Probability.pdf

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Probability In this chapter, only rules with illustrative examples have been given. For basics please consult "Magical Book on Quicker Maths."

4 •••

Theorem: Probability of an event (E) is denoted by P(E) and is defined as P(E) no. of desired events

n(S)

total no. of events (ie no. of sample space)

^ 6

2 =3

(vii) E (a no. greater than 6) = { } , i.e. there is no number greater than 6 in the sample space

Rule 1

n(E)

P

.-. P ( E ) = ^ = 0 Probability of an impossible event = 0 (viii) E (a no. less than or equal to 6) = {l,2,3,4,5,6},n(E) = 6

Illustrative Example Ex.:

A dice is thrown. What is the probability that the number shown on the dice is (i) an even no; (ii) an odd no.; (iii) a no. divisible by 2; (iv) a no. divisible by 3; (v) a no. less than 4; (vi) a no. less than or equal to 4; (vii) a no. greater than 6; (viii) a no. less than or equal to 6. Soln: In all the above cases, S = { 1 , 2 , 3 , 4 , 5 , 6 } , n(S) = 6. (0 E(anevenno.;={2,4,6},n(E) = 3 P(E) =

n(S) n(E) _ 3 _ 1

•'• " 6~2 (iii) E (a no. divisible by 2) = {2,4,6}, n(E) = 3 (

E

)

Probability of a certain event = 1. Note: 0 < P(E) < 1

Exercise 1.

In a simultaneous throw of two dice find the probability of getting a total of 8. a)

(ii) E ( a n o d d n o . ) = { l , 3 , 5 } , n ( E ) = 3

P

P(E) = « = 1 6

=

9

b

>' 336 l

1 c) ' 6

A coin is successively tossed two times. Find the probability of getting 2) at least one head. l)exatly one head 14 12 1 2 4'7 >2'3 2'4 In a box carrying one dozen of oranges, one third have become bad. I f 3 oranges are taken out from the box at random, what is the probability that at least one orange out of the three oranges picked up is good? 2

a)

C )

P(E) =

3 V 6 ~2

(iv) E (ano. divisible by 3) = { 3 , 6 } , n(E) = 2 2-1 .-. P(E) =

6 ~3

(v) E (a no. less than 4 ) = { 1 , 2 , 3 } , n(E) = 3 3

1

(vi) E (a no. less than or equal to 4) = {1,2,3,4} n(E) = 4

d) Data inadequate

54

1 a)

55

b)

55

45 c) ' 55

d

3 d )

5?

(SBI Bank PO Exam 1999) Out of 15 students studying in a class, 7 are from Maharashtra, 5 are from Karnataka and 3 are from Goa. Four students are to be selected at random. What are the chances that at least one is from Karnataka?

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P R A C T I C E B O O K ON Q U I C K E R MATHS

664 12 TI

a )

5.

11 b

)

B

10

1

T5

C)

d )

15

(BSRB Guwahati PO1999) The probability that a teacher will give one surprise test 1 during any class meeting in a week is —. I f a student is absent twice, what is the probability that he will miss at least one test. 4 a)

6.

1

l5~

b)

Ts

91

16

Ys

c)

i2T

d )

14. The odds in favour of an event are 3:5. The probability of occurrence of the event is 1 a)

c)l d) None of these 2 In a simultaneous throw of two coins, the probability of getting at least one head is .

c)

4 b) 77 9

1 c)" ' 5

» 1

c

5 8 3 d) 15. The odds against the occurrence of an event are 5:4. The probability of its occurrence is . 4

>?

4" 16. In a lottery there are 20 prizes and 15 blanks. What is the probability of getting prize? a

(BSRB Mumbai PO Exam 1999) In a throw of a coin, the probabi; y of getting a head is 1

1

?

b)

a)

To

d )

>7

d

>7

17. An urn contains 9 red, 7 white and 4 black balls. A ball is drawn at random. What is the probability that the ball drawn is not red?

3 )

1

2

a>2 8.

>y

b

c

d

1 I

)

a )

3 c)

3

>8 7 Three unbiased coins are tossed. What is the probability of getting at most 2 heads? 3 )

b

1 a)-

d

)

3 b)-

7 0 -

1 d)-

10. A fair coin is tossed 100 times. The probability of getting head an odd number of times is . 1 3 c)~ d) a) 3 : - 4 11. A bag contains 6 black balls and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?

9

TT

b

)

2 l

2 C )

TT

11 d

>20

18. What is the probability that a number selected from the numbers 1,2,3,4,5,..., 16 is a prime number? 1

Three unbiased coins are tossed, what is the probability of getting exactly two heads? 1

9.

3 > i

1

a)

16

b )

"c) I

¥

d

>l6

19. Ticket numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn bears a number which is a multiple of 3? 3 a)

20

b )

10

2 c)5 T "*

1 d)

20. Ticket numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn bears a number which is a multiple of 3 or 7?

2

a)

1 b)~

L5

c)

b )

4 a)y

3 b)-

4 O-

16

b) ' 13

c) ' 26

1. b;

1 d)-

12. A bag contains 8 red and 5 white balls. 2 balls are drawn at random. What is the probability that both are white? a)

Answers

d)

39

13. A bag contains 5 blue and 4 black balls. Three balls are drawn at random. What is the probability that 2 are blue and 1 is black?

Hint: In a simultaneous throw of two dice Sample space = 6 x 6 = 36 Favourable cases are (2,6) (3,5) (4,4) (5,3) (6,2) 5 So, the required probability = —

2. a;

Hint: In tossing a coin 2 times the sample space is 4 ie (H,H),(H,T),(T,H),(T,T) 1) I f A, denotes exactly one head then, A, = { ( H , T ) ( T , H ) } S o , P ( A , ) = 2) I f A denotes at least one head

I a

)

I

b

>7

C

>6

d) None of these

then A = {(H, T), (T, H), (H, H)}

.\) = •

:

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Probability

3.c;

Hint: n ( S ) = C = , 2

^ ^ 3x2

3

= 2x11x10 = 220

10. c; Hint: w(s) = 2

1 0 0

n(E) = no. of favourable ways

No. of selection of 3 oranges out of the total 12

=

, 0 0

C + 1

, 0 0

C +...+ 3

oranges = C =2 « 11 x 10 = 220 , 2

665

, 0 0

C

=2

9 9

, 0 0

- = 2 1

9 9

3

|v C + C + C + ... = 2 " - | n

No. of selection of 3 bad oranges out of the total 4

1

n

3

n

5

,

bad oranges = C = 4 4

3

.-. n(E) = no. of desired selection of oranges = 220-4 = 216

v

4. b;

;

n(s)

220

i<

n(s)

'

K

2

2

m

Note: The given case can be generalised as "If a unbiased coin is tossed times, then the chance that the head will present itself an odd number of times is

55

Hint: Total possible ways of selecting 4 students out of 15 students =

•

15x14x13x12 , „ „ <- = — —-— = 1365 1x2x3x4 4

1

it

2 ' 11. a; Hint: Total no. of balls = (6 + 8) = 14 No. of white balls = 8

The no. of ways of selecting 4 students in which no student belongs to Karnataka =

1 0

C

.-. P(drawing a white ball) =

4

.-. number of ways of selecting at least one student fromKarnataka= Probability = 5. b;

C -

, 5

1 0

C , =115 5.

1155

77

11

1365

91

13

14 ~ 7

12. d; Hint: n(S) = Number of ways of drawing 2 balls out of 13 =

, 3

C =

13x12

7

= 78

n(E) = No. of ways of drawing 2 balls out of 5

Hint: The probability of absenting of the student in = 'C =

2 1 the class " ~ " r 6 3

5x4

2

= 10

n(E) _ 10 _ 5 .-. the probability of missing his test 6. a;

•'• 5*3 ~ 15 "

Hint: Here S = {H, T} and E = {H} n{E)_\

n(s)

2

P

(

E

)

Hint: S== {HH, HT, TT, TH} and E = {HH, HT, TH} Hint:S

... He)8. d;

n{E)

=

3

n(s)~4

Hint: S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTTJand E = Event of getting exactly two heads = {HHT, HTH, THH}

9x8x7

of 9 = C = 9

7.c;

~njs)~ 78 ~ 39

=

13. c; Hint: Let S be the sample space and E be the event of drawing 3 balls out of which 2 are blue and 1 is black. Then, n(S) = Number of ways of drawing 3 balls out 3

3x2x1

n(E) = Number of ways of drawing 2 balls out of 5 and 1 ball out of 4. 5

C + C, 2

9. c;

;

n{s)

8

Hint: n(S) = 8 [See hint of the Q. No. 8] E = Event of getting 0, or 1 or 2 heads = {TTT, TTH, THT, HTT, HHT, HTH, THH} or, n(E) = 7 n(E)

1

( 5x4

4

4M)

14

+ 4 =14

i

•• ^ = ^ ) ^ 6 14. b; Hint: Number of cases favourable to E = 3 Total number of cases = (3 + 5) = 8 '3 .-. P(E) = 8' P

V

= 84

=

=

15. b; Hint: Number of cases favourable to E = 4 Total number of cases = (5 + 4) = 9 4 ••• P(E)= n -

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P R A C T I C E B O O K O N Q U I C K E R MATHS

666

16.c;

Hint: P(getting a prize)

17. d; Hint:P(red) =

1

20

20

4

20 + 15

35

7

9

_9_

9+7+4

20 II

9*

P(not-red) =•

1 20

{Sunday}, n(E)= 1 .-. P ( E ) = y

Exercise 1.

What is the probability that a leap year selected randomly will have 53 Mondays?

~ 20

7 a)-

n(E)

6

3

•• ®=^) r6 ip

19. b;

=

=

53

Hint: S = {1,2,3,... 20} and E - ;3,6,9,12,15,18} a

n{E) •'•

P

(

E

)

=

n(s) " 20 ~ 10

20. c; Hint: Clearly, n(S) = 20 and E = {3,6,9,12,15,18,7, 14}

)

Hint: A leap year has 366 days = 52 weeks + 2 days These 2 days can be (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday) ... or, (Saturday, Sunday). Out of these total 7 outcomes, there are 2 cases favourable to the desired event ie (Sunday, Monday) and (Monday, Tuesday) 2 .-. required probability = — .

2. b;

Hint: An ordinary year has 365 days ie 52 weeks and 1 day. So the probability that this day is a Sunday is 1 7 '

Illustrative Example

Rule 3

Ex.:

.-. P(E) = (ii) When the year is not a leap year, it has 52 complete weeks and 1 more day that can be {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}, n(S) = 7 Out of these 7 cases, cases favourable for one more Sunday is

53

1. a;

Rule 2

(i) What is the chance that a leap year selected randomly will have 53 Sundays? (ii) What is the chance, if the year selected is a not a leap year? Soln: (i) A leap year has 366 days so it has 52 complete weeks and 2 more days. The two days can be {Sunday and Monday, Monday and Tuesday, Tuesday and Wednesday, Wednesday and Thursday, Thursday and Friday, Friday and Saturday, Saturday and Sunday}, i.e. n(S) = 7. Out of these 7 cases, cases favourable for more Sundays are {Sunday and Monday, Saturday and Sunday}, i.e.n(E) = 2

48 d)

c)

Answers

" 20 ~ 5 •

Problems based on leap year. A leap year has 366 days, hence it has 52 complete weeks and 2 more days. When the year is not a leap year, it has 52 complete weeks and 1 more day.

1

1 b)

3^

n(E) _ _8_ _ 2 P(E) =

5

b) c) — d) Data inadequate 7 7 What is the probability that an oridinary year has 53 Sundays?

18. c; Hint: S= {1,2,3,..., 16} andE= {2,3,5,7,11,13}

Problems based on dice Following chart will be helpful in solving the problems based on dice. Chart: When two dice are thrown, we have, S = {(1,1), ( 1 , 2 ) , ( 1 , 6 ) , (2,1), ( 2 , 2 ) , ( 2 , 6 ) , (3,1), ( 3 , 2 ) , ( 3 , 6 ) , ( 4 , 1 ) , ( 4 , 6 ) , ( 5 , 1 ) , ( 5 , 6 ) , (6,1) (6,6)} n(S) = 6><6 = 36 Sum of the no. n(S) of the two dice (!)

(«)

2

12

3 4

Events (0

(ii)

1

[1,1)

{6. 6}

11

2

(1.2), {2, 1)

{6, 5}, {5, 6}

10

3

(1.3), (3,1), {2,2}

{6,4}, {4,6}, {5,5}

5

9

4

(1,4), {4. 1), (2,3), (6,3), {3,6}, |5, 4), (3,2) H, 5}

6

8

S

Hi 5), (5, 1}, (2, 4), {6,2}, {2,6}. (5,3), {4.2}, {3,3} {3, 5}, {4. 4}

7

6

{1,6}, {6, 1}, {2,5|, {5.2}, M, 3}, {3,4}

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Illustrative Example When two dice are thrown, what is the probability that (i) sum of numbers appeared is 6 and 7? (ii) sum of numbers appeared < 8? (iii) sum of numbers is an odd no? (iv) sum of numbers is a multiple of 3? (v) numbers shown are equal? (vi) the difference of the numbers is 2? (vii) Sum of the numbers is at least 5. Soln: (i) Use the above chart:

In a simultaneous throw of two dice, what is the probability of getting a doublet?

Ex.:

1 3 )

3.

For 7, reqd probability

:

7 &

4.

1

Answers

36

6

1. b;

a

6 •••

P ( E ) =

8 ••

P ( E

>=36

2 =

9

(vii) Events; either 2 or 3 or 4 or 5 n(E)=l+2 + 3+4=10 n(S) = 36 n(E)_ •

P

(

E

)

=

10 _ 5

«(S)"36"18'

Exercise In a throw of a die, the probability of getting a prime number is 1

1 b)

c)

6

1 V~4

>6

Hint: Here S = {1,2,3,4,5,6} and E =» {2,3,5}

3 2. a;

r

1 .

i

Hint: In a simultaneous throw of two dice n(S) = 36 Let E = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)} 6

••• 3. b;

5

P

1

^ 3 T 6 -

Hint: Clearly, n(s) = 36 Let(E)= {(4,6), (5,5), (6,4), (5,6), (6,5)} 5 .-. P(E) =

4.c;

36

Hint: Clearly, n(S) = 36 Let E = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

36

37 = 6

(vi) Events = {3,1}, {4,2}, {5,3}, {6,4}, {4,6}, {3,5}, {2,4}, {1,3} orn(S) = 8

I

C )

2

.-. P(E) =

1

2 d) ~' 3

1

37

,.P(E)=

11 = 1 36 ~ 3

33 j _4

d) r j

6

(v) Events = { l , 1}, {2,2}, {3,3}, {4,4}, {5,5}, {6,6}; n(S) = 6

b )

1

=

:

)

5

n

~ 36

26 13 .-. reqd probability = ~ ~r^ 36 18 (iii) Desired sums of the numbers are 3,5,7,9 and 11; n(S) = 2 + 4 + 6 + 4 + 2 = 18 . . . 18 1 .-. reqd probability = — - % 36 2 (iv) Desired sums of the numbers are 3,6,9 and 12; n(S) = 2 + 5 + 4 + l = 12'

c

In a single throw of two dice what is the probability of not getting the same number on both the dice?

(ii) Desired sums of the numbers are 2,3,4,5,6,7 and 8;

.-. reqd probability

)

n(S)

n ( S ) = l + 2 + 3 + 4 + 5 + 6 + 5 = 26

4

In a simultaneous throw of two dice, what is the probability of getting a total of 10 or 11 ?

n(E) _ 5 For 6, reqd probability =

b)

6~

So, P(not-E) = 1 —

_5 6

Rule 4 Problems based on cards Following chart will be helpful to solve the problems based on cards. Chart: A pack of cards has a total of 52 cards. Red suit (26) Diamond (13)

Heart (13)

Black suit (26) Spade (13)

Club (13)

The numbers in the brackets show the respective no. of cards in that category. Each of Diamond, Heart, Spade and Club contains nine digit-cards 2,3,4,5,6,7, 8,9 and 10 (a total of 9

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668

P R A C T I C E B O O K ON Q U I C K E R MATHS x 4 = 36 digit-cards) along with four Honour cards Ace, King, Queen and Jack (a total of 4 x 4 = 16 Honour cards).

(ii) Total no. of Queens = 4 Selection of 1 Queen card out of 4 can be done in 4

Illustrative Examples Ex. 1: A card is drawn from a pack of cards. What is the probability that it is (i) a card of black suit? (ii) a spade card? (iii) an honours card of red suit? (iv) an honours card of club? (v) a card having the number less than 7? (vi) a card having the number a multiple of 3? (vii) a king or a queen? (viii) a digit-card of heart? (ix) ajack of black suit? Soln: For all the above cases n(S) - 52, 26,

26

or,

(OH

26

52.

52

C,

4 ways. He can select the remaining 1 card

=

from the remaining (52 - 4 =) 48 cards. Now, cards in 48

C, =48 ways. n(E) = 4 x 4 8

P(E) =

32

26x51 221 (iii) Total no. of honours card = 16 To have no honours card, he has to select two card: out of the remaining 52 - 16 = 36 cards which he car. do in

3 6

C

= ^

9

.'. P(E) =

:52

^ = 18x35 ways 2 .

18x35

105

26x51

221

16, (iv)P(E) =

(v » C , = n )

4x48

26x51

8x15

20

26x51

221

(v)n(E)= C , x C , = 4 x 4 = 16 4

13

00

s

(iii)

-

4x2

2_

52

13 • ® 26x51 663 Ex. 3: From a pack of 52 cards, 3 cards are drawn. What is the probability that it has (i) all three aces? (ii) no queen? (iii) one ace, one king and one queen? (iv) one ace and two jacks? (v) two digit-cards and one honours card of black suit? ,

5x4 (iv)

52

(vi)

=

(v)

n

5

52

3x4 52

4 1 4 1 (vii) P(a king) = — = — ; P(a queen) = ^ = 1 .-. P(a king or a queen) = 7J

(viii)

+

1 7J

=

13 1

52 26 Ex. 2: From a pack of 52 cards, 2 cards are drawn at random. What is the probability that it has (i) both the Aces? (ii) exactly one queen? (iii) no honours card? (iv) no digit-card? (v) One King and one Queen? ( k )

52^ Soln: For all the above cases, n(S)=

52x51

C =—^—— 2

P

Soln: For all the above cases, n(S)

2

2 52

4

5 2

C

(i)n(E) _

52x51x50 = 26x17x50 3x2 4

C =4 3

=

,P(E) = (ii)n(E)=

4 8

C

,P(E) = 26x51

1

26x17x50 3

=8x47x46 8*47x46

4324

26x17x50

5525

(iii)n(E)= C,x C,x C, =4x4x4 4

(i) Total no. of Aces = 4 nr. VC = :. n(E)= 4

x

3

2

••, P(E):

1 26x51

221

= f.6

5525

,.P(E) =

4

4

4x4x4

16

26x17x50

5525

(iv)n(E)= C , x C , = 4 x 6 4

4

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Probability

4x6

,P(E) =

6 P(E) =

26x17x50 ~ 5525 (v)n(E) = 36P(E) =

:

3. a;

x ° C , =18x35x8

18x35x8

252

26x17x50

1105

m

28

7

n(s)

52

13

Hint: n(S) = Number of ways of drawing 2 cards out of52 =

52x51

5 2

=

1 3 2 6

2x1 n(E) = Number of ways of drawing 2 cards out of 4 2

Exercise One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card? a)

b)

13

1

1

4

d) ' 13

c)

52

One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is either a red card or a king? 27

1 c

>lJ

d)

1 .'. P(E) = 4.c;

1

2

2^1

b )

13

3 C )

2~6

• 5.b;

1

'

a

d) None of these

9 b)— 13

2 c)— 13

g a

)

=

C )

«(5)"52"13-

Hint: There are 13 hearts and 3 more kings

\nswers

52

1 4 _ 1 -,P(F) = - - - a n d

]_ _1_ 4 L3 +

4 52

Rule 5 Theorem: If a bag contains x red, y yellow and z green balls, 3 balls are drawn randomly, then the probability of the balls drawn contain balls of different colour is given by 6xyz {x + y + z)(x + y + z-\)(x + y +

52 * 13

Hint: Clearly n(S) = 52. There are 26 red cards (including 2 kings) and there are 2 more kings. Let (E) be the event of getting either a red card or a king. Then, n(E) = 28

11

P(a spade or a king) = P(E u F) = P(E) + P(F) - P(E n F)

Hint: Clearly, n(S) = 52 and there are 16 face cards. ] 6 _ _4_

~ 13

52

1 P(EnF)=-

d) ~' 13

C )

13 + 3 _ 4

9_ P(neither a heart nor a king) = » 4 = 13 13 ' Hint: Let E and F be the event of getting a spade and that of getting a king respectively. Then £ n F>s the event of getting a king of spade v n(E) = 13, n(F) = 4 and n(E n F) = 1 So, P(E) =

4

a

P(E) =

E

1

A card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a spade or a king? 4 3 2 1 ) 77 b) 13 ' 13 T3 | I . a;

(

P(heart or a king) =

d)None of these c) 26 13 ' 13 A card is drawn from a pack of 52 cards. A card is drawn at random. What is the probability that it is neither a heart nor a king? 4 ) 77 13

P

52

What is the probability of getting a king or a queen in a single drawn from a pack of 52 cards? a)

221"

n(E) _ 8 _ 2

Two cards are drawn at random from a pack of 52 cards. What is the probability that the drawn cards are both aces? a)

1326

Hint: Clearly, n(S) =52, there are 4 kings and 4 queens.

z-2)

C] X ^Cy X C| or

Illustrative Example Ex:

A bag contains 3 red, 5 yellow and 4 green balls. 3 balls are drawn randomly. What is the probability that the balls drawn contain balls of different colours?

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P R A C T I C E B O O K ON Q U I C K E R MATHS

670 Soln: Detail Method: Totalno.ofballs = 3+ 5 + 4 = 12 n(S)

_

12

12x11x10

c, =

3

6x3x5x4 _ 3 the required answer = ——~—— - — . 12x11x10 11

era

A bag contains 3 red, 5 yellow and 4 green balls, j balls are drawn randomly. What is the probability that balls drawn contain exactly two green balls? Sol: Detail Method: Total no. of balls = 3 + 5 + 4 = 1 2 1 2

"

( S ) =

^

12x11x10

C

3

„„„

- ^ 2 -

=

=

2 2

°

2 green balls can be selected from 4 green balls r

24 a)^T

the remaining (12 - 4) = 8 balls in C , ways.

14 b ) -

13 O -

4

21 d ) -

Tl8

35 b )

35

136

C )

l37

2

.-. P(E) =

a )

r7

b

)

6

n

c)

n

8 d )

1.

b)

c)

17

17

d )

3. a

12

12x11x10

55

91

T\9f

a)

Theorem: If a bag contains x red, y yellow and z green balls, 3 balls are drawn randomly, then the probability of the balls drawn contain (i) exactly 2 green balls is given by

15 b

)

9l

10 C )

91

d) Data inadequate

28 C )

?T

27 d)

37

Yl

(iii) What is the probability that balls drawn contain exactly 2 red balls? 54 a)

(ii) exactly 2 yellow balls is given by

3x4x3x8

12x11x10

20

Rule 6

(x + y + z) (x + y + z -1) (x + y + z - 2) or

3x4x(4-l)x8

(ii) What is the probability that balls drawn contain exactly 2 yellow balls?

4.c

3z(z-\)(x + y)

55

20 a)

I7

Answers 2.b

12

A bag contains 4 red, 6 yellow and 5 green balls. 3 ba are drawn randomly. (i) What is the probability that balls drawn contain exactly 2 green balls?

8 17

48 220

Exercise

T7

A bag contains 4 yellow, 5 red and 8 green balls. 3 balls are drawn randomly. What is the probability that the balls drawn contain balls of different colours? a)

6x8 = 48

s

Quicker Method: Applying the above theorem, we have the required answer

163

A bag contains 6 red, 8 yellow and 4 green balls. 3 balls are drawn randomly. What is the probability that the balls drawn contain balls of different colours? 3

2

n(E) = ' C x C,

35 d )

C ways and the rest one ball can be selected frorr 8

A bag contains 5 red, 7 yellow and 6 green balls. 3 balls are drawn randomly. What is the probability that the balls drawn contain balls of different colours?

4

l.a

+ z)

A bag contains 4 red, 6 yellow and 5 green balls. 3 balls are drawn randomly. What is the probability that the balls drawn contain balls of different colours?

3 )

4.

X

Ex:

'

35 3.

W*+$C

Illustrative Example

220 " 11 Quicker Method: Applying the above theorem, we have

2.

o r

(x + y + z) (x + y + z -1) (x + y + z - 2) or

60 _ 3

1.

3x(x-\)(y

4

P(E) =

Exercise

+ y + z-2)

(iii) exactly 2 red balls is given by

C , x C , x C , = 3 x 5 x 4 = 60 5

+ z)

(x + y + z)(x + y + z-\)(x

= 220

3x2

In order to have 3 different coloured balls, the selection of one ball of each colour is to be made. n(E)=

3y(y-\)(x

>cx:c2 2.

455

44 b) ' 455

54

d) None of the«

A bag contains 5 red, 7 yellow and 6 green balls. 3 ba.: are drawn randomly. (i) What is the probability that balls drawn contain exactly 2 green balls?

https://telegram.me/Banking_Zone Probability 14 a) 77 68

b

13 ) — 68

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15 )7TT 91

the balls drawn contain no yellow ball? Soln: Detail Method: Total no. of balls = 3 + 5 + 4 = 12

15 d) 68

C C )

,2-

(ii) What is the probability that balls drawn contain exactly 2 yellow balls? 77 a) —

76

b)

2

?

c)

2

77 2

?

d)

1

76

= 7 balls in C 7

2

?

54

55

1

b )

27 a) 777 91

20 b)' — 91

20

20 91

1.

54 c) - r j j d) Data inadequate

19 b

12 c)

>9l

Answers l.(i)a (ii)c (iii) a I (i) a (ii) a (iii) c

91

2.(i)d

24 33 12 a) 777 b) — c) 77 d) Data inadequate yi 91 65 (ii) What is the probability that the balls drawn contain no red ball? 12 a) 77 65

d) None of these

(ii)a

A bag contains 4 red, 5 yellow and 6 green balls. 3 balls are drawn randomly. (i) What is the probability that the balls drawn contain no yellow ball?

Rule 7

+

(x + y + z)(x + y + z - l ) ( x + y + z - 2 )

or

x+y+

Ts

33 b)

Ti

55

(*«) { z)

3 )

c

24

204

c)

Yi

1

143 408

b )

55 c ) T272 77

d )

55 d)' 208

(ii) What is the probability that the balls drawn contain no red ball?

u) no red ball is given by (y + z)(y + z - l ) ( y + z - 2 ) (x + y + z)(x + y + z - l)(x + y + z - 2)

55 or

( ) x+y+:

c a)

+

(x+y)

y-2)

(x + y + z) (x + y + z -\)(x + y + z - 2)

Or

(r y+ ), +

2~n

55 b

)

2 ^

143 C)

4^8"

143 d )

406~

(iii) What is the probability that the balls drawn contain no green balls?

iii) no green ball is given by (x + y)(x + y-\)(x

d) None of these

5~ A bag contains 5 red, 6 yellow and 7 green balls. 3 balls are drawn randomly. (i) What is the probability that the balls drawn contain no yellow ball? a)

z-2)

24 c)' 91

(iii) What is the probability that the balls drawn contain no green balls?

(iii)c

Theorem: A bag containsx red,yyellow andzgreen balls. I balls are drawn randomly. The probability of the balls Jrawn contain (i) no yellow ball is given by

:

_55_ _55_ _55_ 408 272 204 208 A bag contains 3 red, 5 yellow and 7 green balls. 3 balls are drawn randomly. (i) What is the probability that the balls drawn contain a )

Illustrative Example •jc

33 b)' -91

12

(x + z)(x + z-l)(x

= 35

3x2

n

(iii) What is the probability that balls drawn contain exactly 2 red balls? a)

3

Exercise d) None of these

c)' 91

91

c =

220 " 44 ' Quicker Method: Applying the above theorem, we have 7 x 6 x 5 _ 35 _ 7 the required answer = ,, - rrr - —. 12x11x10 220 44

d )

54

27

7

.-. P(E) =

(ii) What is the probability that balls drawn contain exactly 2 yellow balls? a)

ways.

35 _ 7

55

C )

3

7x6x5

n(E)

406 408 408 480 A bag contains 4 red, 5 yellow and 6 green balls. 3 balls are drawn randomly. (i) What is the probability that balls drawn contain exactly 2 green balls? a )

= 220

3 balls can be selected from 3(red) + 4(green)

(iii) What is the probability that balls drawn contain exactly 2 red balls? 55

12x11x10

A bag contains 3 red, 5 yellow and 4 green balls. 3 balls are drawn randomly. What is the probability that

3.

b )

C )

d )

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P R A C T I C E B O O K ON Q U I C K E R MATHS

672 no yellow ball?

4x3x2 _ 1 = P(girls are together) 8! " ° ' 8 x 7 x 6 ~ 14 .-. P(A11 girls are not together) = 1 - P 4!5!

:

24

25

n

44

8

Yi

v

Ts

9l b) c) d) (ii) What is the probability that the balls drawn contain no red ball? a

)

8 a )

6T

44 b )

^T

24 c )

, 1 13 (All girls are together) = 1 - — = — . Quicker Method: Applying the above theorem, we have

45

91

d)

Y\

5!4! the required answer = 1 - 8!

(iii) What is the probability that the balls drawn contain no green balls? 44 a)

9l

24 b)

~9l

5!4!

|8 c)

2.(i)a

(ii)c

(iii)b

6!4! ) ~m 9!

5!4! b) — ~' 9!

5!4! c) ~' 10!

6!4!

> IT

a

(x + l)!y!

5!5! b

)^T

5!5! c

)Tri

(y + l)!x!

1 a)-

(x + y)!

\x + l)!y!" (x + y)!

20 b ) -

19 0 -

6 a)

(x + y)! C .x!y! y

(x + y )

2.

126

Ex:

There are 4 boys and 4 girls. They sit in a row randomly. What is the chance that all the girls do not sit together? Soln: Detail Method: Total no. of arrangements = n(S) = P - 8! S

Consider all the 4 girls as one, we have 4 boys + 1 girl

"* > 126

126

d)None of these

6^

65 b )

67

1 c

76

t7

)

d)

Ti

(ii) What is the chance that all the boys sit together" 76

J_

d) Data inadequate 77 77 (iii) What is the chance that all the girls do not sit together? a )

67

b)

C )

65

= 5 persons. Which can be arranged in P = 5! ways.

1

7

76

I

5

3 )

But the 4 girls can also be arranged in P = 4! ways 4

2 d ) -

121 c)

b

1 a )

5

ToT

37 25 d) Data inadequate b)T7 c) 42 ' 42 There are 6 boys and 5 girls. They sit in a row random!) (i) What is the chance that all the girls sit together?

Illustrative Example

8

d )

(v) What is the chance that the no two girls sit together"

(y + l)!x!

(v) no two girls sit together (x >y) is given by

6!4!

(iv) What is the chance that all the boys do not sit together?

(iii) all the girls do not sit together is given by

x+\

9

(iii) What is the chance that all the girls do not sit together?

(x + y)! _

(iv) all the boys do not sit together is given by

d) None of these

(ii) What is the chance that all the boys sit together

Theorem: There are 'x' boys and 'y' girls. If they sit in a row randomly, then the chance that

(ii) all the boys sit together is given by

There are 5 boys and 4 girls. They sit in a row randomly (i) What is the chance that all the girls sit together?

a

Rule 8

(i) all the girls sit together is given by

14 '

Exercise 1.

(ii)b (iii)a (ii)b (iii)d

14

8!

65

Answers l.(i)a 3.(i) a

o

4

among themselves. So, in 4! x 5! ways can the persons be arranged so that girls are together />

67

b )

66

C )

77

d )

77

(iv) What is the chance that all the boys do not sit together?

1 a )

7T

65 b)

66

c)

66

76 d )

7T

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(v) What is the chance that the no two girls sit together? .-. Required probability, P(E) =

21 3 19 1 a)— b)— c) d) 22 '22 ' 22 ' 22 There are 5 boys and 5 girls. They sit in a row randomly, (i) What is the chance that all the girls sit together? 1 a) 7 7

41 b) -

3 c) -

3 b)

41 42

3

1 d )

C)

41 39 31 d) None of these a) 742 7 b)' — c) 42 42 (iv) What is the chance that all the boys do not sit together? a)

3_

b)

35

11

c)

42

d) Data inadequate

42

Answers 1. (i)a 2. (i) a 3. (i) a

(ii)b (iii)b (ii)b (iii) a (ii)d (iii) a

(iv)c (iv)d (iv)b

(v)a (v)d

(4 + 3 + 5)(4 + 3 + 5 - l ) 12 + 6 + 20

19

12x11

66

Rule 9

cr

+ycr+zcr

x

given by

(x +

y

+

z

)

(x + y + zfx + y + z- l\x + y +

Case I: If r = 2; then the formula for required probability is x(x-l)+y(y-\)+z{z-\) given by (x + y + z\x + y + z - 1 ) Ex:

A box contains 4 black balls, 3 red balls and 5 green balls. 2 balls are drawn from the box at random. What is the probability that both the balls are of the same colour? Soln: Detail Method: Total no. of balls = 4 + 3 + 5 = 1 2 ,2„ 12x11 „ n ( S ) = ' C = - ^ — = 66

Ex.:

A box contains 5 green, 4 yellow and 3 white marbles. 3 marbles are drawn at random. What is the probability that they are not of the same colour? (SBI Associates PO Exam 1999) Soln: Detail Method: Total no. of balls = 5 + 4 + 3 = 1 2 :

12x11x10

c = 3

1x2x3

C + C + C 3

4

3

3

3

=10 + 4 + 1 = 15 ways

_

15 _ 205 _ 41 220 ~ 220 ~ 44 ' Quicker Method: Applying the above theorem, we have, the required answer 1--

]

5x4x3+4x3x2+3x2x1 12x11x10

60 + 24 + 6 12x11x10

12xllxl0~ 2

2

2

=

4x3 ^7~

+

_

3x2 5x4 7 -y_

220

i.e, 3 marbles out of 12 marbles can be drawn in 220 ways. If all the three marbles are of the same colour, it can be done in

90

2

™ V J-r n(E)= C + C + C

z-2)

Now, P(A11 the 3 marbles of the same colour) + P(all the 3 marbles are not of the same colour) = 1 .-. P(all the 3 marbles are not of the same colour)

Illustrative Example

2

xjx - iXx - 2)+ y(y - jjy - 2)+ z{z -\\z - 2)"

5

where r <x,y ,z

c

Note: The probability that both the balls are not of the same colour is given by 1 - P (Probability of the same colour) Case II: I f r = 3; then the formula for required probability is given by

n(S) =

Theorem: A box contains x black balls, y red balls and z green balls, 'r' balls are drawn from the box at random. The probability that all the balls are of the same colour is

66

4(4 - 1 ) + 3(3 - 1 ) +- 5(5 -1) Required answer =

d) None of these

V5 4i (iii) What is the chance that all the girls do not sit together? V 2

)

19

n(S)

Quicker Method: Applying the above theorem,

(ii) What is the chance that all the boys sit together?

a

n(E)

673

+

= 6 + 3 + 10=19

3 _41 44~44"

Note: The probability that all the balls are not of the same colour is given by 1 - P (Probability of the same colour).

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P R A C T I C E B O O K ON Q U I C K E R MATHS

674

Rule 10

Exercise 1.

A box contains 5 black balls, 4 red balls and 6 green balls. 2 balls are drawn from the box at random. What is the probability that both the balls are of the same colour? 1

30

31_ a)

05

c)

150

b )

74

105

d

Theorem: A hag contains 'x' redand 'y' black balls. If two draws of three balls each are made, the ball being replaced after the first draw, then the chance that the balls were red in the first draw and black in the second draw is given b\

> To?

{x(x-\){x-2)){y(y-\)(y-2)]\ ^C 3

A box contains 3 black balls, 5 red balls and 7 green balls. 2 balls are drawn from the box at random. What is the probability that both the balls are of the same colour? 74_ a)

71

34_

105

105

b )

31 d)

105

C )

49

50 b)

53

C)

+

102

'i 5153 3~

204 b)

1365

K

x

+

y

- \ x + y-2)}

2

\

A bag a contains 5 red and 8 black balls. Two draw; of three balls each are made, the ball being replace: after the first draw. What is the chance that the balls were red in the first draw and black in the second? Soln: Detail Method: Total no. of balls = 5 + 8 = 13 ,»„

11x12x13

Chance that the balls were red in first draw = 1 3 -j and

1635

d) None of these

Chance that the balls were black in the second dm

A box contains 3 green, 5 yellow and 3 white marbles. 3 marbles are drawn at random. What is the probability that all the three marbles of the same colour?

= i £ 3

[ v balls are replaced after first draw] s

a)

4

5~5

b )

3

5?

C )

TS

5?

9_

134

43

A)

8

7c)

143

b )

143

135 d)

47

49

43 b)

90

c)

90

90

d)

143 the required probability =

11 90

3. d;

Hint: See Note Required probability =

4. a 7. b;

5b 6. b Hint: Applying the given rule, we have the required probability *C + C + C 4

5

4

C

4

8

4

1 -

49 77J

=

104 TTfi 153

28 2. 43

15x4x3

90

(13x12x11)' 20160

140

2944656

20449'

A bag a contains 4 red and 7 black balls. Two draws af three balls each are made, the ball being replaced aftr the first draw. What is the chance that the balls were rec in the first draw and black in the second? a)

1 + 15 + 70

20449

Exercise 1.

2.b


(5x4x3)x(8x7x6)

Answers la

140

3

Note: In the above example, the two events are indepe-dent and can occur simultaneously. So, we used mutiplication. Quicker Method: Applying the above theorem, whave,

a box contains 4 black, 6 red and 8 green balls. 4 balls are drawn from the box at random. What is the probability that all the balls are of same colour? a)

c

Required probability = 1 3

52

A box contains 4 green, 5 yellow and A white marbles. 3 marbles are drawn at random. What is the probability that all the three balls are not of the same colour? a)

= 286

102 c)

1365

51

'

r

Illustrative Example

A box contains 4 green, 5 yellow and 6 white marbles. 3 marbles are drawn at random. What is the probability that all the three marbles are of the same colour? a)

y

104 d) ' 153

103

153

{x

Ex.:

105

A box contains 4 black balls, 6 red balls and 8 green balls. 2 balls are drawn from the box at random. What is the probability that both the balls are not of the same colour? a)

{

3

5445

25 b) ' 5448

28 c) " 4554

25 d)

4554

A bag a contains 5 red and 6 black balls. Two draws :•three balls each are made, the ball being replaced after the first draw. What is the chance that the balls were rtz

Probability https://telegram.me/Banking_Zone

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in the first draw and black in the second? 1081 b) d) c) 1089 ' 1089 ' 1089 ' 1089 A bag a contains 7 red and 8 black balls. Two draws of three balls each are made, the ball being replaced after the first draw. What is the chance that the balls were red in the first draw and black in the second? a)

3.

18 a)

845

b

>845

8 c)

Note: From the above example we can see that how the quicker methods for such questions have been derived.

Exercise 1.

A bag contains 4 black and 6 white balls. A ball is drawn out of it and replaced in the bag. Then a ball is drawn again. What is the probability that (i) both the balls drawn were black;

d) None of these

845

4

9

Ys

a)

Answers l.a

675

b )

2

2?

c)

d)

c)

d)

(ii) both were white;

2.b

3.c 3 a)

Rule 11 Theorem: A bag contains x black andy white balls. A ball is drawn out of it and replaced in the bag. Then a ball is drawn again. The probability that

x+ y

f (ii) both the balls drawn were white is given by

\

6 a) 725 7

21 b)' 725 7

c

19 )' 25

d) Data inadequate

A bag contains 6 black and 9 white balls. A ball is drawn out of it and replaced in the bag. Then a ball is drawn again. What is the probability that (i) both the balls drawn were black;

y a )

(Hi) the first ball was white and the second black and vice versa is given by

25

(iii) the first ball was white and the second black;

2. (i) both the balls drawn were black is given by

25

b)

2?

b

^

c)

25

d) None of these

(ii) both were white;

xy

16 a)

25

b

21

>25

C

>25"

d )

2?

(iii) the first ball was white and the second black;

Illustrative Example Ex:

A bag contains 5 black and 7 white balls. A ball is drawn out of it and replaced in the bag. Then a ball is drawn again. What is the probability that (i) both the balls drawn were black; (ii) both were white; (iii) the first ball was white and the second black; (iv) the first ball was black and the second white? Soln: The events are independent and capable of simultaneous occurrence. The rule of multiplication would be applied. The probability that (i) both the balls were black

12 12 ~ 144

7 7 49 (ii) both the balls were white = — x — = 12 12 144 (iii) the first was white and the second black 5 _ 35

~ 12* 12 ~ 144 (iv) the first was black and the second white 5 7 —x— 12 12

35 144

25

b)

c)

25

25

d)

25

Answers l.(i)a

(ii)d (iii)a

2.(i) a

(ii)a

(iii) a

Rule 12 Theorem: A bag contains x red and y white balls. Four balls are drawn out one by one and not replaced. Then the probability that they are alternatively of different colours

5_ 5 _ 25 : X

_ 1_

a)

2x(x-\)y(y-\) is given by

(x + y)(x + y-\)(x

+ y-2)(x

+ y-3)

Illustrative Example Ex. 2: A bag contains 6 red and 3 white balls. Four balls are drawn out one by one and not replaced. What is the probability that they are alternatively of different colours? Soln: Detail Method: Balls can be drawn alternately in the following order: Red, White, Red, White OR White, Red, White, Red

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P R A C T I C E B O O K ON Q U I C K E R MATHS

676 If red ball is drawn first, the probability of drawing the balls alternatively 6

3

5

5

2

= — X — X — X —

— X — X — X

1

5

X — X — X — =

5 1

5

6x3x5x2

l)(6 + 3 - 2 ) ( 6 + 3 - 3 ) x2 =

(x + y)(x +

x2

5

Illustrative Example Ex.:

A bag contains 4 white and 6 red balls. Two draws of one ball each are made without replacement. What is the probability that one is red and other white? Soln: Detail Method: Such problems can be very easily solved with the help of the rules of permutation and combination. Two balls can be drawn out of 10 balls in

42 9x8x7x6 Note: Wherever we find the word AND between two events, we use multiplication. Mark that both also means first and second. On the other hand, i f the two events are joined with OR, we use addition as in the above example.

10!

A bag contains 6 red and 4 white balls. Four balls are drawn out one by one and not replaced. What is the probability that they are alternatively of different colours?

2.

c) "'7

d) Data inadequate

red ball are C , x C 4

6

1

o r 4 x 6 = 24.

The required probability would be No. of cases favourable to the event Total no. of ways in which the event can happen 24

T? 145 165 16i A bag contains 9 red and 7 white balls. Four balls are drawn out one by one and not replaced. What is the probability that they are alternatively of different colours?

45

15

d )

8 65

b) ' 65

c) ' ' 130

^130

A bag contains 5 red and 4 white balls. Four balls are drawn out one by one and not replaced. What is the probability that they are alternatively of different colours? ]0 a

3

The total number of ways of drawing a white and a

14

a)

4! C , or j , , or4 ways.

6 ways.

c)

a )

or 45 ways.

2

6

A bag contains 8 red and 3 white balls. Four balls are drawn out one by one and not replaced. What is the probability that they are alternatively o f different colours? b)

C , or

One red ball can be drawn out of 6 red balls in C , or

1 7

1 0

One white ball can be drawn out of 4 white balls in

4

b)

10x9

2!8! or -

Exercise

>7

y-\)

=

6x3x(6-l)(3-l) (6 + 3)(6 + 3

Theorem: A bag contains 'x' white and 'y' red balls. If two draws of one ball each are made without replacement, then the probability that one is red and the other white is given by

9 8 7 6 9 8 7 6 84 84 42 ' Quicker Method: Applying the above theorem, we have the required probability

a

4.b

2xy (*)

6 3 5 2 3 6 2 5

1.

3.a

Rule 13

....(II)

9 8 7 6 Required probability = (I) + (II) s

2.c

(I)

9 8 7 6 If white ball is drawn first the probability of drawing the balls alternately 3

l.c

2

— X — X — X —

6

Answers

>6T

b)

63

F5 c)

63

d) None of these

Quicker Method: Applying the above theorem, we have the required probability =

2x6x4

10x9 15 Note: The above theorem may be put as given below. "A bag contains x' white and 'y' red balls. If two balls are drawn in succession at random, then the probability that one of them is white and the other 2xy red,isgivenby\j^r J^ ^) y

y

Exercise 1.

A bag contains 8 white and 12 red balls. Two draws of one ball each are made without replacement. What is the

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Probability

Similarly, i f the second basket be chosen, the probability of drawing a white ball =

probability that one is red and other white? 48

24

1

d) None of these 95 95 19 A bag contains 5 white and 5 red balls. Two draws of one ball each are made without replacement. What is the probability that one is red and other white? a)

b )

2.

25 a

3.

>17

b

1

C )

1

5

>3"

C

2

d

>9

16 b) ' 33

33

|

b)

c)

22

d)

22

3_ 14

14

A

7 + 12

19

14

56

56

the required probability

A bag contains 9 white and 3 red balls. Two balls are drawn in succession at random. What is the probability that one of them is white and the other red?

a)

6_

C ~2

Quicker Method: Applying the above theorem, we have,

d) Data inadequate

c) "Ml

1 4

I 8

A bag contains 4 white and 8 red balls. Two draws of one ball each are made without replacement. What is the probability that one is red and other white? a)

C\ X

Since, the two events are mutually exclusive, we use addition, therefore, the probability of drawing a white ball from either basket is

4

>9

677

:

3_

6_

J9

12

14

56

Exercise 1.

_3_ 11

A basket contains 4 white and 10 black balls. There is another basket which contains 5 white and 7 black balls. One ball is to be drawn from either of the two baskets. (i) What is the probability of drawing a white ball?

Answers

89

l.a 2.c 3.b 4. a; Hint: See Note.

a

)

m

59 M

b)

59

89

168

C )

d )

84

(ii) What is the probability of drawing a black ball?

Rule 14 Theorem: A basket contains x white and y x

x

119 a) 168

blackballs. 2.

There is another basket which contains x

2

white and y

2

black balls if one ball is to be drawn from either of the two baskets, then the probability of drawing

f

\

\2

?

and

+

59 168

C )

d)

109 168

A basket contains 5 white and 9 black balls. There is another basket which contains 7 white and 7 black balls. One ball is to be drawn from either of the two baskets, (i) What is the probability of drawing a white ball?

x

(i) a white ball is given by

89 b) 168

a)

7

6 b) "'7

3_ 7

C )

d) None of these

(ii) What is the probability of drawing a black ball? (ii) a black ball is given by ^

Illustrative Example Ex:

A basket contains 3 white and 9 black balls. There is another basket which contains 6 white and 8 black balls. One ball is to be drawn from either of the two baskets. What is the probability of drawing a white ball? Soln: Detail Method: Since there are two baskets, each equally likely to be chosen, the probability of choosing either basket is

x

\

C,

1

A

2

12

8

4 7

C )

A basket contains 6 white and 9 black balls. There is another basket which contains 8 white and 7 black balls. One ball is to be drawn from either of the two baskets, (i) What is the probability of drawing a white ball? 8 a)

TI

7 b )

T5

3 C

)

I

6 d )

Ts"

(ii) What is the probability of drawing a black ball?

fc 2' I f the first basket is chosen, the probability of draw1 ing a white ball = ~

5_ b) "'7

a) 7

a

>T?

b)

c)

15

Answers l.(i) c 3.(i)b

(ii)d (ii)b

2.(i)c

(ii)c

15

d)

15

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P R A C T I C E B O O K ON Q U I C K E R MATHS

678 3.

Rule 15 Theorem: A and B stand in a ring with 'x' other persons. If the arrangement of all the persons is at random, then the probability that there are exactly 'y' persons between A

A and B stand in a ring with 7 other persons. I f the arrangement of the 9 persons is at random, then the probability that there are exactly 2 persons between A and B is 1

2 and B is given by I ~ ~ j J . Where y < x.

3

4.

Illustrative Example Ex:

A and B stand in a ring with 10 other persons. I f the arrangement of the 12 persons is at random, then the probability that there are exactly 3 persons between A and B is. (Provident Fund Exam 2002) Soln: Detail Method: 7 6,

b )

C )

1 3 )

5.

2

3

) l 4" 9 >8 A and B stand in a ring with 11 other persons. I f the arrangement of the 13 persons is at random, then the probability that there are exactly 3 persons between A and B is. a

2

6

b )

d

3

IT

>*

C

4 d

)

9

A and B stand in a ring with 14 other persons. I f the arrangement of the 16 persons is at random, then the probability that there are exactly 6 persons between A and B is. 1 b)

a) 77

c)

14

d)

Answers l.b Let A stand on some point of the ring. Then n(S) = the number of points on which B can stand = 11 If there be exactly 3 persons between A and B, then corresponding to any position occupied, B can take up only two position, the 4th place and the 8th place as counted from A. Thus n(E) = 2 n{E) P(E) = n(S)

11

the required probability =

10 + 1

11

Exercise A and B stand in a ring with 9 other persons. I f the arrangement of the 11 persons is at random, then the probability that there are exactly 4 persons between A and B is. a) 2.

11

1 b) "' 5

1 c)

Rule 16

(n—m)\m\ given by

Illustrative Example

10

10 persons are seated at a round table. What is the probability that two particular persons sit together? Soln: Detail Method: n(S) = no.of ways of sitting 10 persons at round table = (10-1)!=9! Since 2 particular persons will be always together, then the no. of persons = 8 + 1 = 9 ,", 9 persons will be seated in (9 - 1)! = 8! ways at round table and 2 particular persons will be seated themselves in 2! ways. .-. The number of ways in which two persons always sit together at round table = 8! * 2! = n(E) n(E) _ 8!x2! _ 8!x2 _ 2 •"• n(S) ~ 9! ~ 9 x 8 ! ~ 9 Quicker Method: Applying the above theorem, we have,

1 d) "Ml

1 b

>9

5.a

Theorem: If 'n' persons are seated at a around table then the probability that'm' particular persons sit together is

P

A and B stand in a ring with 8 other persons. I f the arrangement of the 10 persons is at random, then the probability that there are exactly 5 persons between A and B is. a)

4. a

Ex.:

Quicker Method: Applying the above theorem, we have,

1.

2.b3.a

C

>9

1 d)y

(

E

)

=

the required probability

:

(10-2)!2!

8! 2!

(10-1)!

9!

Exercise 1.

12 persons are seated at a round table. What is the probability that 4 particular persons sit together?

https://telegram.me/Banking_Zone Probability 4 a)

ToT

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8 b)

9

T6?

C )

4

185

d)

q = probability of not-happening = —

Y5

8 persons are seated at a round table. What is the probability that 3 particular persons sit together? 2 1 d)None of these a) r b) c) 7 ' 1 "'14 10 persons are seated at a round table. What is the probability that 3 particular persons sit together? 1

1

1 b

c)

>9

d)

2.b

7

Exercise 1.

2.

Theorem: If an event is repeated, under similar conditions, exactly 'n' times, then the probability that event happens r

r

r

An unbiased coin is tossed 5 times, find the chance that exactly 3 times tail will appear.

3.

Illustrative Example An unbiased coin is tossed 7 times, find the chance that exactly 5 times head will appear. Soln: Here, n = 7, r = 5 p = probability of happening = —

b )

21 256

21 °^ 64

15 b)

Answers l.a

10 c) ' 64

d) None of these

d

)

D

a

t

a

m

a

d

e

c

l

u

a

t

e

An unbiased coin is tossed 6 times, find the chance that exactly 4 times tail will appear. 15 a) 77

Ex.:

5 b) — " ' 32

An unbiased coin is tossed 9 times, find the chance that exactly 6 times head will appear. 21 a) 7777

provided that

p = probability of happening and q = probability of not happening ie p + q = 1.

\, 2 ,

128

5 a) 77 16

exactly V times is | " C x p xq"~ \,

*

5

21

3.a

Rule 17

7-5

r i V (V

.-. required probability = C xX ,2j

Answers l.a

679

2. a

3.c

128

c)

11 64

15 d)

256