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PROBABILITY
PROBABILITY SOLUTIONS
Exercise I
1.
P( A)  0 , P( B)  0, P( A  B)  0  P( A  B)  P( A).P( B)  A and B are dependent .
2.
P( A1 ) 
2
2
2
, P( A2 ) 
, P( A3 ) 
4
4
4
P( A1  A2 ) 
1
1
1
,  P( A1 ).P( A2 ) , P( A2  A3 )  , P( A3  A1 ) 
4
4
4
P( A1  A2  A3 )  0  P( A1 ).P( A2 ).P( A3 )
Thus A1 , A2 , A3 are pair wise independent but mutually dependent .
3.
P( A) 
3
3
, P( B) 
6
6
P(C ) 
3 3 3 3 1


36
36 2
( OR P( A) 
3 6
3 6
, P( B) 
)
36
36
( (1,1) , (1,3) ,(1,5) ; (3,1), (3,3) , (3,5); (5,1),(5,3),(5,5)
( Similarly 9 cases for both are even )
Note that A , B, C are pair wise independent but P( A  B  C )  0
4.
n
2
  ,
n
n
3
P  A
P  B    , P  A  B 
n
n n n
6 2 3
For independence, we must have      2 
n
n
n
n
n
    
(*)
n     
6 2 3
n
 6 
.
n
Now n = 6k, 6k + 1, 6k + 2, …………….., 6k + 5.
It is easy to see that the two sides are equal only when n = 6k or 6k + 2.
Alternate solution

If n  6k then P( A) 
3k 1
2k 1
k
, P( A  B ) 
 , P( B) 

6k 2
6k 2
6k
P( A  B)  P( A).P( B)
If n  6k  2 then P( A) 
3k  1
2k
, P( B) 
3k  2
6k  2
59
PROBABILITY
P( A  B ) 
k
6k  2
 P( A  B)  P( A) .P( B)  A and B are independent
We can easily verify that if n  6k  1, 6k  3 ,6k  4,6k  5 then
P( A  B)  P( A). P( B) is not satisfied .
5.
The result is true for n  1 since P( A1 )  P( A1 )
Also , it is true for n  2 since P( A1  A2 )  P( A1 )  P( A2 )
(*)
k
 P( A )
Suppose the result is true for n  k then P( A1  A2  .........  Ak ) 
i
i 1
(**)
Now P( A1  A2  .......  Ak  Ak 1 )
 P( B  Ak 1 ),
where B  A1  A2  .......  Ak
 P( B)  P( Ak 1 ) from
6.
If P( A / B)  P( A) then
7.
we must have 0 
k 1
k
 P( A )  P( A
i 1
k 1
i
P( A  B)
 P( A) 
P( B)
) from (**)
  P ( Ai )
i 1
P( A  B )
 P( B) 
P( A)
P ( A / B)  P( B)
1 3p
1 p
1 2 p
1 , 0 
1, 0
1
3
4
2
We must also have 0 
8.
(*) 
1 3p 1 p 1 2 p


1
3
4
2
1
3
3
P  A   , P  A  B   , P  B   follows from P(B) ≤ P(A  B)
3
4
4
5
5
 P  A  B 
And P  B   P  A  B   P  A   P  A  B  
12
12
Alternate solution
P( A) 
1
, P ( A  B)  3 / 4
3
Now P( B)  P( A  B) is certainly true 
Also
3 1
  P( B)  P( A  B) 
4 3
Thus
5
 P ( B)  3 / 4
12
P( B) 
max P( B)  3 / 4
5
 P( A  B) 
12
60
P( B) 
5
12
PROBABILITY
3
3 3
1
. Also P( A  B)  P( A)  P( B)  1    1 
8
4 8
8
9.
P( A  B )  P( B ) 
10.
(i) P  A / ( A  B)  

(ii)
P  A  ( A  B) 
P( A  B)
P( A)
P( A)  P( B)  P( A).P( B)


A , B are independent )
1/ 2
5

1 1 1 1
6
  .
2 5 2 5
P  ( A  B) / ( A  B)   P  ( A  B) / ( A  B)   P(C / C) where A  B  C
Which is obviously zero
11.
RHS  P( A  B)  P( A) P( B )
 1  P( A  B)  P( A) P( B)
= 1  P( A)  P( B)  P( A  B)  (1  P( A)) (1  P( B))
12.
P( A)  P( A / B)

 P( A  B)  P( A).P( B)
 A and B are independent P( A / B) 
1 P( A  B)

4
P( B)
P( B / A) 
(i)
1 1 1
 P( B  A)   
4 2 8
1

2
from (i) P( B)  4 
P( B  A) 1

P( A)
2
1 1

8 2
Now P( A  B)  P( A  B)  P( A)  P(B)  P( A  B)

13.
A and B are independent )
P( A  B)
P( B)

3 1 3 1
  
4 2 4 2
 7/8
P( A  B)  P( A  B)  P( A)  P( B)  P( A  B)  P( A)  P( B)   P( B)  P( A  B) 
P( A)  P( A  B)  1  p1  p3
14.
P( A) .P( B) 
 xy 
2
15
2
1
, 1  P( A)  1  P( B)  
on putting x  P( A) , y  P( B) .
15
6
(i)
(1  x) (1  y) 
1
6
(ii)
The second equation is 1  ( x  y)  xy  1/ 6 
Putting y 
1  ( x  y) 
2
1
29
  x y 
15
6
30
29
 29
 2
 x in (i) we get x   x   The last equation can be easily solved .
30
 30
 15
61
PROBABILITY
17.
We have three equations
a(1  b)(1  c) 
1
8
From (ii) and (iii)
(i) , b(1  a)(1  c) 
b(1  c)
3
c(1  b)
1
4
(ii) ,
c(1  a) (1  b) 
1
12
(iv)


1
1
Putting value of a from (i) in (iii) we get c(1  b) 1 

 8(1  b) (1  c)  12

c(1  b) 
c
 12
8(1  c)
(vi)
From (iv) b  bc  3c  3bc 
Putting b 
b
3c
;`
2c  1
3c 
c
1
3c


in (v) we get c 1 



2c  1
 2c  1  8(1  c) 12
c(1  c)
c
1


2c  1 8(1  c) 12
On multiplying by 96(1  c)(2c  1) we get 96 c(1  c)2  12c (2c  1) 8(1  c)(2c  1)

18.
(a)
24c2  50c2  19c  2  0 . By hit and trial c 1 / 4 is a solution .
Both 1 and P( A)  P( B) are greater than P( A  B) and P( A  B) exceeds
both P( A) and P( B)
(b)
min  P( A) , P( B)   P( A  B) follows from
P( A  B)  P( A) and P( A  B)  P( B)
Or P( A  B)  P( A) P( A / B) or P( A  B)  P( B) P( A / B)
To prove other part , we note that if P( A)  B( B)  1  0 then
max 0 , P( A)  P( B)  1  0 and result follow as probability is non negative .
If P( A)  P( B)  1  0 then the result follows from P( A  B)  1
19.
For n  2 , P( A1  A2 )  P( A1 )  P( A2 )  1
(*) Which follows from P ( A1  A2 ) 1
Let the inequality be true for n  k then
P( A1  A2  ......  Ak )  P( A1 )  P( A2 )  ........  P( Ak )  (k  1)
Now P( A1  A2  .....  Ak  Ak 1 )
(**)
 P( B  Ak 1 ) where B  A1  A2  .......  Ak
 P( B)  P( Ak 1 )  1 from (*)
62
(iii)
PROBABILITY
 P( A1 )  P( A2 )  .......  P( Ak )  P( Ak 1 )  (k  1)  1
 inequality is true for n  k  1 as well
 P( A1 )  P( A2 )  ....  P( Ak 1 )  k
Exercise II
3.
A: Letter has come from Calcutta, B : Letter has come from Tatanagar, C: TA is visible.
1
2
C  1
P  
 A 7
1
2
C  2
P  
B 8
P  A 
P  B 
(Since only one TA is there in 7
(Only two TA)
Consecutive combinations of two letters)
4.
Required probability
1 1

A
4
 
2 7
 P  

 C  1  1  1  2 11
2 7 2 8
A: Card is black,
B: Card is red,
C: 13 Cards drawn are of the same colour.
26
 A  P  B  C  P  same colour is black 
P  


P C 
P  same colour 
C 
C13
B
. Similarly P   
26
25
C13  C13
C 
(a)
C13

52
C13
25
52
C13
C13
26
C13
 A
B
whence P    2  
25
C13  C13
C 
C 
P (head appears twice)
= P(event occurs through fair coin) + P(event occurs through unfair coin)
2 1 1 1
     1 1
3 2 2 3
(b)
26
C13
C13
25
26
7.
52

1
2
The event can occur in only one way i.e., on both occasion a fair coin is selected and a
2 1 1 1 1
   
3 2 2 2 12
1
 1 1
2
3
Required probability


2 1 1 1
    1 1 3
3 2 2 3
tail is thrown. The chance 
(c)
9.
Let Ei denote the event that the bag contains i red balls (i = 1, 2, 3, 4, 5) and R denote the event
that ball drawn on two occasions (with replacement) is red. Then
P  E1  
1
5
P  E2  
1
5
R 1 1
P   
 E1  5 5
 R 2 2
P     etc
 E2  5 5
63
PROBABILITY
Now
1 i i
 
 Ei  P  Ei  R 
5
5 5
P  

1
1
1
1
2
2
1 5 5
R
P
R


 
      ......   
5 5 5 5 5 5
5 5 5
2
2
i
i
.
 2

2
2
1  2  .......  5
55
Now the chance of future event
12
22
1
32 3 C 42 4 C 52 5 C
377
0   5   5 2   5 2   5 2 
55
55 C2 55 C2 55 C2 55 C2 550
10.
P ( same result )
 P( same result is incorrect and they make same mistake )
1 1

1 1 7 11
1
13
8 12
. Required probability 
    

1 1 7 11
1
8 12 8 12 1001
14
   
8 12 8 12 1001
11.
A: First number is smaller then second, B: Third number falls between first two numbers
n
Required probability
13.
15.
B

 A
= P
C3
P  B  A
P 1

 n 3 
C2 3
P  A
n
P2
n
P ( Head appears both times ) = P ( either fair coin is selected after which head appears both times
+ P ( unfair coin is selected and head appears both times )
1 1 1 1 1 1 1 2 2
1 1 4
        

 
3 2 2 3 3 3 3 3 3
12 9 27
1
9
12
Required probability 

1 1 4
29
 
12 9 27
Define the events
E3 : Bag contains 3 white balls
E4 : Bag contains 4 white balls
E5 : Bag contains 5 white balls
E6 : Bag contains 6 white balls
1
( Let us agree P( E3 )  P( E4 )  P( E5 )  P( E6 )  )
4
E : Three balls drawn with replacement are white .
F : Next draw will be white .
3
4
5
C
C
C
P( E / E3 )  9 3 , P( E / E4 )  10 3 , P( E / E5 )  11 3  P( E / E6 ) 
C3
C3
C3
64
6
C3
C3
12
PROBABILITY
1 3 C3

4 9 C3
Now P ( E3 / E ) 
1 3 C3 1 4 C3 1 5 C3 1 6 C3

 
 
 
4 9 C3 4 10 C3 4 11 C3 4 12 C3
With similar expansion for P( E4 / E ) , P( E5 / E ) , P( E6 / E )
0
1
2
3
Then P( F )  P( E3 / E ).  P( E4 / E).
 P( E5 / E ).  P( E6 / E ).
6
7
8
9
16.
The last two digits should be divisible by 4

They can be 12, 16, 24, 28, 32, 36, 48, 52, 56, 64, 68, 72, 76, 84, 92, 96.
Exercise IV
8.
25
36
25 11

= 1
whence the given answer follows.
36 36
P(Not throwing 6 on any dice in any trial)

P(throwing at least one 6)
=
Exercise V
3.
The common difference d can take the values from 1 to n
If d  1 then possible APs are (1,2,3) ; (2,3,4);........(2n 1,2n ,2n  1) which are 2n  1 in
number .
If d  2 then possible APs are (1,3,5); ( 2,4,6) ......... (2n  3,2n  1,2n  1)
Which are 2n  3 in number and so on .
If d  n then we will have only one AP (1, n  1,2n  1) with common different n .
Thus total favorable cases  2n  1  2n  3  .... 1 n2
6.
Total cases  12! , Favorable cases are 8 10!  2 ( ( A and B can take 8 positions )
Required probability 
8  10! 2
.
12!
4!
ways . In any case we will have 5 empty spaces out of which
2!
4!
5
2!  1 .
S will be put at 4 places  Required probability =
8!
4
4!2!
8.
AAIN can be arranged in
9.
(i)
Total cases  (n  1)! .Favorable cases. 2  (n  2)!  Required probability =
(ii)
Total cases  nC3 . Favorable cases  nC3  n  n(n  4) etc.
65
2
.
n 1
PROBABILITY
10.
Total cases  N Pn . Favorable cases 
N n 1
Cn  n!
(arrange N  n identical green balls and n red balls in a row such that no two red balls are
together )
11.
First Method : P ( at least one pair )  1  P( No pair )  1 
Second Method : Total cases  20C4 . Favorable cases 
12.
10
20 18 16 14
. . . .
20 19 19 17
C4 .24 ( See problem 16 )
Total cases  100C2 . Now xi  x j  10 will be satisfied in following cases
(1,2), (1,3), (1,4) .............(1,11) ,
(2,3) , (2,4) ,(2,5) ..............(2,12)
…………………………………………………….
(90,91) , (90,92)
 900 cases
(90,900)
After which will get 9  8  7  6  5  4  3  2  1  45 cases.
n1
16.
(i)
Follow problem 11
P ( exactly one pair )  n .
(ii)
C 2 r 2 2 2 r  2
2n
C2 r
.
( Choose exactly one pair in n ways then select 2r  2 pairs from remaining n  1 pairs
After which select exactly one shoe from each pair to get no pair )
20.

Favorable cases  2  8C4  4 7 C4  6 C4 5 C4  4 C4
6
21.
27.

C2 . u 4
where u4 is the number of no match for 4 objects which is equal to 9.
6!
There are 48 non kings 
q
49
C4 .4! 49!
. Now p is 1  q.
52!
Exercise VI
1.
Take any five consecutive positive integers we will observe that n2  1 is divisible in 2 cases
only. For example if we take 11,12,13,14,15 then 122  1,132  1 are divisible by 5
 Required probability = 2 / 5 .
2.
a , b , c can be chosen is 63  216 ways .
Now b2  4ac  0 is condition for real roots . b can take values from 1 to 6 .
If b  6 then the inequality becomes ac 9 which has 17 solutions
(1,1) ,(1,2),(1,3),(1,4),(1,5),(1,6)
(2,1) (2,2),(2,3),(2,4),
66
PROBABILITY
(3,1) (3,2) (3,3)
(4,1),(4,2), (5,1) , (6,1)
If b  5 then we must have ac 6.25 which has 14 solutions .
Namely (1,1),(1,2) (1,3),(1,4),(1,5),(1,6)
(2,1) (2,2),(2,3), (3,1) (3,2)
(4,1),(5,1),(6,1)
If b  4 then ac  4 has 8 solutions
(1,1),(1,2),(1,3),(1,4),(2,1),(2,2) (3,1)(4,1)
If b  3 then ac  2.25 has 3 solutions
If b  2 then ac 1 has 1 solution.
1
If b  1 then ac  has no solutions.
4
Thus total number of favorable cases = 17  14  8  3  1  43
Required probability
 43 / 216.
n
r
 l  where l is some integer and 0 ≤ r ≤ k – 1. As n  , l  
k
k
l
1
1
lim Pn  lim 
 lim 

( r is finite).
n 
l 
lk  r l  k  r k
l
3.
By division algorithm,
4.
Define the events
A : All drawn numbers are less than 10
B : None of the numbers drawn is 9
7
Required probability = P( A  B)  P( A)  P( A  B)
 9  8
    
 15   15 
7
C2
2  n C2
.
The
probability
of
second
event
=
.
n
n
P2
P2
n
6.
P(b > a) =
7.
P( x  9 / y  0) 
P  ( x  9)  y  0 
P( y  0)
19
since in 19 cases product will be zero .
100
Out of these cases in 2 cases sum will also be 9 namely 90,09
2
2
 Required probability  100  .
19
19
100
Total number of five digit numbers which can be formed by using the digits 0,1,2,3,4,5
 6! 5!  600
The number must be divisible by 3 which will be possible if we select either 1, 2,3, 4,5
(leaving 0) or 0,1,2,4,5 (leaving 3)
But the number should also be divisible by 2
In first case number of even numbers (4  3  2)  2  48 ( fix 2 and 4 in the end successively )
In the IInd case if 0 is fixed at the end then number of even numbers formed  4!  24
If 2 is fixed then 4! 3! 18 ,If 4 is fixed then also 4! 3! 18
Now P( y  0) is clearly
8.
67
PROBABILITY
Total favorable case  48  24  18  18  108
108
9
 Required probability 

600 50
(i)
Define the events
A: last digit in none of the chosen numbers is 0, or 5
B : last digit in none of the chosen numbers is 2,4,6,8

9.
n
 Required probability
 8   4
 P( A  B)  P( A  B)      
 10   10 
n
(ii)
10.
A: last digit in none of the chosen numbers is 0,2,4,6,8
B : last digit in none of the chosen numbers is 5
Required probability = P( A  B) etc.
(iii)
P ( last digit is 0 )
= 1 - P ( last digit is 5 ) - P ( last digit is 2,4,6,8 ) - P ( last digit is 1 , 3 , 5 , 7 )
Pc  c,
Let Pa  a , Pb  b ,
a  b  c.
The tournament can be played in 3!  6 ways
Order of play
Chance of winning
AB  ABC
ABC
ab  (1  a)bc
ACB
AC  ACB
ac  (1  a)bc
BAC
BA  BAC
ba  (1  b)ac
BCA
BC  BCA
bc  (1  b)ac
CAB
CA  CAB
ca  (1  c)ab
CB  CBA
CBA
cb  (1  c) ab
Now, P(I win by playing A first ) = P( ABC  ACB)  ab  2bc  ac  2abc
P(I win by playing A IInd ) = P( BAC  CAB)  2ab  2ac  2abc
P(I win by playing A Third ) = ab  2bc  ac  2abc.
It is evident that probability in the second case is maximum .
 a  b , a  c .
MISCELLANEOUS EXERCISE
1.
Required probability 
2.
P( BBWWWW RR)

1
.
1260
3.
(i)
7! 6!
12!
4.
Required probability

2 3 4 5 13
    .
6 8 6 8 24
2 1 4 3 2 1 3 2 1
       
9 8 7 6 5 4 3 2 1
( consider 6 girls as one object ) (ii)
2  (6!) 2
( There are two major cases )
12!
 .4  (1  .4)(.3)  (1  .4) (1  .3)(.2)  (1  .4)(1  .3)(1  .2)(.1)  .6976.
68
PROBABILITY
6.
P ( N  n)
 P( 1 ace is drawn in first n  1 trials at any drawings )
. P ( another ace is drawn at nth trail )

4
C1 48 Cn  2
3

52
Cn 1
52  (n  1)

(n  1)(52  n)(51  n)
50  49 17 13
Note : P( N  1)  P( N  52)  P( N  51)  0
7.
( why ?)
Note the following
( i)
Formula for P( A  B  C )
(ii)
Upper bound for P( A  B  C ) will come by using P( A  B  C ) 1
(iii)
Lower bound for P( A  B  C ) will come by given in equality P( A  B  C )  .75
(iv)
The data in this question of IIT -83 is incorrect .
Since P( A  B  C ) must be less than P( B  C )
8.
P  A 
25
,
100
P  B 
20
,
100
P  A  B 
8
100
If the event of looking into advertisement is denoted by T, then
 T  30
P
,

 A  B  100
We have to find P(T)

 T  40
P
,

 A  B  100
 
 T  50
P

 A  B  100

P T   P T   A  B   P T   A  B   P T   A  B  
30
40
50
 P  A  B
 P  A  B
100
100
100
30
40
50
  P  A  P  A  B  
  P  B   P  A  B 
 P  A  B
 .139
100
100
100
There are 24  115 possible answers to a multiple answer objective.
 P  A  B
10.
Therefore probability of correct answer

1
15
P ( student will get marks )  P ( he gets at the first trial ) + P ( he gets at the second trial )
 P( he gets at the third trial ) 
11.
1 14 1 14 13 1
1
    

15 15 14 15 14 13 5
Define the events
A : Lot contains 2 defective articles
B : Lot contains 3 defective articles , E : Testing ends at twelfth trial
69
PROBABILITY
Required probability
 P( A  E )  P( B  E ) = P( A) P(E / A)  P( B) P( E / B)
P( A) .4, P( B) .6.
To calculate P( E / A) we note that testing
will end at twelfth trail if 1 defective item is drawn
in first 11 trial and then another defective item is drawn at
XIIth trial
 P( E / A) 
12.
2
3
C1 18 C10 1
C1 17 C10 2
.
Similarly
etc.

P
(
E
/
B
)


20
20
C11
9
C11
9
He will be one step a way in two cases .
(i)
6 steps forward 5 steps backwards (ii) 6 steps backwards 5 steps forward
 Required probability
13.
11
C6 .4 .6 11 C5 .4 .6 .
6
5
5
6
The disjoint cases are
BBB , BWB ,WBB ,WWB
 Required probability
2 3 4 2 2 3 2 2 3
2 1 2
        
  
4 5 6 4 5 4 4 3 4
4 3 2
14.

23
.
30
We will calculate the probability of complimentary events ie P( value  1.50)
Let the number of ten paisa and five paisa coins in the bag be n1 and n2 . n1  n2  N
Letus denote fifty and twenty five paisa coins by F and T respectively.
F
0
1
T
5
4
10 paisa
0
0
5 paisa
0
0
cases.
Not possible
2
C1  5C4
2
3
0
0
2
C2  5C3
2
2
1
0
2
C2  5C2  n1 C1
2
2
0
1
2
C2  5C2  n2 C1

15.
Total favourable cases 10  10  10 N  20  10 N
 Required probability  1 
20  10 N
.
N 7
C5
P( in best of three option)
 3C2 (.4)2 (.6)1  3C3 (.4)3 .352
P( in best of five option)
 5C3 (.4)3 (.6)2  5C4 (.4)4 (.6)1  5C5 (.4)5
70
 .31744
PROBABILITY
Note that first option has a greater probability.
18.
P( correctly answered ) 
19.
P  A 
50
,
100
1 1 1 1  1 1
    1     1 etc.
3 4 6 8  3 6
50
50 50
,
P C  

2
100
100 100
50 50
P  A  B 

 P  A P  B 
100 100
P  B 
P  B  C  = P(B and C both happen) = (both are non-defective)
=
50 50

100 100
= P(B) P(C)
P  C  A = P(both are non-defective)
20.
=
50 50

100 100
= P(C) P(A)
Now, P(A  B  C) = P (impossible event) = 0  P(A) P(B) P(C)
P( India getting at least points )
= P( 7 points )  P( 8 points ) = P( India draws only one match and wins remaining matches )
+ P( India wins all the matches )  4C1 (.05)1 (.50)3  (.5)4
4
100
 .0875
(2,9) (3,6),(6,3),(9,2) are n  4 r  3,4
21.
p
22.
Required probability  P ( Head then 7 or 8 total by pair of dice )
(
 P ( trail then 7
or 8 from the simple lot 2,……11) 
1 11 1 2 193

  
2 36 2 11 792
23.
In any van 3 girls can sit at back on adjacent seats in two major ways
 Number of ways  2  2  3! 11P9

24.
Required probability

2  2  3! 11 P9
.
14
P9
P ( exactly one of A or B occur )
 P( A)  P( B)  2P( A  B) etc
Twice of our target  2P( A  B  C )
 P( A)  P( B)  2P( A  B)  P( B)  P(C )  2P( B  C )  P(C )  P( A)  2P(C  A)
 2 P( A  B  C )
 3 p  2 p2 .
71
PROBABILITY
15
25.
14
C7 1
C
8
P(S1 is the winner) = 16
 P(S1 but not S2 or S2 but not S1) = 2  16 7 
C8 2
C8 15

27.
P( A)  p  (1  p)3 p  (1  p)6 p  ...........
a
p


1  r 1  (1  p)3
P( B)  (1  p) p  (1  p)4 p  ........   (1  p)  
P(C )  1     etc .
28.
We will solve the problem by two methods . The first method is based upon results on pairings.
Note that
4
(i)
Four teams T1 , T2 , T3 , T4 can be paired for first round match in
C2 , 2 C2
 3 ways.
2!
Observe the table for better understanding
T1 T2
T3 T4
T1 T3
T2 T4
T1 T4
T2 T3
Now suppose the strength of the players is in the order T1  T2  T3  T4 then T3 will reach the
next round in only one case ( when the pairing is done as T1 T2 ;T3 T4 )
 P (T3 will reach next round and fight with T1 ) 1 / 3 .
(ii)
When there are eight players P1 , P2 ,..........P8 whose strengths are in the order
P1  P2  ......  P8 then P4 will reach the final if
(a)
In the first round P4 is paired with Pj ( j  4) and one more Pk ( k  4) is paired with
Pl (l  4) so that P4 and Pm reach the second round where Pm is winner of Pk and Pl
(b)
In the second round P4 is paired with Pm and reaches the final round to play with Pl .
Now at the first round total pairing 
8
C2 .6 C2 . 4C2 .2 C2
.
4!
Now lat us calculate all possible favourable pairing at the first round. Suppose P4 is
paired with P5 . Observe the table
P4 P5
P6 P7
remaining four players P1 , P2 , P3 , P8 will give rise to 3 pairings
P7 P8
remaining four players P1 P2 P6 P3 will give rise to 3 pairings
P6 P8
remaining four players P1 P2 P6 P3 will give rise to 3 pairings
72
PROBABILITY
P6 P8
remaining four players P1 P2 P3 P7 will give rise to 3 pairings
Thus there are 9 favorable pairings when P4 is paired with P5 . Similarly when P4 is paired with
P6 we will obtain 9 favorable pairings and so on . Thus total number of favorable pairings
 9  4  36.
In the second round ( from (i))
Total pairings  3, favorable pairings = 1
(For instance if the first round is played as P4 P5 , P6 P7 ; P1 P2 ; P3 P8 then in the second round we
will have P4 , P6 , P1 , P3 and they are paired in 3 ways of which one favors P4 from (i))
Thus required probability  8
36
1 4
  .
2
C2 . C2 . C2 . C2 / 4! 3 35
6
4
Second Method : We can simply construct the sample space with respect to P4 . We note that
P4 . reaches the final if he is paired with Pi (i  4) and one more pairing in the first round is
 P , P )  where
j
k
j , k  4. In the second P4 must be paired with the winner of Pj and Pk .
Thus we just have to choose i , j , k from P5 , P6 , P7 , P8 .
 Number of favorable cases  4C3 ,Total number of cases  7C3 .

29.
Required probability 
4
7
C3
.
C3
a1a2a3……….a9 will be divisible by 11 if alternating sum a1 – a2 + a3 - ……….. + a9 = 11k.
Also a1 + a2 + a3 + ………. + a9 = 1 + 2 + 3 + ……… + 9 = 45. On subtracting
(a2 + a4 + a6 + a8) = 45 – 11k

45 – 11k even 
k is odd.
Now min (a2 + a4 + a6 + a8) = 1 + 2 + 3 + 4 = 10
max (a2 + a4 + a6 + a8) = 6 + 7 + 8 + 9 = 30

20 ≤ 45 – 11k ≤ 60


15
15
k

11
11
k
= 1 or -1
Now for these two values of k we can easily determine the number of favorable cases

m
mk
n
m
m




mn mnk mn mnk mn
30.
Required probability
31.
When n dice are thrown a sample point will be a point of n-dimensions where each co-ordinate
can take 6 values. Thus there will be 6n sample points in all. Let us now evaluate the number of
favourable sample points. These will be sample points having exactly 3 different digits at n
places. Let us assume that these 3 numbers are 1, 2 and 3, then number of sample points of ndimensions having exactly 3 different digits 1, 2, 3 as their co-ordinates must be 3n – 3.2n + 1.
73
PROBABILITY
But the 3 different digits can be any combination of 3 from 1, 2, 3, …………., 6. Thus total
 6C3  3n  3.2n  1
number of favourable sample points
6
Thus the required probability
33.
6n
.
Let denote the event of passing then required probability .
 P( AAA)  (AAA)
34.
=
C3  3n  3  2n  1
 P ( AAA)  P(AAA)
p p
p
 p3  p 2 (1  p)  (1  p) .  p.(1  p)
2 2
2
Define the events
 2 p 2  p3
E4 : exactly four white balls are drawn
E5 : exactly five white balls are drawn
E6 : exactly six white balls are drawn
E : Six balls are drawn with replacement from the bag in the figure.
Now P( E4 ) 
6
6
6
C5 12 C1
C6 12 C0
C4 12 C2
,
P
(
E
)

,
P
(
E
)

5
6
18
18
18
C6
C6
C6
clearly P( E )  P( E4 )  P( E5 )  P( E6 )
P( E4 / E ) 
P( E4  E )
P( E )
P( E5 / E ) 
P( E5 )
 p5
P( E4 )  P( E5 )  P( E6 )
P( E6 / E ) 
P( E6 )
 p6
P( E4 )  P( E5 )  P( E6 )

P( E4 )
 p4
P( E4 )  P( E5 )  P( E6 )
Now P( exactly one white in next two draws )
 p4 .
36.
2
1
C1.10 C1
C1.11 C1

p
.
5
12
12
C2
C2
+ p6  zero etc .
Note that if A and B are independent P( A  B)  P( A) P( B) , P( A  B)  P( A) P(B)
Now P( A  B) P( A  B)  ( P( A)  P( B)) P( A).P( B)
 P( A) P(B) P( A)  P(B) P( A).P(B)
44.
 P( A) P(B)  P(B) P( A)  P(C )
P ( at least one rainy day )  1  P ( No rainy day) 1  (.7)7
74
PROBABILITY
Now P ( at least two rainy days / at least one rainy day )

46.
P( at least two rainy days )
1  (.7)7  7(.7)6 (.3)1

P( at least one rainy day )
1  (.7)7
Consider a favorable configuration of n H and T ( ie n tosses contains no two successive
heads ) . This either ends with a T or ends with H . If ends with T then previous n  1 letters
are favorable cases f n1 for probability un 1 . Again if it ends with H then second last should
essentially be trail and previous cases are favorable cases of n  2 tosses
f n f n 1 f n  2

 n
2n 2n
2
 f n  f n 1  f n  2 
47.
The score n can be obtained in two mutually exclusive ways
(i)
A head is obtained when the score is n  1
(ii)
A tail is obtained when the score is n  2
 Pn  Pn1 .
P1 
P3 
48.
fn
f
1 f
1
1
1
 nn11 .  nn22 .  un  un1  un2
n
2
2
2 2
4
2
4

1
1
 Pn2 .
2
2
1
 Pn  ( Pn1  Pn2 )
2
1
1 1 1
, P2   
 3 / 4 (ie either two heads in succession or directly one tail )
2
2 2 2
1 1 1 1 1
     2  5 / 8 etc . Induction part is simple .
2 2 2 2 2
Let within k trials the number of balls drawn from bag A be n1 and from bag B be n2, then
n1 + n2 = k
and
N – n1 = n – n2


k n N
nN k
, n2 
2
2


AAA.......
BB........B
Required probability
= P
,

 chosen n1 times chosen n2 times 
n1

 n1  n2 !  1 n  1 n
=
49.
k
1
     Ck n N   .
n1 !n2 !  2   2 
2
2
1
2
k
Suppose the required event (Bag A becomes empty and B contains r balls) occurs exactly after k
drawings then we must have

Probability
=

AAA......... A
chosen n times
n
k
1 1
Cn     
2 2
BBB.......
chosen k  n times
k n
. But n – (k – n) = r
2nr

50.
k = 2n – r
Apply (x + y)n – (x – y)n

Required answer =
Cn  2
.
22 n  r
= 2[nC1xn-1y + nC3xn-3y3 + ………. ].
75
(why ?)
PROBABILITY
51.
x + y = 2n. Greatest product xy = n2
3 2
3 

n   P  x  2n  x   n 2 
4 
4 


 n 3n  
= ( P  2 x  n  2 x  3n   0)
 P x  ,  .
 2 2 

3n n
n 1
Now favourable range =
  n . Total range = 2n. Required probability =

2 2
2n 2
Required probability
52.


= P  xy 
The seats are numbered.
In any arrangement n seats are occupied by men and n are occupied by women. Now n seats for
women can be chosen in 2nCn ways. Thus total number of cases = 2nCn.
Now there are only two major ways in which the arrangement is favourable.
Either the pairs occupy
(*)
(1, 2), (2, 3), (3, 4)………………. (2n – 1, 2n) OR
(**)
(2, 3), (4, 5), (6, 7) ……………. (2n, 1)
In case (i) a male is placed in two ways (either at the first co-ordinate or at the second)

The number of ways of seating men and women = 2n
Which is also true for second case. But a case (W M)(W M) …….. (W M) or (*) is same as a case
(M W) (M W) ……. ((MW) of ** and vice-versa.)

Number of favourable case = 2n + 2n – 2 = 2n+1 – 2
Thus required probability
53.
=
2n 1  2
.
2n
Cn
Consider an m  n grid  m  n  which is obtained by drawing m  1 equally spaced
horizontal and n  1 equally spaced vertical lines .
Note that following
(i)
Number of shortest routes (paths ) from Oto N 
condition m  n )
(m  n)!
( This does not require the
m!n!
(m  n  1)!
m!(n  1)!
(ii)
Number of paths from B(0,1) to N (m, n) 
(iii)
A good path from O to N will start from A and will always be below the line y  x .
(iv)
OD is a line which passes throw (1,1), (2,2),(3,3)......... or y  x .
(v)
Any favorable path will start from A(1,0) but this will include non favorable (bad)
(m  n  1)!
paths also . Total paths from A to N 
(m  1)!(n)!
(vi)
Any bad path from A will go to intersect line OD . If we reflect this portion of bad path.
From A we get a path from B to N
76
PROBABILITY
(m  n  1)! (m  n  1)!

(m  1)!n !
m!(n  1)!
 favorable paths from A 
(m  n)!
m !n !

mn
mn
By shifting origin from O to O '(1,0) we can solve another problem for m  n
Note :
Suppose there are 2n children standing in a queue to buy a ticket of Rs 5 . n of these
children are having 5 rupee coin the other n are carrying 10 rupee coin .The box office
has no change to begin . what is the probability that there will not be any problem of
change at any stage. Using one to one correspondence we can easily observe that
number of favorable cases
 Number of paths from O to (n, n) satisfying y  x
 Number of good paths from O ' to (n, n)
The last number can be found from ballot problem described above . To illustrate let us
consider
Illustration consider 4 children only . We form 2  2 grid
For queue ABIDE is a good path. This corresponds
To path from A ' which will not touch line AF
OBJECTIVE EXERCISE
1.
6 1
5
 , P (Not getting a doublet ) =
36 6
6
P (getting a doublet not more than twice ) 1  P (getting doublet all the three times)
P (getting a doublet with a pair of dice ) 
3
2.
1
215
1
1    1 

.
216 216
6
Define the events E : 3 dice show different faces F: At least one face shows 6
1 5 4
  3
P( F  E )
1
 6 6 6

Require probability P( F / E ) 
6 5 4
P( E )
2
666
(Note that P( F  E )  P (six on one dice and different non 6 numbers on remaining dice)
1 5

10
2 9
 P( B / red ) 

.
3.
1 2 1 5
19
  
2 4 2 9
6. Let  be the probability of getting an odd number then the probability of getting even number = 2
We must have   2 1   1/ 3.
1 2 1 5
P(red)    
2 4 2 9
77
PROBABILITY
7.
9.
10.
11.
Now required event will occur if either both out comes are even or both are odd
1 1 2 2 5
 Required probability     
3 3 3 3 9
52 39 26 13
2197
.
 


52 51 50 49 20825
4
C2  5C2 10
 Required probability 

9
21
C4
`
1 23

4! 24
Required probability  P(0,0)  P(1,1)  P(2,2)  P(3,3)
1
2
2
0
3
1
2

1 1  
1 1 
  3 C0        3 C1       ....

 2   2   
 2   2  


C02  C12  C22  C32
5
 .
6
16
2
12.
Make cases with d  1, d  2,....., d  7
14.
P( A)  .5, P( B)  .3 , P( A  B)  0 , P( A  B)  P( A  B)  1  P( A  B)
 1  [ P( A)  P( B)  P( A  B)]  1  [.8]  .2

None of choices (a),(b),(c) are true
1
8
3
P( A  B ) 
8
P( A  B) 
15.
1 
8 

3
 P ( A) P ( B )  
8 
 P ( A) P ( B ) 
which easily yields P( A) 
12
16.
 (d) is true .
A and B are independent
1
1
, P( B) 
2
4
10
C3 .29
110  2 

 
12
9 3
3
4
17.
1 1
Let the coin be tossed n number of times . then n C4    
2 2
n4
7
1 1
 n C7    
2  2
2

18.
P( A) 
n
9
25
1 1
C4  nC7  n  11  P (getting two heads )  11C2     
2048
2 2
3
,
16
P( B) should also be equal to
3
since cases are similar
16
 Choices (a) , (c) are wrong . (b) can be called correct
19.
n 7
P( A  B  C ) 
1 1 1
1
5
   0  0  0
4 4 4
8
8
78
P( A)  P( B) is true .
PROBABILITY
20.
E ( x 2  2)  (02  2) P(0)  (12  2) P(1)  (22  2) P(2)  (32  2) P(3)
3
3
3
3
1
1
1
1
 2 C0 .   .  3  3C1   .  6 3 C2    11  3C3    5
2
2
 3
2
3
2! 8! 2
 .
9!
9
22.
Required probability 
23.
Required probability  .4  (1  .4)(.3)  (1  .4)(1  .3)  .2  (1  .4)(1  .3)(1  .2)  .1  .6976.
24.
p  2q


2
1
p  ,q 
3
3

p  q  1
P (at least four successes)
4
2
5
1
 2 1
 2 1
2
 C4      6C5      6 C6  
 3 3
 3 3
3
6

6
27.
496
729
Total number of cases  6 4
Number of favorable cases  coeff of x13 in
x
1
 x 2  x3  ....  x 6 
4
 x 1  x 6  
4
4
  coeff of x9 in 1  x6  1  x 
 coeff of x in 
 1 x 


4
13


 coeff of x9 in 1  4 x6  6 x12  .... 1  x 
 491 C9  4.
coeff .of
4  31
C3
4
xr in 1  x 
n
n  r 1
is
Cr

 12C3  4. 6C3  220  80  140
 Required probability 
140
35

1296 324

( a ) is correct.
4! 2! 2

5!
5
30.
Required probability 
31.
There are n numbers from 1 to 3n which are of the type 3k , n numbers are of the type 3k  1
and n numbers are of the type 3k  2 . If 3 numbers are selected then their sum will be
divisible by 3 if all three numbers belong to the same variety or they belong to different
variety 
Number of favorable cases  3.n C3 
79
 C
3
n
1
PROBABILITY

n  n  1 n  2 
2
Total cases 3n C3 
32.
 n3 
n 2
 n  3n  2  2n 2 
2
3n  3n  1 3n  2 
6
 Required probability 
3n 2  3n  2
 3n  1 3n  2 
P  X  Y  3  P   X  0  Y  3   P   X  1  Y  2 
 P   X  2  Y  3   P   X  3  Y  0 
Since X and Y are independent P
 X  r   Y  s   P  X  r  P Y  s  etc.
IIT
1.
2.
18
18
9
P( X  Y )  P (first shows even , second shows odd) =
, P(Y ) 
36
36 ,
36
( (2,1),(2,3),(2,5),(4,1),(4,3)(4,5),(6,1),(6,3),(6,5) are 9 cases)
 P( X  Y )  P( X ).P(Y )  X and Y are independent
But as P( X  Y )  0 , (d)is correct answer.
P( X ) 
P  A  .25, P  B   .50, P  A  B   .14




Now P A  B  P A  B  1  P  A  B   1  P  A  P  B   P  A  B   .39
4.
.1 10 5


.42 42 21
Total possible determinants = 16. There are three determinants only whose values are positive
1 0 1 0 1 0
namely
 Required probability = 3/16
,
,
0 1 1 1 1 1
5.
P A/ B 
6.
Define the events. A: all drawn numbers are less then 10 B: None of the drawn numbers is 9.
3.
P (either 1 or 2) = .1 + .32 = .42  P(it is 1) =



P AB
P  B
  P  A  B  1 P  A  B
P  B
P  B
7
7.
9.
7
 9  8
Required probability = P  A  B   P  A   P  A  B       
 15   15 
8!
4!
Total cases =
, Now, AA IN can be arranged in
ways and in the five empty spaces of
4!2!
2!
these four letters, four S can be arranged in 5C4 = 5 ways
4!
5
4!
2!  1
 Total favourable cases = 5 
 required probability =
8!
14
2!
4!2!
P (exactly one of the events occurs)
 P (M  N )  (M  N )  P M  N  P M  N






 P  M   P  M  N   P  N   P  M  N   P  M   P  N   2P  M  N 
80
PROBABILITY
 (a) is correct




Since P M  N  P M  N  1  P  M  N   1  P  M   P  N   P  M  N 
= P  M   1  P  N   2  2P  M   2P  N   2P  M  N 
 LHS of choice (c)
10.
= P  M   P  N   2P  M  N   (c) is also correct.
Let A: maximum number is not more than 10, B: minimum on them is 5
P  B  A
Required probability  P  B / A 
P  A
Now P  A  
10
C2
C2
100
and favourable cases for P(B  A) are (5, 6), (5, 7), (5, 8), (5, 9) and (5, 10)
5 / 100C2
1

10
100
C2 / C2 9
1 3p
1 p
1 2 p
We must have 0 
 1, 0 
 1, 0 
1
3
4
2
1 3p 1 p 1 2 p
We must also have 0 


 1  Events are mutually exclusive 
3
4
2
1
1
On solving the inequalities we easily get  p 
3
2
P((A  B)  (A  C)) = ½
 P(A  B) + P(A  C) – P(A  B  C) = ½
 pq + ½ p – ½ pq = ½  pq + p = 1
Out of the given choices (c) is correct
P(A  B) = .6, P(A  B) = .2 Now P A  P B  2  [ P  A  B   P  A  B ]  1.2
 P  B / A 
11.
12.
13.
14.
15.
16.
17.
 
 
(c) is correct since P(A  B) = P(A) + P(B) – P(A  B)
P(A  B) ≤ 1  P(A) + P(B) – P(A  B) ≤ 1  P(A  B) ≤ P(A) + P(B) – 1
 (a) is correct
Again P(A) + P(B) = P(A  B) + P(A  B) ≥ P(A  B)  (b) is also correct
There are four cases in which the event can occur. These are RRR, RBR, BBR where RBR denotes
the event that a red balls is transferred from urn A to urn B then a black ball is transferred from
urn B to urn A and finally a red ball is drawn from urn A, with similar meanings for RRR, BRR,
BBR, required probability
6 5 6 6 6 5 4 4 7 4 7 6 32
           
10 11 10 10 11 10 10 11 10 10 11 10 55
According to the condition 100 C50 p50 (1  p)50  100C51 p51 (1  p)49 . Solve for p.
E and F are independent  P(E  F) = P(E) . P(F)
Now P  E  F   P  E   P  E  F   P  E   P  E  P  F   P  E  1  P  F    P  E  P  F 
 E and F are independent again


P E  F   P E  F  1 P E  F   1 P E   P F   P E  F 
 1  P  E   P  F   P  E  P  F   1  P  E   1  P  F    P  E  P  F 
18.
 E and F are independent (d) is also true
P(getting a total 5 with a pair of dice) = 4/36 = 1/9
P(getting a total of 7 with a pair of dice) = 6/36 = 1/6
81
PROBABILITY
P(getting neither 5 nor 7) = 26/36 = 13/18
2
3
1 13 1  13  1  13  1
a
1/ 9
2
           ..... 


9 18 9  18  9  18  9
1  r 1  13 5
18
Since A and B are independent P(A  B) = P(A).P(B)  P(A  B) = P(A) + P(B) – P(A).P(B)
 .8 = .3 + P(B) - .3  P(B)  P(B) = 5/7
P  A  B
(a) is correct since P(A/B) =
and P  A  B   P  A  P  B   P  A  B 
P  B
Required probability 
19.
20.
21.
22.
24.
25.
(c) is correct because RHS = 1- (1 – P(A))(1 – P(B)) = P(A) + P(B) – P(A).P(B)
= P(A) + P(B) – P(A  B)
(Since A and B are independent)
= P(A  B)
np = 2, npq = 1 pq = 1  p = ½, q = ½, n = 4
0
4
1
3

1 1
1 1 
P(X > 1) = 1 – [P(0) + P(1)] = 1   4 C6      4 C1       11/16
2 2
 2   2  

3 2 1 1
Required probability    
6 6 6 36
The numbers appearing must be from 2, 3, 4, 5
 Number of favourable cases = 44
Total cases = 64
 required probability 44/64 = 16/81
A white ball is drawn for the 4th time on 7th draw if in the first six draws exactly 3 while balls
appear followed by a white ball at the 7th draw.
3
26.
27.
28.
3
 12   12  12 5

 Required probability = 6C3     
 24   24  24 32
P (getting a number greater than 4) = 2/6 = 1/3, P(Not getting a number greater than 4) = 2/3
The required event must occur either in two tosses or in four tosses or in six tosses and so on
2 1
3
6

2 1 2 1 2 1
a
2
 3 32 
 Required probability            ........ 
3 3 3 3 3 3
1 r
5
2
1  
3
Since A and B and C are mutually independent events we have
P(A  B) = P(A) P(B),
P(B  C) = P(B) P(C),
P(C  A) = P(C) P(A)
P(A  B  C) = P(A) P(B) P(C)
(*)
P(A  (B  C)) = P((A  B)  (A  C)) = P(A  B) + P(A  C) – P(A  B  C)
= P(A)P(B) + P(A)P(C) – P(A) P(B) P(C)
(From *)
= P(A)(P(B) + P(C) – P(B  C)) = P(A)P(B  C)
 A and B  C are independent  statement S1 is true. Again for S2
P(A  (B  C)) = P(A  B  C) = P(A)P(B)P(C)
(From *)
= P(A) P(B  C)
(Form *)
 S2 is also true
P A  .3  P  A  .7, P  A  B   .5  P  A  P  A  B   .5  P  A  B   .2
 
 P(A  B) = P(A) + P(B) – P(A  B) = .7 + .4 - .2 = .9
Now P  A / A  B  

P A A B
P A  B
  P  B  A   B  B   P  B  A
P  A  B
82
P A  B

B  B 
PROBABILITY

29.
30.
31.
32.
P  B  A
P  A  P  B   P  A  B 
1  1 1  1 1 1 1
 Required probability   1     1     
2  2 2  2 2 2 4
The relation P(A  B) = P(A) + P(B) – P(A)P(B)  A and B are independent
 (d) is obviously correct .Again P((A  B)′) = P(A′  B′) = P(A′)P(B′)
(Since A, B are independent  A′, B′ are also independent),
Define the events, A: minimum of chosen number is 3, B: maximum of chosen is 7
P(A  B) = P(A) + P(B) – P(A  B)
7
C
P  A   10 2
( 3 should be there and remaining two numbers must be from 7 numbers)
C3
6
C2
C3
10
3
(3 and 7 should be essentially there and one number must be from 4, 5, 6)
C3
11
 P  A  B 
40
To fulfill the condition of the problem the three black balls must be kept in 8 empty spaces
7
C2 7
created by seven white balls  Required probability 

10! 15
3!7!
The possible cases are WWB, WBW, BWW
3 2 3 3 2 1 1 2 1 13
 Required probability =         
4 4 4 4 4 4 4 4 4 32
PE  F  PE  F  PE  F   PF   PE  F 
LHS of (a) =


 1  (a) is correct
PF 
PF 
PF 
P  A  B 
34.
35.
LHS of (d) =

10
PE  F 
PF 

PE  F 
PE  PE  F  1 PE  F 
36.
37.
.2
1

.7  .6  .5 4
The favourable case are two only. Total number of cases = 6C3
If W denote India’s win then the favourable patterns are W W W , W W W
P  B 
33.

PF 
PF 



PE  PE  F   P E  F
PF 

PE   P E  F  1 P E   P F   P E  F 
PF 

1 P F 
PF 
1
It can be checked that other two choices are wrong.
The two faulty machines will be detected in two test if the first two draws yield either both
faulty or both non faulty
2 1 2 1 1
 Required probability =    
 (a) is correct
4 3 4 3 3
Take an example to disprove choices (a) , (b) ,(c) Suppose a dice is thrown twice .
Let E be the event that first throw is 2 and F be the event that second throw is 3
Then (a) , (b) , (c) are disproved.
83
PROBABILITY
38
Since fifth toss is independent of other tosses . the required probability should be ½
31
NOTE:- By Laplace law of succession the chance can also be
by applying Baye’s
32
Theorem for future event .
75 3
50 1
39.
 1  (1  p)(1  c)(1  m),
  pc(1  m)  pm(1  c)  mc(1  p)  pmc
100 4
100 2
40 2
1
  pc(1  m)  pm(1  c)  mc(1  p), From last two relations pmc  ,
100 5
10
From first relation
3
3
 1  [1  ( p  m  c)  pc  pm  cm  pcm) 
 ( p  c  m)  ( pc  pm  cm)  pcm
4
4
2
2 3 7
Also the last relation is , pc  pm  cm  3mcp   pc  pm  cm   
5
5 10 10
27
We easily get p  m  c 
20
40.
Total ways of choosing m and n  100 100, Now note that ,
71  7  5k  2,72  49  5k  4,73  343  5k  3,74  2301  5k  1
The same sequence will repeat for next four powers i.e. 75  5k  2,76  5k  4,77  5k  3,78  5k  1.
Thus 71 ,75 ,79 ,......797 are of the type 5k  2;72 ,76 ,710 ,.....798 are of the type 5k + 4;
73 ,77 ,711,....799 are of the type 5k  3 and 74 ,78 ,......7100 are of the type 5k + 1.
There will be only four favourable combinations in which 7 m  7 n will be divisible by 5 .
Note the following
7m
7n
5k + 2
5k + 3
5k + 4
5k + 1
5k + 3
5k + 2
5k + 1
5k + 4

Number of favourable cases = 25  25  25  25  25  25  25  25  4  625
4  625 1

Required probability =

10000 4
NOTE:- The problem can be done without any calculations in a pure objective number. Just consider
71 ,72 ,73 ,74 which are four and note that only the ordered combinations.
71  73 ,72  74 ,73  71,74  72 are divisible by 5.
 probability = 4/16 = ¼. The idea is applicable due to two reasons.
(i)
Power of 7 from a cycle of 4 with respect to remainders when they are divided by 5.
(ii)
100 is divisible 4.
84
PROBABILITY
41.
The favourable cases are (1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),  12
12 4
 Required probability 6  .
C2 5
42.
(B  C)  B  ( A  B  C )  ( A  B  C )  P( B  C)  P( B)  P( A  B  C )  P( A  B  C ) etc.
43.
44.
100 
Numbers which are divisible by both 2 and 3 are multiples of 6 which are 
 16
 6 
16
C
4
 Required probability = 100 3 
.
C3 1155
1 should come either in 2 trials or in 4 trials or in six trials and so on
 Required probability
5 1

5 1 5 1 5 1
a
6
6  5  36  5
           ......

2
6 6 6 6 6 6
1 r
36 11 11
5
1  
6
3
45.
(i)
If P (ui )  ki then
5
n

P(ui )  1  k 
i 1
46
47.
2
n(n  1)
1
2
n

P(W )  P(u1 ) 
 P(u2 ) 
 .....  P(un ) 
n 1
n 1
n 1
2 1
1
2 2
2
n2
n
2




 .... 


[12  22  32  .....  n2 ]
2
n(n  1) n  1 n(n  1) n  1
n(n  1) n  1
n(n  1)
2
n(n  1)(2n  1)
2n  1


2
6
3(n  1)
n(n  1)
1
In this part P(ui )  c  nc 1  c  ,
n
1
1
1
2
1
n
Now P(W )  
 
 .....  
n n 1 n n 1
n n 1
1
n

P(un  W ) n n  1
1  2  3  ....  n 1
2
 

 P(un / W ) 


.
n(n  1)
2
P(W )
1/ 2
n 1
1
In this part P(ui )  P(u2 )  P(u3 )  ..........P(un ) 
n
P(W  E )  P (2nd bag is chosen and a white ball is drawn )
 P (fourth bag is chosen and a white ball is drawn )+………..
1
2
1
4
1
6
1
n
 
 
 
 ........  
n n 1 n n 1 n n 1
n n 1
 n  n 
n
2    1
2(1  2  3  .......  )
2  4  6  ...  n
n2
2 2 
2


  

n(n  1)
n(n  2)
n(n  1)
4(n  1)
P( E )  P ( Choosing an even numbered urn )
85
PROBABILITY
1
( Since half the bags are even numbered and half the bags are odd numbered )
2
n2
n2
4(n  1)
 P(W / E ) 

1/ 2
2(n  1)

48.
P(E  F /G) =
c
c
P  Ec  F c  G 

P G 
1 – P(E/G) – P(F/G) = 1 – P(E) – P(F)
P G   P  E  G   P  F  G 
P G 
(E, F, G are pairwise dependent)
= P(Ec) – P(F)
49.
Define the events
A: Indian is seated adjacent to his wife
B: Four Americans are seated adjacent to their wines.
 B 
P A
P  A  B
P  B
4! 2!

5
9!  2
4
5
5! 2!
9!
50.
(b)
51.
(b) :
: P( A  B) 
4 p
2p/5


10 10
10
For unique solution
a b
c d

2p
is an integer 
5
p  5 or 10
 0 where a, b, c, d  {0,1} .Total cases =16
Favourable cases =6 (Either ad=1, bc = 0 or ad = 0, bc =1).
Probability that system of equations has unique solution is
6 3
 and system of equations has
16 8
either unique solution or infinite solution so that probability for system to have a solution is 1
52.
5 5 1 25
P( X  3)    
6 6 6 216
53.
5 1
2
3
4
  
6
6 25
5 1 5 1 5 1
P ( X  3)              ......   

 5  36
6 6 6 6 6 6
1  
6

Option A is correct.
2

54.
Option B is correct.
P(( X  6) / ( X  3)) 
P  ( X  6)  ( X  3)  P( X  6)

P( X  3)
P( X  3)
86
PROBABILITY
5
 5 5  1   5  6  1 

          ....   5 
2
 6   6   6   6 
   6    5 

 
3
P( X  3)
5 6
 
6
3
(Since
55.
4
5
3
5 1 5 1 5 1
5
P( X  3)              ......    )
6 6 6 6 6 6
6

Option D is correct.
(C):
Note that r1 , r2 , r3 have to be distinct positive integers.
Now w3  w6  1, w2  w5 , w  w4
w3  w  w2  0 

w3  w  w5  0 
Note that 3
 Each case will give arise to 3!  6 cases.
w  w4  w2  0 
w3  w4  w5  0 
w6  w  w 2  0 

w6  w  w5  0 

w6  w 4  w 2  0 
w6  w4  w5  0 
56.
(C) :
Thus required probability 
8  3!
2

666 9
Probability of station B receiving the green signal
= P( green signal is correctly received at A and correctly transmitted to B )
 P ( green signal is incorrectly received at A and incorrectly transmitted to B )
 P ( red signal is correctly received at A and wrongly transmitted to B)
 P ( red signal is incorrectly received at A and is wrongly transmitted to B)
=

4 3 3 4 1 1
    
5 4 4 5 4 4
1 3 1 1 1 3
     
5 4 4 5 4 4
required probability by Bayes theorem
4 3 3 4 1 1
    
20
5 5 4 5 4 4
=
=
4 3 3 4 1 1 1 3 1 1 1 1 3
23
           
5 4 4 5 4 4 5 4 4 4 5 4 4
87
 (c) is correct .
PROBABILITY
57.
SOL:- (B)

Following the condition of the problem. Required probability
1 3
2 1 1  3
1 1 6 2

1




1

  
2  5
5 2  2 10
10 3 10 3 

23
30
 (B) is correct.
58.
SOL:- (D)
4
12
Required conditional probability  10 
23 23
30
59.
SOL:- (A,D)
P ( exactly one of events E or F occurs)  P( E )  P( F )  2 P( E  F )
P( E )  P( F )  2 P( E  F ) 
Also
 (D) is correct.
11
25
1   P( E )  P( F )  P( E  F )  
2
(given)
25
Since E and F are independent P( E  F )  P( E ).P( F ) solving for P ( E ) and P ( F ) we get
60.

3
4
4
3
P ( E )  , P( F )  or P ( E )  , P( F ) 
5
5
5
5

Choices ( A) and ( D ) are correct.
1 1  1
P( X )     1  
4 4  4
 1 1 1
 1    
 4 4 4

(E3 not working ) 
1 1 1
 
2 4 4

(E1 not working )
1  1 1
(E2 not working )
 1   
2  4 4
( All are working )
3 3
1
1
8
  

32 32 32 32
32
Now P( exactly two working / X )

P ( exactly two working )
P( X )
3
3
1


P( X  X 1 )
P( X / X 1 ) 
 32 32 32
1
P( X 1 )
2

7
16
Thus (b), (d) are correct . We can show ( C) is wrong .
3
1
1


P( X )
5
Since P ( X  X 2 ) 
 32 32 32  .
1
P( X 2 )
8
4
88

7 / 32
7 / 8
8 / 32
PROBABILITY
61
If D4 shows 1 then there should be at least one 1
On D1 D2 , D3 ie 63  53  91 cases . Similarly D4
 Required probability 
62.
P( X / Y ) 
P( X  Y )
P(Y )
Similarly P( X ) 

can show 2 and so on
6  91 91
.

64
216
1 1/ 6

2 P(Y )
 P(Y ) 
1
3
1
. Now P( X  Y )  P( X )  P(Y )  P( X  Y )
2

1 1 1 2
  
2 3 6 3
Also P( X  Y )  P( X ) .P(Y ) Thus (a) ,(b) are correct .
63.
(a)
Prob that the problem is solved correctly
= 1- prob that problem is not solved correctly by none
1 1 3 7
 1 . . .
2 4 4 8
64.
 1
21
256

of them
235
256
Let P  E1   x, P  E2   y, P  E3   z Since all the three events are independent
 α  x 1  y 1  z  , β  y 1  x 1  z  ,
γ  z 1  x 1  y  and p  1  x 1  y 1  z 
Given relations are
 α  2 β  p  αβ and  β  3γ  p  2 βγ On dividing by αβ and βγ respectively these relations
1 2 1
1 3 2
  ,
 
β α p
γ β p
Now putting the values of α, β and p in the first relation we get
1
2
1



y 1  x 1  z  x 1  y 1  z  1  x 1  y 1  z 
become
1  y 2 1  x 
1
2
1 2

 1 1  2  1   0  x  2 y
y
x
y
x
y x
Similarly from the second relation we get y  3z

Whence x  6 z and hence
P  E1 
P  E3 
6
89
PROBABILITY
65.
P (Out of 2 balls drawn from randomly selected box; 1 is red, 1 is white)
1  1 C  3C1 2 C1  3C1 3 C1  4C1 
P( E )   16
 9
 12

3
C2
C2
C2 
1 2 C1  3C1
 9
3
C2
55
Required conditional probability  1

 (d) is correct.
3
2
3
3
4
C1  C1
C1  C1 181
1 C1  C1
 9
 12
3 6 C2
C2
C2
66.
(a) Required Probability  P (all white) + P (all red) + P (all black)
1 2 3 3 3 4 2 4 5
82
         
 (a) is correct
6 9 12 6 9 12 6 9 12 648
67.
(a) In any favorable or non favorable case there may be some boys between first girl ,
some boys
between first and second girl and some boys after second girl ie it could be
      G      G         .
We must have x1  x2  x3  3,
0
0
3
2
2
1
1
0
0
1
0
3
0
0
1
0
2
1
2
1
0  xi  3 which has 10 solutions.
3
0
0
1
0
2
0
2
1
1
The favorable solutions will be 0,0,3 ; 1,0 ,2; 1 ,2,0 ; 0, 1 ,2;

favorable arrangements  5  3!  2!  60 
68.
90
0,2 ,1
Required probability
60
120

1
.
2
PROBABILITY
x1  x2  x3 will be odd if either
2
C1  3C1  4C1  24 ways)
(i)
all three are odd (
(ii)
any two are even and remaining third number is odd
x1 , x2 even x3 odd
1 2  4  8
x2 , x3
even x1 odd
2  3  2  12
x1 , x3
even x2 odd
1 3  3  9
Total favorable ways  24  8  12  9  53
 3 5  7
Total ways
69 .
 105 .Required probability

53
105
If x1 , x2 , x3 are in AP then the AP will have common differences 0,1, 2,3, 4
If
d  0 then 3 AP’s are possible  (1,1,1) ,(2, 2, 2),(3,3,3) 
If
d 1 then also 3 AP’s are possible 1, 2,3) (2,3, 4)(3, 4,5 
If
d  2: AP,s are (1,3,5) , (2 ,4 , 6) , (3,5,7)
If
d  3:
AP,s are (1,4,7) , (3 ,2 , 1) , ( decreasing AP)
Thus total number of favorable case 11  (c) is the correct answer .
70.
P
(red ) 
n3
n1
1
1

 
2 n1  n2 2 n3  n4
 P ( conditional probability of second bag )
n3
1

2 n3  n4
n3 (n1  n2 )


n3
n1
1
1
n1 (n3  n4 )  n3 (n1  n2 )
.

2 n1  n2 2 n3  n4
This is equal to 1 / 3 for values of n1 , n2 , n3 , n4 in choices (A) and (B)
71.
The probability of drawing a red ball after 1 ball has been transferred to Box II
n1
n1  1
n2
n1
n1
n

.



 n1  1  n2   1
n1  n2 n1  n2  1 n1  n2 n1  n2  1
(n1  n2 )(n1  n2  1)
n1  n2
Which is 1 / 3 for values of n1 and n2 given in choices ( C) and (D)
91
PROBABILITY
AIEEE /IIT MAINS
1.
4 3 3
P ( on both occasion Mr .A selected loosing horse )    .
5 4 5
3 2
 1 
5 5
Required probability
2.
4  3 3 4 7
Required probability   1    1    .
5  4  4  5  20
3.
A = {X is prime number } = { 2 , 3 , 5 , 7 }
P( A)  P( X  2)  P( X  3)  P( X  5)  P( X  7)  0.23  0.12  0.20  0.07  0.62
B {X  4}  {1,2,3}

P( B)  P( X  1)  P( X  2)  P( X  3)  0.15  0.23  0.12  0.5
A  B { X is prime number as well as < 4 }
= {2,3}
P( A  B)  P( X  2)  P( X  3)  0.23  0.12  0.35
Required probability
4.
np  4, npq  2

P( A  B)  P( A)  P( B)  P( A  B)  0.62  0.5  0.35  .077
q
1
1
whence p 
2
2
2

5.

28
256

( a ) is correct.
(
Favourable cases = 3
( when all 3 apply for the same house )
Required probability

each person can apply in 3 ways )
3 1

33 9
P( X  1.5)  1  P( X  1.5) 1  P( X  1)  1  [ P(0)  P(1)] 1 
But m  2 is given
7.
6
p  q  1 and then n  8
Total cases = 33

6.
1 1
P(2 successes )  8C2    
2 2

 P( X  1.5)
5
1
3
P( A  B)  , P( A  B)  , P( A) 
6
4
4
e m m0 e  m m1

1  e  m (1  m)
0!
1!
 1  3/ e2 .

5 3
1
  P( B) 
6 4
4

P( B) 
1
3
 P( A  B)  P ( A) P( B) is satisfied  A and B are independent but are not equally likely .
92
PROBABILITY
 .51 
6
Required probability = P(0)  P(1)  e5 1    5
 1!  e
8.
m=5 
10.
Required probability  1  .3  .2
11.
P  A  B   P  A  P  B   P  A  B 
12.
1
1
2
1 2 1
P  A   , P  A B   , P  B A    P  A  B   P  A .P  B A  . 
4
2
3
4 3 6
 .7  .2
Again P  A  B   P  B  .P  A B  
13.
3 4 1
   1  ( a ) is correct.
6 6 6
1
1
 P  B.
6
2
n
9
1
3
3
Now 1    
  
 4  10
 4  10


 1  P (No success)
P (at least one success)
n
 .14
n  log 4  log3  1  n 
 n log
 P  B 
3
 1  
4
3
 1 
4
1
 (d ) is correct
3
n
n  log3  log 4  1
1
 ( a ) is correct.
log 4  log 3
14.
The truncated sample space is 00, 01,..09,10, 20,30, 40 i.e. truncated sample space
contains 14 cases out of which only 08 is favorable  ( a ) is correct
15.
(b):
Number of ways to select exactly one ball  3C1  4C1  2C1
Number of ways to select 3 balls out of 9 is 9 C3 . Required probability
16.
(c):
d  1, x  17 ; d  2, x  14 ; d  3, x  11
d  4, x  8 ; d  5, x  5 ; d  6, x  2
Clearly statement is wrong . Hence (c) is the correct answer.
17.
P (at least one failure)  1  p (all success)
5
31
1
1
 p5     p  
Now 1  p 
32
2
2
 1
p  0, 
 2
C  D  C  D  C , P C   P  D 
(*)
5
18.
 1  p5
93
3 4  2 2

 9 C3  7
PROBABILITY
Now ( a ) is not correct since P  C D   P  C  if C is independent of D
(d ) is not correct since in general P  C D  
P C  D 
P  D
Now P  C D   P  C  D  is always true  P  C D   P  C 
 (b) is correct , ( c) is false.
20.
Let A be the event that minimum is 3 , B be the event that maximum is 6
 P  A B 
P  A  B  P  Besides 3,6 either 4 or 5 is there 

P( B)
P( B)
2
C
1
 5 3   (d ) is correct
C3 5
8
C3
8
21.
Required probability = P (4 correct answers)  P (5 correct answers)
(c)
4
5
22.
5
1 2
 1  11
C4 .   .  5C5    5
 3 3
 3 3
1
1
 1  P( A  B )  
6
16
(c)
P( A  B ) 
P( A) 
1
3
 P( A) 
4
4
P( A  B) 
5
6
1
… (1), P( A  B)  … (2)
4
… (3)
Now P( A  B)  P( A)  P( B)  P( A  B)

5 3
1
5 1
1
  P ( B )   P( B)    P ( B ) 
6 4
4
3
6 2
A of B are Independent because P( A  B)  P( A) P( B). A and B have different probabilities
There fore they are not equally likely.
94