1. No category

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Probability
ber greater than 6 in the sample space
Rule 1
Theorem: Probability of an event (E) is denoted by P(E)
and is defined as P(E)
0
 P(E) = 6  0
Probability of an impossible event = 0
(viii) E (a no. less than or equal to 6)
= {1, 2, 3, 4, 5, 6}, n(E) = 6
n( E )
no. of desired events
= n( S )  total no. of events (ie no. of sample space)
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.
(i) E (an even no.) = {2, 4, 6}, n(E) = 3
n( E ) 3 1
 P(E) = n( S )  6  2
6
 P(E) = 6  1
Probability of a certain event = 1.
Note: 0  P(E)  1
Exercise
1.
2 1

6 3
(v) E (a no. less than 4) = {1, 2, 3}, n(E) = 3
2.
2 1
1 4
1 2
1 
,
b) ,
c) ,
d) ,
3 4
4 5
2 3
2 4
In a box carrying one dozen of oranges, one third have
become bad. If 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?
a)
3.
 P(E) =
1
54
45
3
b)
c)
d)
55
55
55
55
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?
a)
4.
3 1
 P(E) = 
6 2
(vi) E (a no. less than or equal to 4)
= {1, 2, 3, 4} n(E) = 4
4 2
 P(E) = 
6 3
(vii) E (a no. greater than 6) = {}, i.e. there is no num-
2
5
1
b)
c)
d) Data inadequate
9
36
6
A coin is successively tossed two times. Find the probability of getting
1) exatly one head
2) at least one head.
a)
(ii) E (an odd no.) = {1, 3, 5}, n(E) = 3
n( E ) 3 1
 P(E) = n( S )  6  2
(iii) E (a no. divisible by 2) = {2, 4, 6}, n(E) = 3
3 1
 P(E) = 6  2
(iv) E (a no. divisible by 3) = {3, 6}, n(E) = 2
In a simultaneous throw of two dice find the probability
of getting a total of 8.
12
11
10
1
b)
c)
d)
13
13
15
15
The probability that a teacher will give one surprise test
a)
5.
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1
during any class meeting in a week is . If a student is
5
absent twice, what is the probability that he will miss at
least one test.
4
1
91
16
b)
c)
d)
15
15
25
125
In a throw of a coin, the probability of getting a head is
_______.
a)
6.
1
1
b)
c) 1
d) None of these
2
4
In a simultaneous throw of two coins, the probability of
getting at least one head is _______.
a)
7.
2
1
1
3
b)
c)
d)
3
3
2
4
Three unbiased coins are tossed, what is the probability
of getting exactly two heads?
a)
8.
1
2
3
3
b)
c)
d)
3
3
8
4
Three unbiased coins are tossed. What is the probability of getting at most 2 heads?
a)
9.
3
7
1
1
b)
c)
d)
8
8
4
2
10. A fair coin is tossed 100 times. The probability of getting head an odd number of times is _______.
a)
2
1
1
3
a)
b)
c)
d)
3
4
2
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?
4
4
1
3
b)
c)
d)
7
5
8
4
12. A bag contains 8 red and 5 white balls. 2 balls are drawn
at random. What is the probability that both are white?
The probability of its occurrence is _______.
4
4
1
1
b)
c)
d)
5
9
5
4
16. In a lottery there are 20 prizes and 15 blanks. What is the
probability of getting prize?
a)
1
2
4
2
b)
c)
d)
10
5
7
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?
a)
9
11
1
2
b)
c)
d)
20
20
11
11
18. What is the probability that a number selected from the
numbers 1, 2, 3, 4, 5, ..., 16 is a prime number?
a)
1
5
3
7
b)
c)
d)
16
8
8
16
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?
a)
3
3
2
1
b)
c)
d)
20
10
5
2
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?
a)
a)
b)
1
2
c)
2
5
d)
7
20
Answers
1. b;
a)
Hint: In a simultaneous throw of two dice
Sample space = 6 × 6 = 36
Favourable cases are (2, 6) (3, 5) (4, 4) (5, 3) (6, 2)
5
36
Hint: In tossing a coin 2 times the sample space is 4 ie
(H, H), (H, T), (T, H), (T, T)
So, the required probability =
2. a;
5
2
3
5
b)
c)
d)
16
13
26
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?
1) If A1 denotes exactly one head
a)
1
2
1
b)
c)
d) None of these
3
5
6
14. The odds in favour of an event are 3:5. The probability
of occurrence of the event is ________.
1
15
then, A1 = {(H, T) (T, H)} So, P( A1 ) =
2 1

4 2
2) If A denotes at least one head
a)
then A = {(H, T), (T, H), (H, H)}  P(A) 
3. c;
12  11 10
 2  11  10  220
3 2
No. of selection of 3 oranges out of the total 12
Hint: n(S) =
12
C3 
3
3
1
1
b)
c)
d)
5
8
3
5
oranges = 12 C3 = 2 × 11 × 10 = 220
15. The odds against the occurrence of an event are 5:4.
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3
4
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
No. of selection of 3 bad oranges out of the total 4
4
bad oranges = C3  4
 n(E) = no. of desired selection of oranges
= 220 - 4 = 216
4. b;
nE  216 54
 P E   nS   220  55
Hint: Total possible ways of selecting 4 students out
15  14  13  12
 1365
1 2  3  4
The no. of ways of selecting 4 students in which no
of 15 students =
15
C4 
student belongs to Karnataka =
10
C4
 number of ways of selecting at least one student
from Karnataka =
5. b;
15
10
C4  C4  115 5.
1155 77 11


 Probability =
1365 91 13
Hint: The probability of absenting of the student in
6. a;
1 1 1
 the probability of missing his test = 5  3  15 .
Hint: Here S = {H, T} and E = {H}
7. c;
n E  1
 P E   nS   2
Hint: S = {HH, HT, TT, TH} and E = {HH, HT, TH}
8. d;
n E  3
 P E   nS   4
Hint: S = {HHH, HHT, HTH, THH, TTH, THT, HTT,
TTT}and
E = Event of getting exactly two heads
= {HHT, HTH, THH}
 P E   n E   3
nS  8
9. c;
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  7
 P(E) = nS   8
100
C1 
100
C3  ... 
100

nE  2 99
1

 .
nS  2100 2
Note: The given case can be generalised as “If a unbiased
coin is tossed ‘n’ times, then the chance that the head
will present itself an odd number of times is
1
”.
2
11. a; Hint: Total no. of balls = (6 + 8) = 14
No. of white balls = 8
 P(drawing a white ball) =
8 4

14 7
12. d; Hint:
n(S) = Number of ways of drawing 2 balls out of 13
13 12
 78
2
n(E) = No. of ways of drawing 2 balls out of 5
=
13
C2 
5 4
 10
2
nE  10 5
 P(E) = nS   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
9 8 7
 84
3  2 1
n(E) = Number of ways of drawing 2 balls out of 5 and
1 ball out of 4.
 5 4

5
4
 4   14
= C 2  C1  
 2 1

nE  14 1
 P(E) = nS   84  6 .
14. b; Hint: Number of cases favourable to E = 3
Total number of cases = (3 + 5) = 8
9
of 9 = C3 
3
 P(E) = 8 .
15. b; Hint: Number of cases favourable to E = 4
Total number of cases = (5 + 4) = 9
4
 P(E) = 9 .
10. c; Hint: nS   2100
n(E) = no. of favourable ways
=
C1  n C3  nC5  ...  2 n 1
 P E  
5
= C2 
2 1
the class = 
6 3
n
16. c; Hint: P(getting a prize) =
20
0 4

 .
20  15 35 7
C99  21001  299
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17. d; Hint: P(red) =
9
9

9  7  4 20
1
 P(E) = 7
9  11

 P(not-red) = 1  20   20


18. c; Hint: S = {1, 2, 3, ..., 16} and E = {2, 3, 5, 7, 11, 13}
Exercise
nE  6 3
 P(E) = nS   16  8 .
19. b; Hint: S = {1, 2, 3, ... 20} and E = {3, 6, 9, 12, 15, 18}
a)
nE  6
3
 P(E) = nS   20  10
20. c; Hint: Clearly, n(S) = 20 and E = {3, 6, 9, 12, 15, 18, 7,
14}
1.
What is the probability that a leap year selected randomly will have 53 Mondays?
2
5
1
b)
c)
d) Data inadequate
7
7
7
What is the probability that an oridinary year has 53
Sundays?
2.
a)
1. a;
2. b;
Illustrative Example
c)
2
7
d)
48
53
1
.
7
Ex. :
2
 P(E) = 7
(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
{Sunday}, n(E) = 1
1
7
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 = .
7
Hint: An ordinary year has 365 days ie 52 weeks and
1 day. So the probability that this day is a Sunday is
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
b)
Answers
nE  8 2
 P(E) = nS   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.
53
365
Rule 3
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.
of the two dice
(i)
(ii)
2
12
3
n(S)
Events
(i)
(ii)
1
{1, 1}
{6, 6}
11
2
{1, 2}, {2, 1}
{6, 5}, {5, 6}
4
10
3
{1, 3}, {3, 1}, {2, 2}
{6, 4}, {4, 6}, {5, 5}
5
9
4
{1, 4}, {4, 1}, {2, 3},
{3, 2}
{6, 3}, {3, 6}, {5, 4},
{4, 5}
6
8
5
{1, 5}, {5, 1}, {2, 4},
{4, 2}, {3, 3}
{6, 2}, {2, 6}, {5, 3},
{3, 5}, {4, 4}
7
6
{1, 6}, {6, 1}, {2, 5}, {5, 2}, {4, 3}, {3, 4}
Illustrative Example
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Ex. :
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:
ability of getting a doublet?
1
2
1
3
b)
c)
d)
6
3
4
4
In a simultaneous throw of two dice, what is the probability of getting a total of 10 or 11?
a)
3.
5
1
7
1
b)
c)
d)
36
6
12
4
In a single throw of two dice what is the probability of
not getting the same number on both the dice?
a)
4.
n( E ) 5
For 6, reqd probability = n(S )  36
6 1

36 6
(ii) Desired sums of the numbers are 2, 3, 4, 5, 6, 7 and
8;
n(S) = 1 + 2 + 3 + 4 + 5 + 6 + 5 = 26
26 13

 reqd probability =
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 + 1 = 12
For 7, reqd probability =
12 1

 reqd probability =
36 3
(v) Events = {1, 1}, {2, 2}, {3, 3}, {4, 4}, {5, 5}, {6, 6};
n(S) = 6
6 1

 P(E) =
36 6
(vi) Events = {3, 1}, {4, 2}, {5, 3}, {6, 4}, {4, 6}, {3, 5},
{2, 4}, {1, 3}
or n(S) = 8
8 2
 P(E) = 36  9
(vii) Events; either 2 or 3 or 4 or 5
n(E) = 1 + 2 + 3 + 4 = 10
n(S) = 36
nE  10 5
 P(E) = nS   36  18 .
Exercise
1.
In a throw of a die, the probability of getting a prime
number is
1
2
1
3
b)
c)
d)
3
3
2
4
In a simultaneous throw of two dice, what is the prob-
a)
2.
a)
1
6
b)
2
3
c)
5
6
d)
1
3
Answers
1. b;
Hint: Here S = {1, 2, 3, 4, 5, 6} and E = {2, 3, 5}
2. a;
3 1
 P(E) = 6  2 .
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)}
3. b;
4. c;
6 1
 P(E) = 36  6 .
Hint: Clearly, n(s) = 36
Let (E) = {(4, 6), (5, 5), (6, 4), (5, 6), (6, 5)}
5
 P(E) = 36
Hint: Clearly, n(S) = 36
Let E = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
 P(E) =
6 1

36 6
 1 5
So, P(not - E) = 1   
 6 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
× 4 = 36 digit-cards) along with four Honour cards
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Ace, King, Queen and Jack (a total of 4 × 4 = 16 Honour
cards).
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) a jack of black suit?
Soln:
For all the above cases n(S) =
26 1 
(i) 52  2 or ,

(ii)
26

4
1

52 13
(vi)
3 4 3

52
13
(vii) P(a king) =

(iii)
(v)
4 2 2

52
13
5 4 5

52
13
4
1
4
1
 ; P(a queen) =

52 13
52 13
 P(a king or a queen) =
1 1
2


13 13 13
9
2
1

(ix)
52
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?
(viii)
Soln: For all the above cases, n(S) =
52
(i) Total no. of Aces = 4
43
4
 n(E) = C 2  2  6
4
52  51
C2 
 26  51
2
C1 = 4 ways. He can select the remaining 1 card
from the remaining (52 – 4 =) 48 cards. Now, cards in
48
C1  48 ways.
4  48
32
 P(E) = 26  51  221
 n(E) = 4 × 48
(iii) Total no. of honours card = 16
To have no honours card, he has to select two cards
out of the remaining 52 – 16 = 36 cards which he can
do in
36
C2 =
36  35
 18  35 ways
2
18  35
105
 P(E) = 26  51  221
C1  52

26

 n C1  n 
52
C1 52

C1
13 1

52 4
(iv)
52
Selection of 1 Queen card out of 4 can be done in
16
(iv) P(E) =
C2
8 15
20


26  51 26  51 221
(v) n(E) = 4 C1  4C1  4  4  16
16
8
 P(E) = 26  51  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?
Soln: For all the above cases, n(S)
=
52
C3 
52  51  50
 26  17  50
3 2
(i) n(E) = 4 C 3  4
 PE  
(ii) n(E) =
48
4
1

26  17  50 5525
C 3  8  47  46
 PE  
8  47  46 4324

26  17  50 5525
(iii) n(E) = 4 C1  4C1  4C1  4  4  4
 PE  
444
16

26  17  50 5525
6
1
(iv) n(E) = 4 C1  4C 2  4  6

26  51 221
(ii) Total no. of Queens = 4
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 PE  
(v) n(E) =
36
46
6

26  17  50 5525
C 2  8C1  18  35  8
 PE  
3. a;
18  35  8
252

26  17  50 1105
52  51
 1326
2 1
n(E) = Number of ways of drawing 2 cards out of 4
=
Exercise
1.
One card is drawn at random from a pack of 52 cards.
What is the probability that the card drawn is a face
card?
4
9
1
1
b)
c)
d)
13
52
13
4
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?
6
7
27
1
b)
c)
d)
13
13
52
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)
3.
2
3
1
b)
c)
d) None of these
13
26
221
What is the probability of getting a king or a queen in a
single drawn from a pack of 52 cards?
a)
4.
4. c;
5. b;
nE  8
2
 P(E) = nS   52  13 .
Hint: There are 13 hearts and 3 more kings
 P(heart or a king) =
Answers
1. a;
Hint: Clearly, n(S) = 52 and there are 16 face cards.
16 4

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
 P(E) =
2. c;
13  3 4

52
13
4
9
 .
13 13
Hint: Let E and F be the event of getting a spade and
that of getting a king respectively.
Then E  F is the event of getting a king of spade
 n(E) = 13, n(F) = 4 and n(E  F) = 1
 P(neither a heart nor a king) = 1 
6. a;
So, P(E) =
13 1
4
1
 , P(F) =

and
52 4
52 13
1
52
 P(a spade or a king) = P(E  F)
= P(E) + P(F) - P(E  F)
1  4
1 1
=    .
 4 13 52  13
P(E  F) =
a)
6.
43
6
2
6
1

.
1326 221
Hint: Clearly, n(S) =52, there are 4 kings and 4 queens.
1
1
2
b)
c)
d) None of these
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
9
2
4
b)
c)
d)
13
13
13
13
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
a)
b)
c)
d)
13
13
13
13
C2 
 P(E) =
a)
5.
52
4
= C2 
a)
2.
nE  28 7
 P(E) = nS   52  13 .
Hint: n(S) = Number of ways of drawing 2 cards out
of 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


6 xyz


(
x

y

z
)
(
x

y

z

1
)
(
x

y

z

2
)


x
y
z
C1  C1  C1
or
x  y  z  C
3
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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Soln: Detail Method: Total no. of balls = 3 + 5 + 4 = 12
12  11 10
 220
3 2
In order to have 3 different coloured balls, the selection of one ball of each colour is to be made.
n(S) =
12
 x z  C . y C
3 y ( y  1) ( x  z )
1
2
or  x  y  z 
( x  y  z ) ( x  y  z  1) ( x  y  z  2)
C3
C3 
(iii) exactly 2 red balls is given by
 y z  C .x C
3x( x  1) ( y  z )
1
2
or  x  y  z 
( x  y  z ) ( x  y  z  1) ( x  y  z  2)
C3
n(E) = 3 C1 5 C1 4 C1  3  5  4  60
60
3

220 11
Quicker Method: Applying the above theorem, we
have
 P(E) =
the required answer =
6  3 5  4 3
 .
12  11 10 11
Illustrative Example
Ex:
A bag contains 3 red, 5 yellow and 4 green balls. 3
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 = 12
12  11 10
 220
3 2
2 green balls can be selected from 4 green balls in
n(S) =
Exercise
1.
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?
4
48 12
 P(E) = 220  55
Quicker Method: Applying the above theorem, we
have
the required answer
35
35
35
35
b)
c)
d)
138
136
134
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?
4
3
6
8
b)
c)
d)
17
17
17
17
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)
4.
a)
5
17
b)
2
17
c)
4
17
d)
8
17
Answers
1. a
2. b
3. a
4. c
Rule 6
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
C2 ways and the rest one ball can be selected from
n(E) = 4 C2 8 C1  6  8  48
a)
3.
C3 
the remaining (12 – 4) = 8 balls in 8 C1 ways.
24
14
13
21
b)
c)
d)
91
91
91
91
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?
a)
2.
12
=
3  4  (4  1)  8 3  4  3  8 12


.
12  11  10
12  11  10 55
Exercise
1.
A bag contains 4 red, 6 yellow and 5 green balls. 3 balls
are drawn randomly.
(i) What is the probability that balls drawn contain
exactly 2 green balls?
20
15
10
b)
c)
d) Data inadequate
91
91
91
(ii) What is the probability that balls drawn contain
exactly 2 yellow balls?
a)
20
28
27
37
b)
c)
d)
91
91
91
91
(iii) What is the probability that balls drawn contain exactly 2 red balls?
a)
54
44
54
b)
c)
d) None of these
455
455
91
2. A bag contains 5 red, 7 yellow and 6 green balls. 3 balls
are drawn randomly.
(ii) exactly 2 yellow balls is given by
(i) What is the probability that balls drawn contain
exactly 2 green balls?
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x  y  C .z C
3 z ( z  1) ( x  y )
1
2
or  x  y  z 
( x  y  z ) ( x  y  z  1) ( x  y  z  2)
C3
a)
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14
13
15
15
a)
b)
c)
d)
68
68
91
68
(ii) What is the probability that balls drawn contain
exactly 2 yellow balls?
the balls drawn contain no yellow ball?
Soln: Detail Method: Total no. of balls = 3 + 5 + 4 = 12
12  11 10
12
 220
n(S) = C 3 
3 2
3 balls can be selected from 3(red) + 4(green)
77
76
77
76
b)
c)
d)
272
272
271
271
(iii) What is the probability that balls drawn contain
exactly 2 red balls?
= 7 balls in 7 C3 ways.
a)
3.
n(E) = 7 C3 =
35
7

.
220 44
Quicker Method: Applying the above theorem, we
have
765
35
7


the required answer =
.
12  11  10 220 44
55
54
55
55
a)
b)
c)
d)
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?
27
20
54
b)
c)
d) None of these
91
91
91
(ii) What is the probability that balls drawn contain
exactly 2 yellow balls?
a)
 P(E) =
Exercise
1.
20
27
54
b)
c)
d) Data inadequate
91
91
455
(iii) What is the probability that balls drawn contain
exactly 2 red balls?
20
91
b)
19
91
c)
12
91
a)
12
33
24
b)
c)
d) None of these
65
91
91
(iii) What is the probability that the balls drawn contain
no green balls?
d) None of these
a)
Answers
1. (i) a
3. (i) a
(ii) c (iii) a
(ii) a (iii) c
2. (i) d
(ii) a
(iii) c
x z C


( x  z ) ( x  z  1) ( x  z  2)
3

 or  x  y  z 
(
x

y

z
)
(
x

y

z

1
)
(
x

y

z

2
)
C3


(ii) no red ball is given by
yz C


( y  z ) ( y  z  1) ( y  z  2)
3

 or  x  y  z 
(
x

y

z
)
(
x

y

z

1
)
(
x

y

z

2
)
C


3
(iii) no green ball is given by
x  y  C


( x  y ) ( x  y  1) ( x  y  2)
3

 or  x  y  z 
(
x

y

z
)
(
x

y

z

1
)
(
x

y

z

2
)
C


3
12
33
24
1
b)
c)
d)
65
91
91
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)
Rule 7
Theorem: A bag contains x red, y yellow and z green balls.
3 balls are drawn randomly. The probability of the balls
drawn contain
(i) no yellow ball is given by
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?
24
33
12
b)
c)
d) Data inadequate
91
91
65
(ii) What is the probability that the balls drawn contain
no red ball?
a)
a)
765
 35
3 2
2.
55
143
55
55
b)
c)
d)
204
408
272
208
(ii) What is the probability that the balls drawn contain
no red ball?
a)
55
55
143
143
b)
c)
d)
282
272
408
406
(iii) What is the probability that the balls drawn contain
no green balls?
a)
143
55
55
55
b)
c)
d)
408
272
204
208
Illustrative Example
3. A bag contains 3 red, 5 yellow and 7 green balls. 3 balls
Ex:
A bag contains 3 red, 5 yellow and 4 green balls. 3
are drawn randomly.
balls are drawn randomly. What is the probability that
(i) What is the probability that the balls drawn contain
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no yellow ball?
4!5!
4  3 2 1

= P(girls are together) =
8!
8  7  6 14
 P(All girls are not together) = 1 - P
=
24
25
44
8
b)
c)
d)
91
91
91
65
(ii) What is the probability that the balls drawn contain
no red ball?
a)
1 13

.
14 14
Quicker Method: Applying the above theorem, we
have
5!4!
the required answer = 1
8!
5!4!
1 13
 1

= 1
.
8!
14 14
(All girls are together) = 1 
8
44
24
45
b)
c)
d)
65
91
91
91
(iii) What is the probability that the balls drawn contain
no green balls?
a)
a)
44
91
b)
24
91
c)
18
65
d)
8
65
Exercise
Answers
1. (i) a
3. (i) a
1.
(ii) b (iii) a
(ii) b (iii) d
2. (i) a
(ii) c
(iii) b
There are 5 boys and 4 girls. They sit in a row randomly.
(i) What is the chance that all the girls sit together?
6!4!
5!4!
5!4!
b)
c)
d) None of these
9!
9!
10!
(ii) What is the chance that all the boys sit together?
a)
Rule 8
Theorem: There are ‘x’ boys and ‘y’ girls. If they sit in a
row randomly, then the chance that
 ( x  1)! y!
(i) all the girls sit together is given by  ( x  y )!  .


6!4!
5!5!
5!5!
6!4!
b)
c)
d)
9!
9!
10!
10!
(iii) What is the chance that all the girls do not sit together?
 ( y  1)! x!
(ii) all the boys sit together is given by  ( x  y )!  .


(iii) all the girls do not sit together is given by
1
20
19
2
b)
c)
d)
21
21
21
21
(iv) What is the chance that all the boys do not sit together?
a)
a)
 ( x  1)! y!
1 

( x  y )! 

5
6
121
b)
c)
d) None of these
126
126
126
(v) What is the chance that the no two girls sit together?
a)
 ( y  1)! x!
(iv) all the boys do not sit together is given by 1  ( x  y )! 


x 1
(v) no two girls sit together (x > y) is given by
C y .x! y!
x  y !
2.
.
1
65
1
76
b)
c)
d)
66
66
77
77
(ii) What is the chance that all the boys sit together?
a)
Illustrative Example
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) =
5
37
25
b)
c)
d) Data inadequate
42
42
42
There are 6 boys and 5 girls. They sit in a row randomly.
(i) What is the chance that all the girls sit together?
a)
8
P8 = 8!
Consider all the 4 girls as one, we have 4 boys + 1 girl
= 5 persons. Which can be arranged in 5 P5  5! ways.
But the 4 girls can also be arranged in 4 P4  4! ways
among themselves.
So, in 4! × 5! ways can the persons be arranged so
that girls are together
1
1
76
b)
c)
d) Data inadequate
66
77
77
(iii) What is the chance that all the girls do not sit together?
a)
65
1
7
76
b)
c)
d)
66
66
77
77
(iv) What is the chance that all the boys do not sit together?
a)
1
1
65
76
b)
c)
d)
77
66
66
77
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(v) What is the chance that the no two girls sit together?
3.
n (E ) 19
 Required probability, P(E) = n (S)  66
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?
3
1
41
b)
c)
d) None of these
35
42
42
(ii) What is the chance that all the boys sit together?
a)
3
3
41
1
b)
c)
d)
35
42
42
42
(iii) What is the chance that all the girls do not sit together?
a)
41
39
31
b)
c)
d) None of these
42
42
42
(iv) What is the chance that all the boys do not sit together?
Quicker Method: Applying the above theorem,
Required answer =
12  6  20 19

12  11
66
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: If r = 3; then the formula for required probability
is given by
=
 x x  1x  2  y  y  1 y  2  z  z  1z  2 

x  y  z x  y  z  1x  y  z  2  .

a)
a)
3
35
b)
41
42
c)
1
42
d) Data inadequate
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
Rule 9
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
x
given by
C r  y Cr  z Cr
x  y  z Cr where r < x, y , z
Illustrative Example
Case I:If r = 2; then the formula for required probability is
 x x  1  y  y  1  z  z  1 
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 = 12
12 11
12
 66
n(S) = C 2 
2
4  3 3 2 5 4
4
3
5


n(E) = C 2  C 2  C 2 
2
2
2
= 6 + 3 + 10 = 19
4(4  1)  3(3  1)  5(5  1)
(4  3  5) (4  3  5  1)
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 = 12
12  11 10
 220
1 2  3
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
n(S) =
12
C3 
5 C  4 C  3 C  10  4  1  15 ways
3
3
3
Now, P(All 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)
15
205 41


.
220 220 44
Quicker Method: Applying the above theorem, we
have,
the required answer
= 1
= 1
5  4  3  4  3  2  3  2 1
12  11 10
= 1
60  24  6
12  11 10
90
3
41
1

.
12  11 10
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).
= 1
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Exercise
1.
Rule 10
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?
2.
31
30
1
74
b)
c)
d)
105
150
105
105
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?
3.
74
34
71
31
a)
b)
c)
d)
105
105
105
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)
49
50
103
104
b)
c)
d)
153
153
153
153
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)
4.
5.
Theorem: A bag contains ‘x’ red and ‘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 by
 x( x  1) ( x  2)y ( y  1) ( y  2)

2  or
 ( x  y ) ( x  y  1) ( x  y  2) 
9
134
8
135
b)
c)
d)
143
143
143
143
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)
47
90
b)
43
90
c)
49
90
d)
41
90
Answers
2. b
3. d;
Hint: See Note Required probability = 1 
4. a
7. b;
5. b
6. b
Hint: Applying the given rule, we have
the required probability
4
C4  6C4  8C4
C3

2
.
5
Chance that the balls were red in first draw =
8
=
18

C3
C3
[ balls are replaced after first draw]
C3 
13
5
8
C3
140


Required probability = 13
C3 13 C3 20449
C3
Note: In the above example, the two events are independent and can occur simultaneously. So, we used multiplication.
Quicker Method: Applying the above theorem, we
have,
(5  4  3)  (8  7  6)
the required probability =
=
(13  12  11) 2
20160
140

.
2944656 20449
Exercise
49 104

.
153 153
A bag a contains 4 red and 7 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?
28
25
28
25
b)
c)
d)
5445
5448
4554
4554
A bag a contains 5 red and 6 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
a)
2.
1  15  70 43

15  4  3 90
C4
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=
C3
13
and
Chance that the balls were black in the second draw
1.
1. a
x y
A bag a contains 5 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?
Soln: Detail Method:
Total no. of balls = 5 + 8 = 13
11  12  13
13
 286
n(S) = C 3 
1 2  3
a)
7.

Ex. :
a)
6.
C3  y C3
Illustrative Example
102
204
102
a)
b)
c)
d) None of these
1365
1365
1635
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?
51
4
3
52
b)
c)
d)
55
55
55
55
A box contains 4 green, 5 yellow and 4 white marbles. 3
marbles are drawn at random. What is the probability
that all the three balls are not of the same colour?
x
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in the first draw and black in the second?
9
8
1081
7
b)
c)
d)
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.
a)
9
845
18
845
b)
c)
8
845
Note: From the above example we can see that how the
quicker methods for such questions have been derived.
Exercise
1.
d) None of these
4
9
b)
25
25
(ii) both were white;
a)
Answers
1. a
2. b
3. c
2
 x 
 .
(i) both the balls drawn were black is given by 
 x y
2
 y 
 .
(ii) both the balls drawn were white is given by 
 x y
(iii) the first ball was white and the second black and vice
xy
6
21
19
b)
c)
d) Data inadequate
25
25
25
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;
a)
2.
4
6
b)
25
25
(ii) both were white;
a)
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
5 5
25
 
12 12 144
7 7
49
 
12 12 144
(iii) the first was white and the second black
7 5
35


=
12 12 144
(iv) the first was black and the second white
=
5 7
35
 
12 12 144
3
5
c)
9
25
d) None of these
a)
Illustrative Example
(ii) both the balls were white =
d)
9
16
21
4
b)
c)
d)
25
25
25
25
(iii) the first ball was white and the second black;
.
(i) both the balls were black =
2
5
3
3
2
9
b)
c)
d)
25
5
5
25
(iii) the first ball was white and the second black;
Theorem: A bag contains x black and y 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 2
c)
a)
Rule 11
versa is given by
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;
a)
6
25
b)
19
25
c)
4
25
d)
9
25
Answers
1. (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


2 x ( x  1) y ( y  1)
is given by  ( x  y ) ( x  y  1) ( 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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If red ball is drawn first, the probability of drawing the
balls alternatively
6 3 5 2
=    ..... (I)
9 8 7 6
If white ball is drawn first the probability of drawing
the balls alternately
6 3 5 2
=    .... (II)
9 8 7 6
Required probability = (I) + (II) ..... (*)
6 3 5 2 3 6 2 5 5
5
5


=        
.
9 8 7 6 9 8 7 6 84 84 42
Quicker Method: Applying the above theorem, we
have
the required probability


6  3  (6  1) (3  1)
=  (6  3) (6  3  1) (6  3  2) (6  3  3)   2


5
 6 35 2 
=
  2  42 .
9876
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, if the two events are
joined with OR, we use addition as in the above example.
Answers
1. c


2 xy
by  ( x  y ) ( x  y  1)  .


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
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?
3.
9
6
9
8
b)
c)
d)
65
65
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?
a)
4.
5
a)
63
10
b)
63
15
c)
63
C 2 or
4!
C1 or 1! 3! or 4 ways.
One red ball can be drawn out of 6 red balls in 6 C1 or
6 ways.
The total number of ways of drawing a white and a
2
4
1
b)
c)
d) Data inadequate
7
7
7
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 of different
colours?
7
7
14
7
b)
c)
d)
15
145
165
165
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?
10
10!
10  9
or
or 45 ways.
2!8!
2
One white ball can be drawn out of 4 white balls in
4
a)
4. b
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
red ball are 4 C1  6 C1 or 4 × 6 = 24.
The required probability would be
a)
2.
3. a
Rule 13
Exercise
1.
2. c
No. of cases favourable to the event
= Total no. of ways in which the event can happen
24 8

45 15
Quicker Method: Applying the above theorem, we
have
26 4 8

the required probability =
.
10  9
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
=


2 xy
red, is given by  x  y  x  y  1  .”


Exercise
d) None of these
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 that one is red and other white?
Similarly, if the second basket be chosen, the probability of drawing a white ball =
48
24
1
b)
c)
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)
2.
1 6 C1 1 6
3
 14   
2
C1 2 14 14
Since, the two events are mutually exclusive, we use
addition, therefore, the probability of drawing a white
ball from either basket is
25
1
5
4
b)
c)
d)
81
3
9
9
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)
3.
17
16
3
b)
c)
d) Data inadequate
33
33
11
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?
1 3 7  12 19



8 14
56
56
Quicker Method: Applying the above theorem, we
have,
1  3 6  19
the required probability =    =
.
2 12 14 
56
a)
4.
a)
9
22
b)
3
22
c)
6
22
d)
Exercise
1.
3
11
Answers
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?
89
59
59
89
b)
c)
d)
168
84
168
84
(ii) What is the probability of drawing a black ball?
a)
1. a
2. c
3. b
4. a; Hint: See Note.
Rule 14
Theorem: A basket contains x1 white and y1 black balls.
2.
There is another basket which contains x2 white and y 2
black balls if one ball is to be drawn from either of the two
baskets, then the probability of drawing
x2 
1  x1
(i) a white ball is given by 2  x  y  x  y  and
2
2 
 1 1
y2 
1  y1
(ii) a black ball is given by 2  x  y  x  y  .
2
2 
 1 1
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
1
.
2
If the first basket is chosen, the probability of draw1 3 C1 1 3 1
ing a white ball = 2  12  2  12  8
C1
119
89
59
109
b)
c)
d)
168
168
168
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?
a)
4
6
3
b)
c)
d) None of these
7
7
7
(ii) What is the probability of drawing a black ball?
a)
3
5
4
8
b)
c)
d)
7
7
7
15
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?
a)
3.
8
7
3
6
b)
c)
d)
15
15
5
15
(ii) What is the probability of drawing a black ball?
3
8
7
9
a)
b)
c)
d)
15
15
15
15
a)
Answers
1. (i) c
3. (i) b
(ii) d 2. (i) c
(ii) b
(ii) c
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Rule 15
3.
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
 2 
 . Where y < x.
and B is given by 
 x 1 
4.
Illustrative Example
Ex:
A and B stand in a ring with 10 other persons. If 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
8
6
4
a)
10
2
A 1
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
Quicker Method: Applying the above theorem, we
have,
the required probability =
2
2
 .
10  1 11
Exercise
A and B stand in a ring with 9 other persons. If the
arrangement of the 11 persons is at random, then the
probability that there are exactly 4 persons between A
and B is.
1
1
2
1
b)
c)
d)
5
10
11
11
A and B stand in a ring with 8 other persons. If the
arrangement of the 10 persons is at random, then the
probability that there are exactly 5 persons between A
and B is.
a)
2.
a)
2
7
b)
2
9
2
15
b)
1
7
c)
3
14
d)
3
7
Answers
1. b
2. b 3. a
4. a
5. a
Rule 16
Theorem: If ‘n’ persons are seated at a around table then
the probability that ‘m’ particular persons sit together is
 (n  m)!m!
given by  (n  1)!  .


Illustrative Example
Ex.:
n( E ) 2
 P(E) = n( S )  11
1.
5.
9
3
1
4
2
3
b)
c)
d)
6
9
11
12
A and B stand in a ring with 14 other persons. If the
arrangement of the 16 persons is at random, then the
probability that there are exactly 6 persons between A
and B is.
a)
B
5
A and B stand in a ring with 7 other persons. If the
arrangement of the 9 persons is at random, then the probability that there are exactly 2 persons between A and B
is
2
3
1
3
a)
b)
c)
d)
9
8
4
4
A and B stand in a ring with 11 other persons. If the
arrangement of the 13 persons is at random, then the
probability that there are exactly 3 persons between A
and B is.
c)
1
9
d)
1
7
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!2! 8! 2 2
 P(E) = n( S )  9!  9  8!  9
Quicker Method: Applying the above theorem, we
have,
(10  2)!2! 8!2! 2
the required probability = (10  1)!  9!  9 .
Exercise
1.
12 persons are seated at a round table. What is the probability that 4 particular persons sit together?
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4
8
9
4
b)
c)
d)
165
165
185
55
8 persons are seated at a round table. What is the probability that 3 particular persons sit together?
q = probability of not-happening =
a)
2.
2
1
3
b)
c)
d) None of these
7
7
14
10 persons are seated at a round table. What is the probability that 3 particular persons sit together?
5
1 1
 required probability = C5      
2 2
2
b)
9
1
a)
12
1
c)
4
1
d)
6
=
2. b
1.
3. a
An unbiased coin is tossed 5 times, find the chance that
exactly 3 times tail will appear.
5
5
10
b)
c)
d) None of these
16
32
64
An unbiased coin is tossed 9 times, find the chance that
exactly 6 times head will appear.
a)
Rule 17
2.
Theorem: If an event is repeated, under similar conditions,
exactly ‘n’ times, then the probability that event happens
 C p

 q n r , provided that
r
p = probability of happening and
q = probability of not happening ie p + q = 1.
exactly ‘r’ times is
21
.
128
Exercise
Answers
1. a
7 5
7
a)
3.
1
2
n
r
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
a)
Ex.:
1
p = probability of happening =
2
21
21
21
b)
c)
d) Data inadequate
128
256
64
An unbiased coin is tossed 6 times, find the chance that
exactly 4 times tail will appear.
a)
15
32
b)
15
128
c)
15
64
d)
15
256
Answers
1. a
2. a
3. c
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