Explanation / Important formulas:
Random Experiment: An experiment in which all possible outcomes are know and the exact output cannot be predicted in advance, is called a random experiment.
Examples:
 Rolling an unbiased dice.
 Tossing a fair coin.
 Drawing a card from a pack of wellshuffled cards.
 Picking up a ball of certain colour from a bag containing balls of different colours.
Details:
 When we throw a coin, then either a Head (H) or a Tail (T) appears.
 A dice is a solid cube, having 6 faces, marked 1, 2, 3, 4, 5, 6 respectively. When we throw a die, the outcome is the number that appears on its upper face.
 A pack of cards has 52 cards.
 It has 13 cards of each suit, name Spades, Clubs, Hearts and Diamonds.
 Cards of spades and clubs are black cards.
 Cards of hearts and diamonds are red cards.
 There are 4 honours of each unit.
 There are Kings, Queens and Jacks. These are all called face cards.
Sample Space: When we perform an experiment, then the set S of all possible outcomes is called the sample space.
Examples:
 In tossing a coin, S = {H, T}
 If two coins are tossed, the S = {HH, HT, TH, TT}.
 In rolling a dice, we have, S = {1, 2, 3, 4, 5, 6}.
Event: Any subset of a sample space is called an event.
Probability of Occurrence of an Event: Let S be the sample and let E be an event. Then, P(E) = n(E) / n(S)
Results on Probability:
For any events A and B we have: P(A∪B) = P(A) + P(B) – P(A∩B)
If A denotes (notA), then P(A) = 1 – P(A)
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Probability  Test
Probability  Question and Answers
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Question 1

A

18/145

B

18/29

C

36/135

D

36/145

= ^{12}C_{1} / ^{30}C_{1} = 12 / 30 = 2 / 5
Since, the ball is not replaced; hence the number of balls left in bag is 29
Hence, the probability the second ball is black:
= ^{18}C_{1} / ^{29}C_{1 }= 18 / 29
Required probability,
= (2 / 5) × (18 / 29)
= 36 / 145
Question 2

A

1 / 13

B

3 / 13

C

1 / 4

D

9 / 52

Total no .of cards = 52
So probability = 12 / 52 = 3 / 13
Question 3

A

10 / 21

B

11 / 21

C

2 / 7

D

5 / 7

Let S be the sample space
Then, n(S) = Number of ways of drawing 2 balls out of 7
= 7C_{2} = (7 x 6) / (2 x 1) = 21
Let E = Event of drawing 2 balls, none of which is blue
n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls
= ^{5}C_{2}
= (5 x 4) / (2 x 1)
= 10
Therefore P(E) = n(E) / n(S) = 10 / 21
Question 4

A

1 / 6

B

1 / 8

C

1 / 9

D

1 / 12

Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}
Therefore, n(E) / n(S) = 4 / 36 = 1/ 9
Question 5

A

1 / 15

B

25 / 57

C

35 / 256

D

1 / 221

Then, n(S) = ^{52}C_{2} = (52 x 51) / (2 x1) = 1326
Let E = event of getting 2 kings out of 4
n(E) = ^{4}C_{2 }= (4 x 3) / (2 x1) = 6
P(E) = n(E) / n(S) = 6 / 1326 = 1 / 221
Question 6

A

1 / 216

B

1 / 36

C

4 / 216

D

3 / 216

simultaneously is:
= 6×6×6×6 = 6 ^{4 }
n(S) = 6 ^{4}
Let X be the event that all dice show the same face
X={(1,1,1,1,),(2,2,2,2),(3,3,3,3),(4,4,4,4),(5,5,5,5),(6,6,6,6)}
n(X) = 6
Hence required probability,
= n(X) / n(S)
= 6 / 6 ^{4} = n(X) / n(S) = 6 / 6^{4} = 1 / 216
Question 7

A

1 / 10

B

2 / 5

C

2 / 7

D

5 / 7

Question 8

A

1 / 3

B

3 / 4

C

7 / 19

D

8 / 21

Let E = event that the ball drawn is neither red nor green
n(E) = 7
Therefore P(E) = n(E) / n(S) = 7 / 21 = 1 / 3
Question 9

A

4

B

5

C

9

D

10

No. of students those did not complete any task: 50  45 = 5
Question 10

A

30

B

35

C

40

D

60

200 = 125 + 115 – Both
Both = 240  200 = 40
Question 11

A

3 / 4

B

1 / 4

C

3 / 8

D

7 / 8

Let E = event of getting at most two heads
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}
Therefore, P(E) = n(E) / n(S) = 7 / 8
Question 12

A

1 / 22

B

3 / 22

C

2 / 91

D

2 / 77

Then, n(S)= number of ways of drawing 3 balls out of 15
= ^{15}C_{3}
(15 x 14 x 13) / (3 x 2 x 1) = 455
Let E = event of getting all the 3 red balls
Therefore, n(E) = ^{5}C_{3} = ^{5}C_{2} = (5 x 4) / (2 x 1) = 10
Therefore, P(E) = n(E) / n(S) = 10 / 455 = 2 / 91
Question 13

A

21 / 46

B

25 / 117

C

1 / 50

D

3 / 25

Then, n(S)= Number ways of selecting 3 students out of 25
= ^{25}C_{3}
=> (25 x 24 x 23 ) / (3 x 2 x1) = 2300
n(E) = (^{10}C_{1} x ^{15}C_{2})
=> [10 x [(15 x 14) / (2 x1)]] = 1050
Therefore, P(E) = n(E) / n(S) = 1050 / 2300 = 21/ 46(3 x 2 x1)
Question 14

A

6 / 13

B

5 / 26

C

5 / 13

D

7 / 26

=> ^{6}C_{2} / ^{13}C_{2}
=> 5 / 26
Question 15

A

1 / 2

B

3 / 4

C

3 / 8

D

5 / 16

Then, E = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(E) = 27
Therefore, P(E) = n(E) / n(S) = 27 / 36 = 3 / 4
Question 16

A

1 / 13

B

2 / 13

C

1 / 26

D

1 / 52

Let E = event of getting a queen of club or a king of heart
Then, n(E) = 2
Therefore, P(E) = n(E) / n(S) = 2/ 52 = 1/ 26
Question 17

A

3 / 20

B

29 / 34

C

47 100

D

13 / 102

Then, n(S) = ^{52}C_{2} = (52 x 51) / (2 x 1) = 1326
Let E = event of getting 1 spade and 1 heart
n(E) = number of ways of choosing 1 spade out of 13 and 1 heart out of 13
=(^{13}C_{1} x ^{13}C_{1})
=> (13 x 13)= 169
P(E) = n(E) / n(S) = 169 / 1326 = 13 / 102
Question 18

A

1 / 13

B

3 / 13

C

1 / 4

D

9 / 52

Therefore, P(Getting a face card) = 12 / 52 = 3 / 13
Question 19

A

1 / 2

B

2 / 5

C

8 / 15

D

9 / 20

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}
Therefore P(E) = n(E) / n(S) = 9 / 20
Question 20

A

22

B

32

C

33

D

44

(P(AUBUC) = P(A) + P(B) + P(C)  P(A n B) – P(A n C) – P(B n C) + P(A n B n C))
76 = 53 + 46 + 39  22  23  x + 15
x = 108  76
x = 32
Question 21

A

3 / 4

B

4 / 7

C

1 / 8

D

3 / 7

Number of white balls = 8
P(drawing a white ball) = 8 / 14 = 4 / 7
Question 22

A

31

B

15

C

21

D

25

Total no.of students = 10 + 20 + 5  4  4  2 + 2 = 31
Question 23

A

55/58

B

40/58

C

41/58

D

39/58

Given that, we can move the balls between the boxes
From the above arrangement, Probability to choose a green ball =Probability of choosing box A x Probability of choosing green ball from box A + Probability of choosing box B x Probability of choosing green ball from box B ...(1)
Since there are only two boxes, the probability of choosing box A = probability of choosing box B = ½
Probability of choosing a green ball from box A = number of green balls in box A / total number of ballsin box A = 1/1 = 1
Probability of choosing a green ball from box B = number of green balls in box B / total number of ballsin box B = 26/(26+23) = 26/49 = 26/49
Required probability (from eqn(1)) = 1/2 x 1 + 1/2 x 26/29 = 1/2 (1 + 26/29) = (1/2) x (55/29) = 55/58
Hence the required maximum probability = 55/58
Question 24

A

1/35

B

1/34

C

1/17

D

1/68

When they occupy alternate position the arrangement would be like:
B G B G B G B
Thus, total number of possible arrangements for boys = (4×3×2)
Total number of possible arrangements for girls =(3×2)
Required probability = (4×3×2×3×2) /7! = 1/35
Question 25

A

1/35

B

3/7

C

34/35

D

4/7

But, Probability of there being all women in the committee + Probability of there being at least one man = Probability of formation of a committee with no restriction of gender = 1
Probability of there being at least one man = 1  1/35 = 34/35
Question 26

A

4/15

B

69/91

C

11/15

D

22/91

=> Selection of 4 marbles out of 15 marbles <br>
=> 15C_{4 }= (15 x 14 x 13 x 12) divided by 1 x 2 x 3 x 4 = 1365 <br>
When no marble is blue, favourable number of cases n(E) <br>
=> Selection of 4 marbles out of 11 marbles <br>
=> 11C_{4} = 11 x 10 x 9 x 8 divided by 1 x 2 x 3 x 4 = 330 <br>
Probability that atleast one ball is blue + Probability that no ball is blue = 1 <br>
Therefore, Probability that atleast one ball is blue = 1  Probability that no ball is blue <br>
Therefore, Probability that atleast one ball is blue = 1  n(E) / n(S) <br>
= 1  330/1365 = 1  22/91 = 69 /91
Question 27

A

70/8281

B

140/20449

C

25/5445

D

35/5448

Total ways of drawing 3 balls, N(S) = 14C3 = 364
No of ways to draw 3 black balls = N(E1) for black balls = 8C3 = 56
Probability of all balls being black = P(E1) = N(E1) / N(S) = 56/ 364 = 14/91
No of ways to draw 3 white balls = N(E2) for white balls = 6C3 = 20
Probability of all balls being white = P(E2) = 20 / 364 = 5/91
Probability of drawing 3 black balls in first draw and drawing 3 white balls in the second draw = P(E1) x P(E2)
Therefore P(E) = 14/91 x 5/91 = 70 /8281
Question 28

A

2/9

B

1/13

C

2/13

D

1/26

Total number of arrangements = n(S) = (14 – 1)! = 13 !
Taking three persons as a unit, total persons = 12 (in 4 units)
Therefore no. of ways for these 12 persons to around the circular table = (12  1)! = 11!
In any given unit, 3 particular person can sit in 3!
Hence total number of ways that any three person can sit = n(E) = 11! X 3!
Therefore P (E) = probability of three persons sitting together = n(E) / n(S) = 11! X 3 ! divided by 13!
3 x 2 divided by 13 x 12 = 1/26
Question 29

A

1663200

B

8316000

C

3326400

D

4158000

Total number of rearrangements = (Number of letters)! / (Number of 1st repetitive letter)! x (Number of 2nd repetitive letter)!
In the word ECHRONICLEE, E occurs thrice. C occurs twice
Hence denominator of the above formula will become 3! x 2!
Therefore Total number of rearrangements = 11!/ 3! 2!
= 3326400
Question 30

A

4/7

B

4/9

C

5/7

D

2/7

If some person are selected at random from the group, the expected value of the ratio of boys and girls will be 4 : 3
If the leader is chosen at random from the selection, the probability of him being a boy = 4/7
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