58.

A missile target may be at a point P with probability or at a point Q with probability we

have 20 shells each of which can be fired either at point P or Q. Each shell may hit the target

independently of the other shoot with probability . Then number of shells must be fired at
point P to hit any target with maximum probability is
Ans. 11
Sol. Let A be the event that target is hit E1, E2 be the events to select target P and Q respectively

Then P(E1) =

where x number of shells used to select target P

Then

32–x = 3x–20, x = 11

59. A coin is tossed (m + n) times (m > n). Show that the probability of atleast m consecutive heads

is
Sol. HH .... m heads  Let it is one unit say X
 X.... rest n places
On rest n places either H or T may come for which probability for each place is 1

 Probability for above case (n times)

…(1)
 Now further to avoid repetition TX comes together and rest are n – 1 places

 Probability = nC1 . …(2)

From equation (1) and (2), we get is the required probability

134. If a, b, c are determined by throw of a single dice thrice, given a 2 + b2 + c2 ≤ ab + bc + ca. the
probability that a point (a, b, c) determined in 3 dimensional co-ordinate system lies inside the
tetrahedron formed by co-ordinate planes and x + y + z = 12 is

(A) (B) (C) (D)

Sol. (B)
 For real numbers a, b, c (a – b) 2 + (b – c)2 + (c – a)2 ≥ 0 ⇒ a2 + b2 + c2 ≥ ab + bc + ca and
equality holds iff a = b = c. Therefore Sample Space (a, b, c) ∈ {(1, 1, 1), …….(6, 6, 6)} of
which 3 cases satisfy a + b + c < 12 to lie inside the specified tetrahedron. Required
probability 1/2

135. Let v and w be distinct, randomly chosen roots (real or complex) of the equation z 9 – 1 = 0.
The probability that 1 ≤ |v + w|

(A) (B) (C) (D)

Sol. (B)
Consider unit circle |z| = 1 with a regular polygon of 9 vertices inscribed in it with one of the

vertices as 1. 'v ' and 'w' are any two distinct vertices such that .

This implies that . Thus, the total number of pairs v and w

can be 27 out of
138. Four dice are thrown simultaneously, given that 4 and 2 has appeared on any two of them, the
probability that 3 has appeared on the remaining dice is

(A) (B) (C) (D)

Sol. (D)
The Sample Space reduces to . Favourable outcomes are all possible

orientations of 4, 2, 3, 3 i.e. . Required probability is

139. Consider a function f : {1, 2, 3, ....... 13} → {1, 2, 3, ....... 9}. Given that function is surjective
and nondecreasing, the probability that f(7) = 4 is

(A) (B) (C) (D)

Sol. (D)

If the function is non-decreasing then, , all the values of set

co-domain are to be images of some numbers of domain. This implies that exactly 4 equalities

should hold in the inequality , which can happen in 12C4

ways. However, if f(7) = 4 exactly 3 equalities should hold before f(7) and exactly one

equality should hold after f(7). This can occur in ways.

Therefore required probability is .

140. Two numbers a and b are chosen at random from the set of first 30 natural numbers.
Probability that a2 – b2 is divisible by 3 is
 (A) (B) (C) (D)
Sol. (B)
Numbers are of 3n, 3n + 1, 3n + 2 form, 10 of each
Either a and b both are of 3n form → 10C2 or any two are out of 3n + 1, 3n + 2 form → 20C2

is the required probability

158. Let A = set of all 3 × 3 determinant having entries + 1 or –1, if a determinant A from the set A
is chosen randomly, then the probability that the product of the elements of any row column
of A is –1 is
(A) 1/32 (B) 1/8 (C) 1/16 (D) none of these
Sol. (A)

202. Consider f(x) = x3 + ax2 + bx + c. parameters a, b, c are chosen respectively by throwing a

dice three times. Then probability that f(x) is an increasing functions is

(A) (B) (C) (D)

Sol. (C)
f’(x) = 3x2 + 2ax + b
y = f(x) is increasing  x  R
f’(x)  0  x  R
So (2a)2 – 4 × 3 × b  0
a2 – 3b  0
This is true for exactly 16 ordered pairs (a, b) I  a, b  6

So required probability

203. Let L1 : 2x + 3y + p – 3 = 0 and L2 : 2x + 3y + p + 3 = 0 be two lines and p i z. Let C : x2 +

y2 + 6x + 10y + 30 = 0. If it is given that at least one of the lines L 1 L2 is chord of C the
probability that both are chords of C is

(A) (B) (C) (D)

Sol. (B)
Center (–3, –5)
Radius  2
For L1 to be a chord its distance from center must be less than radius
Since P is an integer p can take value
S1 = {17, 18, 31}
L2 to be chord of a circle, P can take value from the set
S2 = {11, 12, … 25}
S1  S2 has 21 elements while
S1  S2 has 9 elements
 Required probability or

204. There were 3 playerS2 = {11, 12, … 25}s from town B in a cricket camp. Every player in the
camp bowls one over to every other player. The number of overs that players from town A
bowled among themselves exceeded by 98 to the number of overs bowled when player from
town A and B are involved. The number of players from town A is
(A) 9 (B) 16 (C) 14 (D) 11
Sol. (C)

205. If a is an integer lying in [–5, 30], then the probability that graph of y = x 2 + 2(a + 4)x – 5a +
64 is strictly above the x-axis is
(A) 1/6 (B) 7/36 (C) 2/9 (D) 3/5
Sol. (D)
D<0
 (a + 16)(a – 3) < 0
 a {–15, –14, …2}
In given internal
 a  {–5, –4, –3, –1, 0, 1, 2}

Probability
277. If n different objects are distributed among n+2 persons, then (n+1)!
(n  1)!
n 1
(A) Probability that exactly 2 persons will get nothing is 2.(n  2)
n2
 
C3 (n  1) n C2 (n  2)!

(B) Probability that exactly 3 persons will not get anything is (n  2)n
n  (n  1) 2  (n  1)
(C) Probability that exactly 3 persons will not get anything is 12(n  2) n 1
n 2  (n  1) 2
n 1
(D) Probability that exactly 2 persons will get nothing is ( n  2)
Sol. (AB)
Probability that exactly 2 persons will get nothing
n! (n  2)(n  1)n! ( n  1)!
 n  2 C2  
(n  2) n 2  (n  2) n 2( n  2)n 1
Probability that exactly 3 persons will get nothing
n 1
n2 C1 n C2  (n  2)!
 C3 
(n  2) n

278. If p and q are chosen randomly from the set {1, 2,3, 4,5,6,7,8,9,10} with replacement, then

the probability that the roots of the equation

x 2  px  q  0
 33 19
(A) are real is 50 (B) are imaginary is 50
3 3
(C) are real and equal is 100 (D) are real and distinct is 5
Sol. (BC)

For real roots

p 2  4q Possible selections are as follows
62 31
 Probability for rcal roots  
100 50
 For equal and real roots p 2  4q
  p, q    2,1 ,  4,4  ,  6,9 
3
Probability 
100
3 19
 Probability for imaginary roots  1  
50 50
62 3 59
 Probability for real and distinct roots   
100 100 100 ,
p q

1 -

2 1

3 1,2

4 1, 2,3, 4

5 1,2,6

6 19

7 110

8 1.10

9 1.10

10 1.10

Total  62

279. A bag contains four tickets marked with numbers 112, 121, 211, 222. One ticket is drawn at
random from the bag. Let Ei (i =1, 2, 3) denote the event that ith digit on the ticket is 2. Then
(A) E1 and E2 are independent (B) E2 and E3 are independent
(C) E3 and E1 are independent (D) E1, E2 and E3 are independent
Sol. (ABC)
1
We have P  Ei   for i  1,2,3
2
 
 For i  j , P Ei  E j 
1

4
 
 P  Ei  P E j

 Also, P  E1  E2  E3 
1
  P  E1  P  E2  P  E3 
4
 E1 , E2 , E3 are not independent

280. From an urn containing x white and y  x black balls, n balls have been lost. then
probability of dra a white ball
x x
(A) before the loss is y (B) after the loss is y for n  1
x x
(C) after the loss is y for n  2 (D) afier the loss is y for n  3
Sol. (ABCD)
Let w be the event that the ball drawn is white
x
P ( w) 
Then y
x x 1 y  x x x
P ( w)   
When, ball is lost y y 1 y y 1 y
x
C2 x  2 x C1 y  x C1 x  1 y  x C2 x x
P( w)   y  y 
When 2 balls is lost
y
C2 y  2 C2 y  2 C2 y  2 y
x
n  3, P ( w) 
Similarly for y

xn
x n 1  (2n  1)  (2n  3)x n 1  n  0
344. If | A |{ f (A)} is the probability that the equation n!

has exactly one real root then

f (A) n  x n  2
has
(A) two real roots and two imaginary roots
(B) two real roots and one imaginary roots
(C) one real root and two imaginary roots
(D) one real root and three imaginary roots
Sol: (C)
343.- 344

A  adjA   A I n
A•  f  A  A1
A•  f ( A) n A1  f  A   f ( A) n A1
 1
But A1 
A
1
 f  A   f ( A) n 
A
 A  f ( A) n 1
 n 1
adjA | A |n 1   f ( A) n 1 
2
 f ( A) ( n 1)
…(1)
2
adjadjA | A |( n 1)
…(2)
A  A  f  A
T
But from definition
 A  f ( A) n 1  A•  f  A 
 n 1  1
n2
From (1) and (2)
A  adjA
5x2
 f  x   x3   6x  2
2
f   x   3x 2  5 x  6
D0
 f   x   0x  R
Graph for f  x 
 f  x  has only one real root
 p 1
 A 1
But given A  f  A 
But we can take A  f ( A) n 1
 f ( A) n  x n  2 , f ( A) n 1  x n 1
 A  x n 1
 x n 1  1
Since n  2
 x3  1
One real and two imaginary roots
399. Die A has 2 white and 4 red faces whereas die B has 4 white and 2 red faces. A coin is flipped
once. If it shows a head, the game continues by throwing die A, if it shows tail, then die B is
32
to be used. If the probability that die A is used is 33 when it is given that red turns up every
time in first n throws. Then find the value of n_____
Sol. [5]
 n n
R  2  R  1
P     , P     ,
A 3  B   3  in n throws
n n
12 11
P(R)      
2 3 2  3  in n throws
n
12
2n 32  
R 2 3
P   P(A)  n n
 n

A A 12 11 2  1 33
P          
R P(R) 2 3 23
⇒n=5

400. Two dice are rolled and at the same time two coins are tossed. Given that one of the dice had
shown a six and one of the coins has shown a heads, the probability that others will also show
p
the same combination is 33 , then p is equal to_____
Sol. [1]
3 × 11 = 33
3 for HH, HT and TH and 11 for the cases when atleast one 6 appears
1
So, required probability will be 33 (as there is only one way in which the required condition
is satisfied)

401. f is any function from A to A where A = 1, 2, …..10}. The discrete points on the graph of f(x)
are joined with line segments the probability that f is onto and the resulting graph has only
2k  2
10
one local minima and no local maxima is 10 then k is______
Sol. [9]
9 9 9 9
Total number of function satisfying the given condition is C1  C 2   C8  2  2
29  2

⇒ Probability 1010

402. If z1, z2, z3 are unimodular complex numbers then the greatest value of |z 1 – z2|2 + |z2 – z3|2 + |
z3 – z1|2 equals to_____
Sol. [9]
2 2 2
z1  z 2  z 2  z 3  z 3  z1

 2
 2 z1  z 2  z3
2 2
  z z  z z
1 2 1 2  z 2 z3  z2 z 3  z 3 z1  z3z1 
 6   z1 z2  z1z 2  z 2 z3  z2 z 3  z 3 z1  z3z1 
...(1)
2
z1  z 2  z3  0
Now
2 2 2
 z1  z 2  z 3  z1 z2  z1z 2  z 2 z3  z2 z 3  z3 z1  z3 z1  0
   z1 z2  z1z 2  z 2 z3  z2 z 3  z 3 z1  z3z1   3
...(2)
From (1) and (2)
 Maximum value = 6 + 3 = 9
444. A man has 3 coins A, B and C, A is unbiased, the probability that a head will result when B is
2 1
tossed is 3 , the probability that a head will result when C is tossed is 3 . One of the coin
chosen at random and is tossed 3 times, giving a total of two heads and one tail.
Column I Column II
(A) The probability that the chosen coin was A (p) 32
75
(B) The probability that the chosen coin was B (q) 1
2
(C) The probability that the forth toss will result in head (r) 209
1296
(D) The probability that the forth toss will result in tail (s) 9
25
(t) 215
1296
Sol. (A)  (s); (B)  (p); (C)  (q); (D)  (q)
Let E be the event of getting 2 head and 1 tail
1
P(A)  P(B)  P(C) 
3
2 2
1 3  2 1 4
P(E / A)  3C 2    P(E / B)  3C2     
 3 8 ,  3 3 9
2
3 1  2 2
P(E / C)  C 2     
 3  3 9
9 32
P(A / E)  , P(B / E) 
So by Baye's theorem 25 75