610

2. (d) Total number of determinants of order 2with 1 or 0 as entries =22 2 2=16.A denote the set of determinants among them with(+)ve values.

A =

1101

,1011

,1001

P(A) =163

3. (c) Probability =n2

!n2!)1n(

4. (b) Probability = P(one club and a carddifferent from club)+P(two clubs)

=3415

C52C13

C523913

2

2

2

5. (b) P(A) = 0.55 P(B) = 0.25 P(C) = 0.20P(D/A) = 0.02 P(D/B) = 0.04 P(D/C) = 0.06Required proof = P(C/D)

=)A/D(P.)A(P

)C/D(P).C(P

=

06.020.004.025.002.055.006.020.0

=

114

6. (b) 1p

12p9

p6

p5

p3

p1

1p

36 p = 36 ;

P(X 4) =32

3624

p9

p6

p5

p3

p1

.

7. (a) Probability = (1 – 0.05)10 = (0.95)10

8. (c) P(x = r) = 6C r pr q6 – r = 6C r

r

21

r6

21

=64C6 r

Probability of getting at least 4 tails= P (4) + P (5) + P (6)

=3211)1615(

641

.

9. (b) Mean np = 6 ; variance =npq = 222

31

npnpq

31q

32p , n = 6

P(x) =x9x

x9xnx

xn

31

32CqpC

10. (d)

11. (c)

From the graph, feasible region in unboundedwith 3 vertices.

12. (b)

The feasible region is OABC with vertices areO(0, 0), A(4, 0), B(3, 1), C(0, 2) Maximum value is 21 at B(3, 1)

13. (a) Draw the graph, then the feasible region have

following vertices (1, 0),

0,23 , (1, 1),

23,0 ,

(0, 1) Minimum value is 2. at ( 1, 0)

14. (d) Contra positive of p q is ~ q ~ p.contra positive of (p ~ q) r is~ r ~ (p ~ q)ie ~ r (~ p q)ie ~ r (p q)

15. (a)

p q ~p ~q p~q ~pq (p~q) (~pq)

T T F F F F F

T F F T T F F

F T T F F T F

F F T T F F F

From the above truth table it is clear that(p ~ q) (~ p q) is a contradiction

CB

AO



Additional Practice Exercise

1. (c) Required probability = 1 –3231

21 5

2. (c) Total number of cases =3

C100

Number of favorable cases =3

C22

Probability =105

19899100

202122

3C1003C22

3. (b) 4 questions which the candidate hasanswered correctly can be selected in6C4 or 15 ways. Remaining 2 questions,candidate can mark in 33 = 9 ways.Total no : of ways = 46

Probability =66 4

1354

915

4. (d) P(X = x) = 5Cx

5

21

P( 2) = 1– [P(X = 0) + P( = 1)]

= 1 –

55

215

21

= .1613

5. (c) Let the events be A, B, C

P(A) =31 , P(B1) =

54 , P(B) =

51

Since, P(A) = P(B) + P(C) = 1

P(C) = 1 –157

51

31

6. (a) Probability =225

8155

154

156

7. (b) A : card drawn is a picture cardB : card is a green

P (B/A) =31

5212524

)A(P)AB(P

8. (d) P(C) =81 P(S) =

84

P(B) =82 P(T) =

81

103

1071C

LP ; 101

1091S

LP

105

211B

LP ; 106

521T

LP

Required probability = LSP

=

CLP.CPS

LP).S(P

=

106.

81

105.

82

101.

84

103.

81

101.

84

=234

610434

9. (d) n C3 .8n21Cn

21 n

5

n

Probability = 8 C2647

21 8

10. (c) Variance 2 = E (x2) 2)x(E 4 = E (x2) 9 E (x2) = 13.

11. (d) log1 + loga2 + loga3 + loga4 +loga5 = 1loga (2345) = 1loga120 = 1 a = 120

12. (d) P(2 < y 20) = P(1<4)=0.2+0.3+0.1=0.6

13. (b) Mean =E(y) = E(a b)= aE(x) – b = 10a b

10a b= 0 b = 10 a1 = Var (y) =Var(ax b)=a2 Var(x) = a225

a2 =251 a = 0.2

b = 10a = 2

14. (d) np = 32 npq = 16, 3216

npnpq

21q ; p =

21 ; n = 64

p (X = 0) = 64C0

0640

21

21

=64

21

15. (c) Favourable cases are { 1, 3, 5 }Sample space {1, 2, 3, 4, 5, 6 }Probability of getting an odd number in

a single throw is21

63

The given problem corresponds to a

Binomial distribution with n = 8, p =21 , q =

21

variance = npq = 221

218

16. (c) E (X) = 2

434

413

21

= 1 + 343 =

419 .

17. (a)

18. (d) Feasible region is unbounded and having only 3vertices



19. (a) There are four vertices.

20. (d) Vertices are (20, 0), (30, 0), (12, 12) and values ofz = 2x + y are 40, 60, 36.

21. (c) 8x + 3y = 0 (Profit line provide no gain no loss inz = 0)

22. (a) The vertices are (2, 0), (4, 0), (4, 2), (1, 5), (0, 5),(0, 2). Maximum value is 14 at (4, 2)

23. (a) Feasible region have vertices (6, 0), (3, 1)(0, 4) and minimum is 5, at (3, 1)

24. (b) Vertices are (2, 0), (3, 0), (0, 3), (0, 2),

32,

32

Minimum value is38 .at

32,

32

25. (d) Vertices (0, 0), (40, 0), (40, 20), (0, 20)Maximum value is 160 at (40, 20)

26. (d) Vertices are (0, 0),

0,23 , (1, 1),

23,0

Maximum value = 5 at (1, 1)

27. (b) ~ (p ~ q) ~ ([p ~ q] [~ q p]) ~ [p ~ q] ~ [~ q p] ~[~ p ~ q] ~[q p] (p q) ~ (p q) ~ (p q) (p q) (p q) (p q)

28. (a) Inverse of P Q is ~P ~Q Inverse of (p q) r is~ r ~ (p q)i.e; ~ r ~ (~ p q)~ r (p ~ q)r (p ~ q)

29. (b) Since r is false and (p (q r) r is trueclearly p (q r) is false P is true and q r is falsesince r is false and q r is falseq is also false p ~ q is T T = T

p q is T F = Fq p is F T = Tq ~ p is F F = T

p q is false is correct

30. (d) The converse of p q is q pThe inverse of p q is ~p ~qThe contrapositive of p q is ~q ~pFrom statement (3), option (a) is true

In (b),Inverse of p q is ~p ~q The converse of inverse of p q is ~q ~p (1)Again the converse of p q is q p Inverse of the converse of p q is ~q ~p (2)from (1) and (2) option (b) is true

In (c),The contrapostitive of p q is ~q ~p the converse of contrapositive of p q is ~p ~q itis the inverse of p q (c) is true

(In (d)The inverse of p q is ~p ~q and contrapositive ofinverse of p q is ~(~q) ~(~p)i.e; q p, it is not logically same as p q

m1210

Documents