PROBABILITY 1

EXERCISE - 3 : ADVANCED OBJECTIVE QUESTIONS

Objective Questions I [Only one correct option]

2 5
(a) (b)
1. A box has four dice in it. Three of them are fair dice but the 3 8
fourth one has the number five on all of its faces. A die is
3 4
chosen at random from the box and is rolled three times (c) (d)
4 5
and shows up the face five on all the three occassions.
The chance that the die chosen was a rigged die, is Ans. (d)

216 215 1 1 1 1
(a) (b) Sol. P (Head)      1 1
217 219 2 2 2 2

216
(c) (d) none 1
219 4
P (D.H.)  2 
Ans. (c) 1 1 5

2 8
3
3 1 1
P 5    1 4. A box contains 5 red and 4 white marbles. Two marbles are
4  6  4
Sol.
drawn successively from the box without replacement and
the second drawn marble drawn is found to be white.
1 Probability that the first marble is also white is
4 1 72
P  Required     3 1
3 1 1 1
   1 73 (a) (b)
4 63 4 72 8 2

2. On a Saturday night 20% of all drivers in U.S.A. are under 1 1

(c) (d)
the influence of alcohol. The probability that a driver under 3 4
the influence of alcohol will have an accident is 0.001. The Ans. (a)
probability that a sober driver will have an accident is
0.0001. If a car on a saturday night smashed into a tree, the 4 3
probability that the driver was under the influence of 
9 8 3
alcohol, is Sol. P 
4 3 5 4 8
  
(a) 3/7 (b) 4/7 9 8 9 8
(c) 5/7 (d) 6/7
5. Events A and C are independent. If the probabilities relating
Ans. (c)
A, B and C are P (A) = 1/5;
20 80 P (B) = 1/6 ; P (A  C) = 1/20 ; P (B  C) = 3/8 then
Sol. P (Car Accident)   0.001   0.0001
100 100 (a) events B and C are independent
(b) events B and C are mutually exclusive
20  0.001
P (Required) = (c) events B and C are neither independent nor mutually
20  0.001  80  0.0001
exclusive
20 5 (d) events B and C are equiprobable
 
28 7 Ans. (a)
3. A box contains a normal coin and a double headed coin. A
coin selected at random and tossed twice, fell headwise on
both the occasions. The probability that the drawn coin is
a double headed coin is
 2 PROBABILITY

Ans. (b)
1/ 20 1
Sol. P C    Sol. Since, three dice are rolled.
1/ 5 4
Total number of cases in S = 6 × 6 × 6 = 216
1 1 3
and   P  B  C   and the same number appear on each of them = 6C1 =6
6 4 8
6 1
Required probability  
1 1 3 469 1 216 36
PB C     
6 4 8 24 24
9. It is given that the event A and B are such that
= P(B) . P(C)
1 A 1
P(A)  , P    and P    . Then P (B) is
B 2
6. Assume that the birth of a boy or girl to a couple to be 4 B 2 A 3
equally likely, mutually exclusive, exhaustive and
independent of the other children in the family. For a couple 1 1
having 6 children, the probability that their "three oldest (a) (b)
2 6
are boys" is
1 2
20 1 (c) (d)
(a) (b) 3 3
64 64
Ans. (c)
2 8
(c) (d) Sol. From the definition of independence of events
64 64
Ans. (d) P  A  B
P  A / B 
Sol. For this to happen first child, second child and third child P  B
should be boys
Then P(B) . P(A/B) = P(AB) ....(1)
1 1 1 1 8 Interchanging the role of A and B in (1)
P     (D)
2 2 2 8 64 P(A) P(B/A) = P(BA) ..(2)
7. A and B are two events such that P(A) = 0.2 and As AB = BA, we have from (1) and (2)

P  A  B  = 0.7. If A and B are independent events then P(A) P(B/A) = P(B)P(A/B)

P(B) equals 1 2 1 1 2 1
 .  P  B  .  P  B   . .2  .
(a) 2/7 (b) 7/9 4 3 2 2 3 3
(c) 5/8 (d) none of these
10. If A and B are two independent events such that
Ans. (c)
P (A) > 0, and P (B) 1, then P (A / B) is equal to
Sol. P (AB) = P(A) + P(B) – P(A) P(B)
0.2 + P(B) - 0.2 P (B) = 0.7 (a) 1  P  A / B  (b) 1  P (A / B)
0.8 P (B) = 0.5
1  P (A  B) P (A)
5 (c) (d)
P B  P (B) P (B)
8
Ans. (b)
8. Three identical dice are rolled. The probability that the
same number will appear on each of them, is Sol.  
Since, P A / B  P  A / B   1
1 1
(a)
6
(b)
36  
 P A / B  1  P A / B . 
1 3
(c) (d)
18 28
 PROBABILITY 3

11. Two cards are drawn from a well shuffled pack of 52 playing
25 1 75 2
cards one by one. If P  3    
100 6 100 6
A : the event that the second card drawn is an ace and
B : the event that the first card drawn is an ace card. 25 1

then which of the following is true?  Fair die  100 6
P  
 3   1 25   75 2 
4 1  6  100    100  6 
(a) P (A) = ; P (B) =    
17 13

25 1
1 1  
(b) P (A) = ; P (B) = 175 7
13 13
13. An instrument consists of two units. Each unit must
1 1 function for the instrument to operate. The reliability of
(c) P (A) = ; P (B) =
13 17 the first unit is 0.9 & that of the second unit is 0.8. The
instrument is tested & fails. The probability that "only the
16 4 first unit failed & the second unit is sound" is :
(d) P (A) = ; P (B) =
221 51 (a) 1/7 (b) 2/7
Ans. (b) (c) 3/7 (d) 4/7
Ans. (b)
4
C1
1
Sol. P  B   52  Sol. P(Fails) = 0.9 × 0.2 + 0.1 × 0.8 + 0.1 × 0.2
C1 13
0.1  0.8
Req-Probability 
4 3 48 4 0.9  0.2  0.1 0.8  0.1 0.2
C1 C1 C1 C1
P  A  52
 51
 52
 51
C1 C1 C1 C1 8 8 2
  
18  8  2 28 7
1
 14. Mr. Dupont is a professional wine taster. When given a
13
French wine, he will identify it with probability 0.9 correctly
12. A purse contains 2 six sided dice. One is a normal fair die, as French, and will mistake it for a Californian wine with
while the other has 2 ones, 2 threes, and 2 fives. A die is probability 0.1. When given a Californian wine, he will
picked up and rolled. Because of some secret magnetic identify it with probability 0.8 correctly as Californian, and
attraction of the unfair die, there is 75% chance of picking will mistake it for a French wine with probability 0.2. Suppose
the unfair die and a 25% chance of picking a fair die. The that Mr. Dupont is given ten unlabelled glasses of wine,
die is rolled and shows up the face 3. The probability that three with French and seven with Californian wines. He
a fair die was picked up, is randomly picks a glass, tries the wine, and solemnly says :
"French". The probability that the wine he tasted was
1 1
(a) (b) Californian, is nearly equal to
7 4
(a) 0.14 (b) 0.24
1 1 (c) 0.34 (d) 0.44
(c) (d)
6 24 Ans. (c)
Ans. (a)
7 3
Sol. P (French)   0.2   0.9
10 10

Sol. 7  0.2
P (Required) 
7  0.2  3  0.9
 4 PROBABILITY

1.4 14 3 2 2 1
  P (D) =   
4.1 41 5 5 5 5
15. Box A contains 3 red and 2 blue marbles while box B
6 2 8
contains 2 red and 8 blue marbles. A fair coin is tossed. If P(D) =  
25 25 25
the coin turns up heads, a marble is drawn from A, if it
turns up tails, a marble is drawn from bag B. The probability 17. A bowl has 6 red marbles and 3 green marbles. The
that a red marble is chosen, is probability that a blind folded person will draw a red marble
on the second draw from the bowl without replacing the
1 2 marble from the first draw, is
(a) (b)
5 5
2 1
3 1 (a) (b)
(c) (d) 3 4
5 2
1 5
Ans. (b) (c) (d)
2 8
Sol. Event A {red marbles is chosen}
Ans. (a)
B1 = {Box 1 is chosen}
B2 = {Box 2 is chosen} 6 5 3 6
Sol. P  Red     
By formula for total probability 9 8 9 8
P(A) = P(A/B1) P(B1) + P(A/B1) P(B2)
6 2
 
1 3 9 3
P(B1) = P(B2) = , P(A/B1) =
2 5 18. Two aeroplanes I and II bomb a target in succession.
The probabilities of I and II scoring a hit correctly are 0.3
2 1 and 0.2, respectively. The second plane will bomb if the
P(A/B2) =  first misses the target. The probability that the target is
10 5
hit by the second plane, is
1 3 1 1 1 2 (a) 0.2 (b) 0.7
So, P(A) =     
2 5 2 5 10 5 (c) 0.06 (d) 0.14
Ans. (d)
16. Lot A consists of 3G and 2D articles. Lot B consists of 4G
and 1D article. A new lot C is formed by taking 3 articles
from A and 2 from B. The probability that an article chosen
Sol. 
P(I) = 0.3, P I = 1 – 0.3 = 0.7
at random from C is defective, is
 
P(II) = 0.2, P II = 1 – 0.2 = 0.8
1 2
(a) (b)
3 5
  
Required probability = I  II  P I P  II 
8
(c) (d) none = (0.7)(0.2) = 0.14
25
19. Let A and E by any two events with positive
Ans. (c) probabilities:
Sol. P (D /A) denotes that a defective articles is from A Statement 1 : P(E/A) P(A/E) P(E)

2 1 3 2 Statement 2 : P(A/E)  P(AE)

P (D /A) = P (D /A) = , P (A) = , P(B) = (a) Both the statements are false
5 5 5 5
P (D) = P (A) P (D /A) + P (B) P (D /A) (b) Both statements are True
(c) Statement–1 is true, Statement–2 is false
(d) Statement–1 is false, Statement–2 is true
Ans. (b)
 PROBABILITY 5

 E  P  E  A P  A  B = p 
Sol. P   ..(1)  Correct
 A P  A P  A  p  B  =1.p 

 A  P A  E (D) Sample space

P   ..(2)
E PE {1.2.3.4.5 , 1.2.3.4.6 , 1.2.3.5.6, 1.2.4.5.6, 1.3.4.5.6, 2.3.4.5.6}
Every no. is divisible by 6
Dividing eq. (1) by eq (2), we get
21. A examination consists of 8 questions in each of which
P  E / A PE one of the 5 alternatives is the correct one. On the
 assumption that a candidate who has done no preparatory
P A / E P  A
work chooses for each question any one of the five
Now, since 0 < P(A) 1 alternatives with equal probability, the probability that he
gets more than one correct answer is equal to :
There fore.
(a) (0.8)8 (b) 3 (0.8)8
P  E / A (c) 1  (0.8)8 (d) 1  3 (0.8)8
 P(E )
P A/ E Ans. (d)
Sol. 1 - P (X = 0) - P (X = 1)
P(E/A) P(A/E). P(E)
Statement (1) is true 0 8 7
1 4 14
 1  8C0      8C1    
Now, since again 0 < P (E) 1, 5 5
    55
thus from (2), P(A/E)  P(AE) Statement (2) is true.
= 1 - (0.8)8 - 2 (0.8)8
20. Indicate the correct order sequence in respect of the
following : = 1 - 3 (0.8)8

I. If the probability that a computer will fail during the first 22. A number is chosen at random from the numbers 10 to 99.
hour of operation is 0.01, then if we turn on 100 computers, By seeing the number a man will laugh if product of the
exactly one will fail in the first hour of operation. digits is 12. If he choose three numbers with replacement
then the probability that he will laugh at least once is
II. A man has ten keys only one of which fits the lock. He
tries them in a door one by one discarding the one he has 3 3
3  43 
tried. The probability that fifth key fits the lock is 1/10. (a) 1    (b)  
5  45 
III. Given the events A and B in a sample space. If
P(A) = 1, then A and B are independent. 3 3
 4   43 
IV. When a fair six sided die is tossed on a table top, the (c) 1    (d) 1   
bottom face can not be seen. The probability that the  25   45 
product of the numbers on the five faces that can be seen
Ans. (d)
is divisible by 6 is one.
Sol. Number marked from 10 to 99 = 90 numbers
(a) FTFT (b) FTTT
(c) TFTF (d) TFFF These are two digit numbers. A product of obtained when
(2, 6), (6, 2), (3, 4), (4, 3) are obtained
Ans. (b)
Sol. A) No, its only a probablity. May be all computer get fail or 4 2
In one probability of success = 
no computer get fail. 90 45
Probabitliy of at least one laughter (Success) in there trials
 9 C4  4! 1 C1  1 = 1 – probability of failure in the three trials
B) 4 failure & 1 success = 10
C5  5! 10
3
 43 
C) Let P(B) = p (obviously B will be subset of A) =1–  
 45 
 6 PROBABILITY

23. A fair die is tossed eight times. Probability that on the eighth Ans. (a)
throw a third six is observed is,
1 3p 1 p 1 2 p
5 7 5 Sol. 0   1
8 5 C 2 .5 3 4 2
(a) C3 (b)
68 68 0  4 + 12p + 3 - 3p + 6 - 12p  12
7
C 2 .55 0  13 - 3p  12
(c) (d) none of these
67 13 1
 p
Ans. (b) 3 3
Sol. On the eight throw third six must appear, which implies
that on first seven throws there must be two six. 1– 2p
and 1  0
Probability of getting 6 is 1/6 and probability of not getting 2
6 is 5/6. 2p 1
We must get two six in seven throws, so the probability is
7
C2(1/6)2(5/6)5. 1
0 p
2
Now prob that the eight throw is six is 1/6. so the final
required probability is 7C2(1/6)2(5/6)51/6. 1 p
24. From an urn containing six balls, 3 white and 3 black ones, and 0  1
4
a person selects at random an even number of balls (all the
01-p4
different ways of drawing an even number of balls are
considered equally probable, irrespective of their number). -1  - p  3
Then the probability that there will be the same number of 1 p  - 3
black and white balls among them
1 3p
4 11 and 0  1
3
(a) (b)
5 15
0  1 3p  3
11 2
(c) (d)
30 5 2
0P
Ans. (b) 3

1 3C1  3C1 1 3 C 2  3C 2 1 6 C6 1 1 
Sol. P      6 So, p   , 
3 6
C2 3 6
C4 3 C6 3 2

26. The probabilities of events, A  B, A, B & A  B are

1 9 9  respectively in A.P. with probability of second term equal
    1
3  15 15  to the common difference. Therefore the events A and B
are
11 (a) compatible

15 (b) independent
(c) such that one of them must occur
(1  3p) (1  p) (1  2p) (d) such that one is twice as likely as the other
25. If , & are the probabilities of three
3 4 2
Ans. (d)
mutually exclusive events defined on a sample space S,
Sol. P(A B), P(A), P(B), P(A B) A.P
then the true set of all values of p is
0, d, 2d, 3d
1 1  1  P(A) = d = P(B) - P(A)
(a)  ,  (b)  ,1
3 2  3 
P(B) = 2 P(A)
1 1 1 1
(c)  ,  (d)  , 
 4 3 4 2
 PROBABILITY 7

27. A number x is chosen at random from the set {1, 2, 3, 4..... 3 4 2

100}. Define the event: A = the chosen number x satisfies 1 1 1 2 1  2
       
3 3 3 3 3  3
 x  10  x  50 
 0. Then P(A) is:
 x  30  2 3 2
1 2 1 2
...            ...
(a) 0.71 (b) 0.70 3  3 3  3
(c) 0.51 (d) 0.20
Ans. (a) 1 2
9 27 1 2 5
    
 x  10  x  50  1
2
1
2 7 3  7 21
Sol. 0
 x  30  9 9

30. A and B play a game of tennis. The situation of the game is

as follows; if one scores two consecutive points after a
deuce he wins; if loss of a point is followed by win of a
Solution : x = {10, 11, .....29} {50, 51......... 100} point, it is deuce. The chance of a server to win a point is 2/
n(A) = 20 + 51 = 71 3. The game is at deuce and A is serving. Probability that A
Total number of elememt in the given set = 100 will win the match is, (serves are changed after each pt)
(a) 3/5 (b) 2/5
71
 P  A   0.71 . (c) 1/2 (d) 4/5
100
Ans. (c)
28. A die is tossed 5 times. Getting an odd number is Sol. P(A win)
considered a success. Then the variance of distribution
of success is    
 P  A  .P  A   P A .P B .P  A  P  A   ....
(a) 8/3 (b) 3/8
(c) 4/5 (d) 5/4 2 2 1 1 2 2
 .  . . .  ...
Ans. (d) 3 3 3 3 3 3

3 3 4
Sol. n  5, p  , q 
6 6 9 4 1
  
1 8 2
Remember Mean = np & Variance = npq 1
9
33 5
 Variance  5  31. A child throws 2 fair dice. If the numbers showing are
66 4
unequal, he adds them together to get his final score. On
29. A fair die is tossed repeatedly. A wins if two consecutive the other hand, if the numbers showing are equal, he throws
outcomes {1, 2} & B wins if two consecutive outcomes 2 more dice & adds all 4 numbers showing to get his final
{3,4,5,6}. Find probability that A wins if the die is tossed score. The probability that his final score is 6 is:
repeatedly.
145 146
(a) (b)
1 5 1296 1296
(a) (b)
3 21
147 148
(c) (d)
1 2 1296 1296
(c) (d)
4 5 Ans. (d)
Ans. (b) Sol. As we know that first 2 dice can land in 36 ways with equal
Sol. P(A Wins) = P(WW) + P (WLWW) + P(WLWLWW)+...+
P(LWW) + P (LWLWW)+ ...
 8 PROBABILITY

probability. we also know that in six cases, we have to

throw the dice 2 more times but the dice doesn't know that.
N
Cn n! N! n! N!
P2    n 
Therefore all 36 possibilities will be equally likely. Following N n
n! N – n ! N n ! N – n  !
are the possibilities after the first two throws:
4/36 we have a total of 6 without a double and we stop. 26/ (one gift coupon in each n envelope arbitary)
36 we don't have a total of 6 or a double and we stop.  (ii) & (iii)
6/36 we have a double and must continue. We need to look 33. A bag contains 3 R & 3 G balls and a person draws out 3 at
more carefully. random. He then drops 3 blue balls into the bag & again
1/36 we have a double 1, we throw 2 more dice and have a draws out 3 at random. The chance that the 3 later balls
3/36 chance of getting a total of 6. being all of different colours is

1/36 we have double 2, we throw 2 more time and have a 1/ (a) 15% (b) 20%
36 chance of getting 6. (c) 27% (d) 40%

4/36 we have a double 3 or more, we throw again but we Ans. (c)

can't get 6. Sol. There are 3 red and 3 green balls in a bag and a person
Thus overall a successful outcome has probability of draws 3 balls. The possibilities are:

4/36 + 1/36 × 3/36 + 1/36 × 1/36 = 148/1296. A 3 Red, 0 Green 0R + 3G + 3B

32. There are n different gift coupons, each of which can B 2 Red, 1 Green 1R + 2G + 3B
occupy N(N > n) different envelopes, with the same C1 Red, 2 Green 2R + 1G + 3B
probability 1/N D 0 Red, 3 Green 3R + 0G + 3B
P1: The probability that there will be one gift coupon in each The balls that will be in the bag.
of n definite envelopes out of N given envelopes
We need to find probability of 3 balls being difference
P2: The probability that there will be one gift coupon in each colors. So A,D cases are not possible.
of n arbitrary envelopes out of N given envelopes
3
Consider the following statements C2 . 3C 1 9
Probability of with drawing 2R, 1G = 
6 20
C3
n!
(i) P1 = P2 (ii) P1 = n
N Probability of withdrawing 1R, 1G, 1B from B case
N! 2
(iii) P2 = C1  1C1  3C1 6
N n (N-n)! 
6 20
C3

n! N! Similarly for the case C

(iv) P2 = (v) P1 =
n
N (N-n)! Nn
9 6 27
Now, which of the following is true So, chance = 2   
20 20 100
(a) Only (i) (b) (ii) and (iii)
34. One purse contains 6 copper coins and 1 silver coin ; a
(c) (ii) and (iv) (d) (iii) and (v)
second purse contains 4 copper coins. Five coins are drawn
Ans. (b) from the first purse and put into the second, and then 2
coins are drawn from the second and put into the first. The
n!
Sol. P1  (one gift coupon in n definite envelope) probability that the silver coin is in the second purse is
Nn
N n 1 4
(a) (b)
2 9
1, 2, 3 … n
N N N….. N 5 2
(c) (d)
[‘n’ coupons going into N different envelopes] 9 3
Ans. (c)
 PROBABILITY 9

(a) M & N are mutually exclusive

6 Copper 4 Copper
Sol. (b) M & N are independent
1 Silver
(c) M & N are independent
let in first five coin there is 4 copper coin & 1 silver coin

(d) P M N + P M N = 1 
6
C4  1C1
Probability  7 Ans. (b,c,d)
C5
Sol. M and N are independent
8
8 Copper C
P 9 2 M and N are independent
1 Silver C2

II - box M  M 
P   P N  1
N   
6
C4  1C1 8
C2 5
 Required Probability=   38. A bag initially contains one red & two blue balls. An
7 9 9
C5 C2 experiment consisting of selecting a ball at random, noting
its colour & replacing it together with an additional ball of
Objective Questions II [One or more than one correct option] the same colour. If three such trials are made, then :
35. If E 1 and E 2 are two events such that P(E 1) = 1/4, (a) probability that atleast one blue ball is drawn is 0.9
P(E2/E1) =1/2 and P(E1/ E2) = 1/4
(b) probability that exactly one blue ball is drawn is 0.2
(a) then E1 and E2 are independent
(c) probability that all the drawn balls are red given
(b) E1 and E2 are exhaustive that all the drawn balls are of same colour is 0.2
(c) E2 is twice as likely to occur as E1 (d) probability that atleast one red ball is drawn is 0.6.
(d) Probabilities of the events E1  E2 , E1 and E2 are in Ans. (a,b,c,d)
G.P.
1 2 3 9
Ans. (a,c,d) Sol. a P  1   
3 4 5 10
1 1 1 1
Sol. P  E1  E2      P  E2  
4 2 8 4 2 1 2 1 2 2 1 2 2
b  P         
3 4 5 3 4 5 3 4 5
1
 P  E2  
2 1

5
1 1 1 5
P  E1  E2     
4 2 8 8 1 2 3
 
1
36. Let 0 < P(A) < 1 , 0 < P(B) < 1 & c P  1 2 3 3 4 2 5 3 4 
     5
P(A  B) = P(A) + P(B)  P(A). P(B), then : 3 4 5 3 4 5
(a) P(B/A) = P(B)  P(A)
(b) P(AC  BC) = P(AC) + P(BC) 2 3 4 3
d  P  1   
(c) P((A  B)C) = P(AC). P(BC) 3 4 5 5
(d) P(A/B) = P(A)
39. For two given events A & B, P (A  B) is :
Ans. (c,d)
(a) not less than P(A) + P(B) – 1
(b) not greater than P(A) + P(B)
Sol.      A
   
P A  B  P A  B  P A .P B and P    P  A 
B (c) equal to P(A) + P(B) – P (A  B)
(d) equal to P(A) + P(B) + P (A  B)
37. If M & N are independent events such that
0 < P(M) < 1 & 0 < P(N) < 1, then :
 10 PROBABILITY

Ans. (a,b,c)
P(E  F) P(E  F)
= + 1
Sol.  P  A  B  P  A   P  B – P  A  B P(F) 1 - P(F)
On rearranging we get Therefore, option (c) is incorrect.
P  A  B  p  A   P  B – P  A  B
E  E  P(E  F) P(E  F)
(d) P   + P   = +
Which is option C F F P(F) P(F)

 0  P  A  B  1
P(E  F) + P(E  F)
=
 P  A   P  B  –1 P  A  B   P  A   P  B  P(F)

Hence B and C are also correct

P(F)
= =1
Now if P  A  B   P  A   P  B   P  A  B  P(F)

 P  A   P  B  0 Therefore, option (d) is correct.

41. Two real numbers, x & y are selected at random. Given that
Which is not necessary for the events A and B hence the
0  x  1 ; 0  y  1. Let A be the event that y2  x ; B be the
correct option are A,B and C
event that x2  y, then :
40. If E and F are the complementary events of E and F
respectively and if 0 < P (F) < 1, then 1
(a) P (A  B) =
3
(a) P (E/F) + P( E /F) = 1
(b) A & B are exhaustive events
(b) P (E/F) + P(E/ F ) = 1 (c) A & B are mutually exclusive
(c) P ( E /F) + P(E/ F ) = 1 (d) A & B are independent events.
Ans. (a,b)
(d) P (E/ F ) + P( E / F ) = 1
Ans. (a,d) 2
Sol. A  P  A    x dx 
3
E  E  P(E  F) P(E  F)
Sol. (a) P   + P   = +
F F P(F) P(F) 1 1
P  B    x 2 dx 
0 3

=
P(E  F) + P(E  F) P(F)
P(F)
=
P(F)
=1 
P  A  B    y – y 2 dy   1
3
0

Therefore, option (a) is correct.  P(A) + P(B) = 1 (Exhaustive events)

42. If A & B are two events such that P(B)  1, BC denotes the
E  E  P(E  F) P(E  F)
(b) P   + P   = + event complementry to B, then
F F P(F) P(F)

P(E  F) P(E  F)
 C
(a) P A B =  P (A )  P (A  B)
1  P (B)
= + 1
P(F) 1 - P(F)
(b) P (A  B)  P(A) + P(B)  1
Therefore, option (b) is incorrect.
(c) P(A) > P A B if P A B > P(A)
C
 
E 
 E P EF P EF   
(c) P   + P   =
F F P(F)
+
P F   C C C
 
(d) P A B + P A B = 1 
 PROBABILITY 11

Ans. (a,b,c,d)


  1
1 P  B

P Bc
Sol. (a)  A
P  
A B

 
P  A  P  A  B 
(True) PB  PB 
c c
B P B  
1 P  B

True.
(b) P (A B) = P(A) + P (B) - P (A B)
43. For any two events A & B defined on a sample space ,
Now
P (A B) 1 P(A) + P(B)-1
(a) P(A/B)  , P (B)  0 is always true
–P (A B) –1 P(B)
P(A) + P(B) – P (A B) P(A) + P(B) – 1
(b) P  A  B  = P (A) - P (A  B)
 P (A B) P(A) + P (B) – 1 (True)
(c) P (A  B) = 1 - P (Ac). P (Bc), if A & B are independent
 A 
(c) Given P    P  A (d) P (A  B) = 1 - P (Ac). P (Bc), if A & B are disjoint
 Bc 
Ans. (a,b,c)



P A  Bc   P  A Sol.
 A  P  A  B
(a) P   
P B   c B P  B

Now,
P  A  P  A  B 
  P  A P  A  B  P  A   P  B – P  A  B
1 P  B

 P(A) – P(A B) > P(A) – P(A) P(B) 

 P  A   p  B – P  A  B – P A  B  P A  B   
 – P(A B) > – P(A) P (B) Thus,
 P(A B) < P(A) . P(B)
P  A  B  p  A   p  B – 1  P A  B  
P  A  B
  P  A
P  B  P  A  B  P  A   P  B –1

 A
P    P  A
(Using the relation P  A  B   P A  B  1  

B
 A  P  A   P  B  –1
True. Hence, p    .
 B P  B

 A   Ac  So, (a) is true.

(d) P c   P  c 
B  B 
 
(b) P  A  B   P  A  – P A  B So b is true.



P A  Bc   P A  B  c c

     
(c) P  A  B   1 – P A  B  1 – P A P B if A and B
PB  PB 
c c
are independent. So, (c) is true.
(d) Unless A and B are mentioned as independent,
 P  A  P  A  B    1  P  A  B 
  
P A  B cannot be written as P A P B .    
P Bc  
So, (d) is not true.
Numerical Type Valued Questions
 12 PROBABILITY

44. A bag contains n + 1 coins. It is known that one of these P  r  .P  s / r 

coins has heads on both sides, whereas the other coins P r / s 
P  r  .P  s / r   P  r '  .P  s / r ' 
are fair. One coin is selected at random and tossed. If the
probability that the toss results in heads is 7/12, find n.
25 24
Ans. (5) 
Probability p  100 100
25 24 75 25
n 1 1 7   
Sol. PH     1  100 100 100 100
n 1 2 n 1 12
6n + 12 = 7n + 7 24
p
n=5 99
45. 7 persons are stopped on the road at random and asked
33p = 8
about their birthdays. If the probability that 3 of them are
born on Wednesday, 2 on Thursday and the remaining 2 Assertion Reason Type
K
on Sunday is , then K is equal to (A) If both assertion and reason are correct and reason is the
76 correct explanation of assertion.
Ans. (30) (B) If both assertion and reason are true but reason is not the
Sol. Given, 7 stopped at road and asked their birthdays. correct explanation of assertion.

Probability that 3 people out of 7 born on Wednesday (C) If assertion is true but reason is false.

= Selecting 3 people among 7 divided by total ways (D) If assertion is false but reason is true.

7 47. Let A and B are two events such that P(A) > 0.
C3
= 3 Assertion : If P (A) + P (B) > 1, then
7
P (B/A)  1 – P (B’)/P(A)
Probability that 2 people out of remaining 4, born on
Reason : If P (A/B’)  P (A), then P(A)  P(A/B).
Thursday = selecting 2 people from remaining 4 people
(a) A (b) B
4
C (c) C (d) D
 22
7 Ans. (b)
The required probability Sol. Assertion  P(A B) 1
P(A) + P(B) – P(A B) 1
 7C   4 C   2 C 
  33    22    2 2  P  B '
B
 7   7   7  P    1
A
  P  A
7K = 7C3 × 4C2 × 2C2
 7K = 35 × 6 × 1 P  A  B '
Reason  
 K = 30 P  B '
46. Two integers r and s are chosen one after the other
P(A)P(A) – P(A B)  P (A) – P(A) P(B)
without replacement from the numbers 1, 2, 3,... 100. Let
p be the probability that r  25 given that s  25. Find the P(A B) P(A) . P(B)
value of 33p. 48. From an urn containing a white and b black balls,
Ans. (8) k (< a, b) are drawn and laid aside, their colour unnoted.
Sol. p(r) = 25/100 p(r’) = 75/100

p  s / r   24100 P  s / r '   25100

 PROBABILITY 13

Then another ball, that is, (k + 1)th ball is drawn. (a) A–Q; B-P; C-S; D-R
Assertion : Probability that (k + 1)th ball drawn is white
(b) A–P; B-Q; C-S; D-R
a
is . (c) A–Q; B-S; C-P; D-R
ab
Reason : Probability that (k + 1)th ball drawn is black is (d) A–Q; B-P; C-R; D-S
Ans. (a)
a
ab Sol. Since, determinanat is of order 2 × 2 and each element is 0
or 1 only.
(a) A (b) B
 n(S) = 24 = 16
(c) C (d) D
And the determination is positive are,
Ans. (c)
Sol. Consider the problem as arrenging a number of white & b 1 0 1 1 1 0
, ,
number of black balls in a row & finding probability that 0 1 0 1 1 1
(k + 1)th ball is white.
 n(E) = 3
 a  b !
Total ways = ; 3
a!b! Thus the required probability 
16
W 50. A ten digit number N is formed by using the digits
Favourable ways = 
   
k a  b 1 k   0 to 9 exactly once. The probability that N is divisible by
It is arranging (a + b – 1) balls keeping (k + 1)th ball white (A) 4 (P) 1
(B) 5 (Q) 20/81
 a  b  1!

 a  1 ! b! (C) 45 (R) 17/81
(D) 12 (S) 2/81
 a  b  1! b!a! a The correct matching is
Required Probability  
 a  1!b!  a  b ! a  b
(a) A–Q; B–R; C–R; D–Q
Match the Following (b) A–R; B–Q; C–R; D–Q

Each question has two columns. Four options are given (c) A–Q; B–R; C–Q; D–R
representing matching of elements from Column-I and
Column-II. Only one of these four options corresponds (d) A–R; B–R; C–Q; D–Q
to a correct matching.For each question, choose the option Ans. (a)
corresponding to the correct matching. Sol. a) the total number of ways to form 10 digit number using
digits 0 to 9 exactly once = 10!-9!=10×9!-9!=(10-1)9!=9×9!
49. A determinant  is chosen at random from the set of all
the number is divisible by 4, therefore the last two digits
determinant of order two with elements 0 and 1 only. can be
Value of  Probability
(A) 1 (P) 5/8
(B) 0 (Q) 3/16
(C) 2 (R) 3/8 but the digits are used only once, therefore we can not
(D) non zero (S) 0 take 44 and 88 as the last digits.

The correct matching is now when 0 is used in the last two digits, (i.e.
04,08,20,40,60,80=6 cases)
 14 PROBABILITY

the number of ways to fill remaining 8 places from 8 digits 51. Events S and T are:
= 8! (a) mutually exclusive
therefore the number of ways to form 10 digits number in (b) independent
the above 6 cases = 6×8! (c) mutually exclusive and independent
we have the remaining (24-2-6) = 16 cases (d) neither mutually exclusive nor independent
so the number of ways (when 0 is not used in last two Ans. (b)
digits) to fill remaining 8 places with 8 digits (0 included)
S
=8!-7!=(8-1) 7!=7×7! Sol. P S   P  
T 
therefore the number of ways to form 10 digits number in
the above 16 cases = 16×(7×7!) =8×2×(7×7!)=14×8! independent
therefore the total number of required ways 52. The value of P(S and T)
= 6×8!+14×8!=(6+14)8!=20×8! (a) 0.3450 (b) 0.2500
(c) 0.6900 (d) 0.350
20  8! 20  8! 20
hence the required probability =   Ans. (a)
9  9! 9  9  8! 81
Sol. P (S T) = P(S).P(T) = 0.5 × 0.69 = 0.345
b) Conditions of forming such a number-
53. The value of P(S or T)
1) All the digits have to be used at once
(a) 0.6900 (b) 1.19
2) 0 should not come at first place (c) 0.8450 (d) 0
3) Last digit should be either 0 or 5 Ans. (c)
Case 1. Number ending with 0 = 9! Sol. P (S T) = P(S) + P(T) – P(S) . P(T) 0.5 + 0.69 - 0.5 × 0.69
Case 2. Number ending with 5 = (9! – 8!) = 0.845
{numbers starting with 0 are subtracted 4
Using the following passage, solve Q.54 to Q.56
total 10 digit numbers that can be formed = 10! – 9!
A JEE aspirant estimates that she will be successful with
17 an 80 percent chance if she studies 10 hours per day, with
P
81 a 60 percent chance if she studies 7 hours per day and with
c) for ‘45’ number should be divisible by ‘3’ & ‘5’ both as a 40 percent chance if she studies 4 hours per day. She
all the number arrangement from 0 to 9 is divisible by ‘3’ further believes that she will study 10 hours, 7 hours and 4
we need to check only for ‘5’ which is calculated in part hours per day with probabilities 0.1, 0.2 and 0.7, respectively
54. The chance she will be successful, is
17
(b)  P  (a) 0.28 (b) 0.38
81
(c) 0.48 (d) 0.58
d) for ‘12’ number should be divisible by ‘3’ & ‘4’ both as
Ans. (c)
all the number arrangement from 0 to 9 is divisible by ‘3’
we need to check only for ‘4’ which is calculated in part (a) Sol. A : She get a success

20 T: She studies 10 hrs : P(T) = 0.1

P
81 S : She studies 7 hrs : P(S) = 0.2
F : She studies 4 hrs : P(F) = 0.7
Paragraph Type Questions

Passage

Using the following passage, solve Q.51 to Q.53

Let S and T are two events defined on a sample space with
probabilities
P(S) = 0.5, P(T) = 0.69, P(S/T) = 0.5
 PROBABILITY 15

In a game of card, each card is worth an amount of points.

A A A
P    0.8; P    0.6;    0.4 Each numbered card is worth its number (e.g. a 5 is worth 5
T S F points) ; the Jack, Queen and King are each worth 10 points
Now, ; and the Ace is worth your choice of either 1 point or 11
points. The object of the game is to have more points in
P  A   P  A  T   P  A  S  P  A  F  your set of cards than your opponent without going over
21. Any set of cards with sum greater than 21 automatically
A A A loses.
 P  T  .P    P  S .P    P  F  .P  
T S F Here's how the game played. You and your opponent are
each dealt two cards. Usually the first card for each player
= (0.1)(0.8) + (0.2)(0.6) + (0.7)(0.4)
is dealt face down, and the second card for each player is
= 0.08 + 0.12 + 0.28 = 0.48 dealt face up. After the initial cards are dealt, the first player
55. Given that she is successful, the chance she studied for 4 has the option of asking for another card or not taking any
hours, is cards. The first player can keep asking for more cards until
either he or she goes over 21, in which case the player
6 7 loses, or stops at some number less than or equal to 21.
(a) (b)
12 12 When the first player stops at some number less than or
equal to 21, the second player then can take more cards
8 9
(c) (d) until matching or exceeding the first player's number without
12 12
going over 21, in which case the second player wins, or
Ans. (b) until going over 21, in which case the first player wins.
We are going to simplify the game a little and assume that
 F  P  F  A   0.7  0.4  0.28 7
Sol. P      all cards are dealt face up, so that all cards are visible.
A P A 0.48 0.48 12 Assume your opponent is dealt cards and plays first.
56. Given that she does not achieve success, the chance she 57. The chance that the second card will be a heart and a Jack,
studied for 4 hour, is is

18 19 4 13
(a) (b) (a) (b)
26 26 52 52

20 21 17 1
(c) (d) (c) (d)
26 26 52 52

Ans. (d) Ans. (d)

Sol. P = first card is not heart and jack and second card is heart

Sol.
 F  P FA
P   
 
P  F – P  F  A  and jack
A P A  
0.52
51 1 1
  
52 51 52
 0.7  – 0.28 0.42 21
   58. The chance that the first card will be a heart or a Jack, is
0.52 0.52 26
13 16
Using the following passage, solve Q.57 to Q.60 (a) (b)
52 52
Read the passage given below carefully before attempting 17
these questions. (c) (d) none
52
A standard deck of playing cards has 52 cards. There are
Ans. (b)
four suit (clubs, diamonds, hearts and spades), each of
which has thirteen numbered cards (2, ....., 9, 10, Jack, Queen, Sol. P = P (H J)
King, Ace) = P(H) + P(J) - P (H J)
 16 PROBABILITY

Getting 21 by drawing three Aces.

13 4 1 16
    There are 4*3*2 = 24 ways to get three aces, and there are
52 52 52 52
(48*47*46) ways to pick three cards. Hence, the
59. Given that the first card is a Jack, the chance that it will be probability of picking three aces is 24/(48 × 47 × 46)
the heart, is
Hence, P(win) = 4/48 + ( 4 × 4)/(47 × 48) + (4 × 4)/(47 × 48)
1 4 + (4 × 3 × 2)/(47 × 48 × 46) = .0907.
(a) (b)
13 13 Text
1 1 61. Six boys and six girls sit in a row at random. Find the
(c) (d)
4 3 probability that
Ans. (c) (a) the six girls sit together
(b) the boys and girls sit alternatively
1
P  Heart  Jack  52 1 1 1
Sol. P(Heart/Jack)   
P  Jack  4 4 Ans. (a)
132
(b)
462
52
Sol. (a) The total number of arrangements of six boys and six
60. Your opponent is dealt a King and a 10, and you are dealt girls = 12!
a Queen and a 9. Being smart, your opponent does not
take any more cards and stays at 20. The chance that you 6!  7! 1
Required probability = (12)! = 132
will win if you are allowed to take as many cards as you
need, is
(since, we consider six girls as one person)
(a) 0.771 (b) 0.088
(c) 0.0797 (d) 0.0907 2  6!  6! 1
(b) Required probability = = (12)!
=
462
Ans. (d)
Sol. (v) To win, you have to get a 21. There are several ways to 62. In a test an examinee either guesses or copies or knows
get 21: draw a three in one card, draw a two and an Ace in the answer to a multiple choice question with four
two cards, and draw three Aces in three cards.
1
Getting 21 by drawing a three: choices. The probability that he make a guess is and
3
Since there are four threes in the deck, and 48 cards
remaining after the first four cards are dealt, the chance of 1
the probability that he copies the answer is . The
getting a 3 is P(get a 3) = 4/48. 6
Getting 21 by drawing an Ace first and a 2 second: probability that his answer is correct given that he copied
P(get Ace first and a 2 second) = P(get a 2 second/ get an 1
Ace first) × P (get an Ace first) = (4 / 47) × (4/48) it, is . Find the probability that he knew the answer to
8
The 4/48 is determined as follows. Since there are four the question given that he correctly answered it.
aces in the deck, and 48 cards remaining after the first four
cards are dealt, the chance of getting an Ace first is 4/48. 24
Ans.
The 4/47 is determined as follows. Since there are four 25 29
in the deck, and 47 cards remaining after the first five Sol. Let E1, E2, E3 and A be the events defined as
cards are dealt, the chance of getting a 2 on the next card
E1 = the examinee guesses the answer
is 4/47.
E2 = the examinee copies the answer
Getting 21 by drawing a 2 first and an Ace second.
E3 = the examinee knows the answer
By similar logic,
and A = the examinee answer correctly
P(get 2 first and Ace second) = P(get Ace second / get 2
first) × P(get 2 first) = (4 / 47) × (4/48) 1 1
We have, P(E1 ) = , P(E2 ) =
3 6
Since, E1, E2, E3 are mutually exclusive and exhaustive
 PROBABILITY 17

events. number on the two chosen tickets is not more than 10.
 P(E1) + P(E2) + P(E3) = 1 The minimum number on them is 5 with probability....

1 1 1 1
 P(E 3 ) = 1- - = Ans. ( )
3 6 2 9

If E1 has already occurred, then the examinee guesses. Sol. Let A be the event that the maximum number on the two
Since, there are four choice out of which only one is chosen tickets is not more than 10, and B the event that
correct, therefore the probability that he answer the minimum number on them is 5
correctly given that he has made a guess is 1/4. 5
C
 P(A  B) = 100 1
1 C2
i.e. P(A/E1 ) =
4
10
C
1 and P(A) = 100 2
It is given that, P(A/E2 )= C2
8
and P(A/E3) = probability that he answer correctly given
 B  P  A  B
that he know the answer = 1 Then P   =
A P A
By Baye’s theorem, we have

P  E 3  .P  A/E 3 
5
C1 1
P(E 3 /A)= = =
 P  E1  .P  A/E 1  + P  E 2  .P  A/E 2  + P  E 3  .P  A/E 3  10
C2 9

65. If two events A and B are such that P (Ac) = 0.3,

1 P(B) = 0.4 and P (A  B c ) = 0.5 then
1
2 24 P[B / (A Bc)] =....
 P(E3 /A) = =
 1 1   1 1   1  29
   +    +   1 1
3 4 6 8  2 
Ans.
4
Fill in the blanks Sol. P(AC) = 0.3 [given]

63. For a biased die the probabilities for the different faces P(A) = 0.7
to turn up are given below P(B) = 0.4 [given]
Face 1 2 3 4 5 6 P(BC) = 0.6 and P(A BC) = 0.5 [given]
Probability 0.1 0.32 0.21 0.15 0.05 0.17 Now, P(A BC) = P(A) + P(BC) – P(A BC)
This die is tossed and you are told that either face 1 or = 0.7 + 0.6 – 0.5 = 0.8
face 2 has turned up. Then, the probability that it is face
1, is....
 P[B/A  B ] = C 
P{B  A  BC } 
Ans. (
5
)

P AB C

21

0.1 0.1 5
=

P{  B  A   B  BC }  =
P{  B  A    }
=
P B  A
Sol. Probability (face 1) = = = =
0.1 + 0.32 0.42 21 0.8 0.8 0.8

64. A box contains 100 tickets numbered 1, 2, ..., 100. Two

1 0.7 - 0.5 0.2 1
tickets are chosen at random. It is given that the maximum [P(A) - P(A  BC )] = = =
0.8 0.8 0.8 4