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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 well-shuffled 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 (not-A), then P(A) = 1 – P(A)
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Probability - Question and Answers
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Question 1 |
A bag contains 12 white and 18 black balls. Two balls are drawn in succession without replacement. What is the probability that first is white and second is black?
18/145 | |
18/29 | |
36/135 | |
36/145 |
Question 1 Explanation:
The probability that first ball is white:
= 12C1 / 30C1 = 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:
= 18C1 / 29C1 = 18 / 29
Required probability,
= (2 / 5) × (18 / 29)
= 36 / 145
= 12C1 / 30C1 = 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:
= 18C1 / 29C1 = 18 / 29
Required probability,
= (2 / 5) × (18 / 29)
= 36 / 145
Question 2 |
One card is drawn at random from a pack of 52cards. What is the probability that the card drawn is a face card (jack, queen, king only)?
1 / 13 | |
3 / 13 | |
1 / 4 | |
9 / 52 |
Question 2 Explanation:
Total no. of face cards (jack, queen and king only) =12
Total no .of cards = 52
So probability = 12 / 52 = 3 / 13
Total no .of cards = 52
So probability = 12 / 52 = 3 / 13
Question 3 |
A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?
10 / 21 | |
11 / 21 | |
2 / 7 | |
5 / 7 |
Question 3 Explanation:
Total number of balls = (2 + 3 + 2) = 7
Let S be the sample space
Then, n(S) = Number of ways of drawing 2 balls out of 7
= 7C2 = (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
= 5C2
= (5 x 4) / (2 x 1)
= 10
Therefore P(E) = n(E) / n(S) = 10 / 21
Let S be the sample space
Then, n(S) = Number of ways of drawing 2 balls out of 7
= 7C2 = (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
= 5C2
= (5 x 4) / (2 x 1)
= 10
Therefore P(E) = n(E) / n(S) = 10 / 21
Question 4 |
What is the probability of getting a sum 9 from two throws of a dice?
1 / 6 | |
1 / 8 | |
1 / 9 | |
1 / 12 |
Question 4 Explanation:
In two throws of a dice, n(S) = (6 x 6) = 36
Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}
Therefore, n(E) / n(S) = 4 / 36 = 1/ 9
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 |
From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?
1 / 15 | |
25 / 57 | |
35 / 256 | |
1 / 221 |
Question 5 Explanation:
Let S be the sample space
Then, n(S) = 52C2 = (52 x 51) / (2 x1) = 1326
Let E = event of getting 2 kings out of 4
n(E) = 4C2 = (4 x 3) / (2 x1) = 6
P(E) = n(E) / n(S) = 6 / 1326 = 1 / 221
Then, n(S) = 52C2 = (52 x 51) / (2 x1) = 1326
Let E = event of getting 2 kings out of 4
n(E) = 4C2 = (4 x 3) / (2 x1) = 6
P(E) = n(E) / n(S) = 6 / 1326 = 1 / 221
Question 6 |
Four dice are thrown simultaneously. Find the probability that all of them show the same face?
1 / 216 | |
1 / 36 | |
4 / 216 | |
3 / 216 |
Question 6 Explanation:
The total number of elementary events associated to the random experiments of throwing four dice
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 / 64 = 1 / 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 / 64 = 1 / 216
Question 7 |
In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?
1 / 10 | |
2 / 5 | |
2 / 7 | |
5 / 7 |
Question 7 Explanation:
P (getting a prize) = 10 / (10 + 25) = 10 / 35 = 2 / 7
Question 8 |
In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?
1 / 3 | |
3 / 4 | |
7 / 19 | |
8 / 21 |
Question 8 Explanation:
Total number of balls = (8 + 7 + 6) = 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
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 group of 50 students were required to clear 2 tasks, one in rock-climbing and the other in bridge crossing during an adventure sports expedition. 30 students cleared both the tasks. 37 cleared bridge crossing, 38 student cleared rock-climbing. How many students could not clear any task?
4 | |
5 | |
9 | |
10 |
Question 9 Explanation:
No. of student those completed at least one task: 37 + 38 - 30 = 45
No. of students those did not complete any task: 50 - 45 = 5
No. of students those did not complete any task: 50 - 45 = 5
Question 10 |
200 pupil total, out of 125 like pizza, 115 like burger, then many like both?
30 | |
35 | |
40 | |
60 |
Question 10 Explanation:
n(A U B) = n(A) + n(B) – n(A n B)
200 = 125 + 115 – Both
Both = 240 - 200 = 40
200 = 125 + 115 – Both
Both = 240 - 200 = 40
Question 11 |
Three unbiased coins are tossed. What is the probability of getting at most two heads?
3 / 4 | |
1 / 4 | |
3 / 8 | |
7 / 8 |
Question 11 Explanation:
Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
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
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 bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is?
1 / 22 | |
3 / 22 | |
2 / 91 | |
2 / 77 |
Question 12 Explanation:
Let S be the sample space
Then, n(S)= number of ways of drawing 3 balls out of 15
= 15C3
(15 x 14 x 13) / (3 x 2 x 1) = 455
Let E = event of getting all the 3 red balls
Therefore, n(E) = 5C3 = 5C2 = (5 x 4) / (2 x 1) = 10
Therefore, P(E) = n(E) / n(S) = 10 / 455 = 2 / 91
Then, n(S)= number of ways of drawing 3 balls out of 15
= 15C3
(15 x 14 x 13) / (3 x 2 x 1) = 455
Let E = event of getting all the 3 red balls
Therefore, n(E) = 5C3 = 5C2 = (5 x 4) / (2 x 1) = 10
Therefore, P(E) = n(E) / n(S) = 10 / 455 = 2 / 91
Question 13 |
In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is?
21 / 46 | |
25 / 117 | |
1 / 50 | |
3 / 25 |
Question 13 Explanation:
Let S be the sample space and E be the event of selecting 1 girl and 2 boys
Then, n(S)= Number ways of selecting 3 students out of 25
= 25C3
=> (25 x 24 x 23 ) / (3 x 2 x1) = 2300
n(E) = (10C1 x 15C2)
=> [10 x [(15 x 14) / (2 x1)]] = 1050
Therefore, P(E) = n(E) / n(S) = 1050 / 2300 = 21/ 46(3 x 2 x1)
Then, n(S)= Number ways of selecting 3 students out of 25
= 25C3
=> (25 x 24 x 23 ) / (3 x 2 x1) = 2300
n(E) = (10C1 x 15C2)
=> [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 bag contains 6 red, 5 blue and 2 green marbles. If 2 marbles are picked at random, What is the probability that both are red?
6 / 13 | |
5 / 26 | |
5 / 13 | |
7 / 26 |
Question 14 Explanation:
The required probability i.e P(Both are red),
=> 6C2 / 13C2
=> 5 / 26
=> 6C2 / 13C2
=> 5 / 26
Question 15 |
Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?
1 / 2 | |
3 / 4 | |
3 / 8 | |
5 / 16 |
Question 15 Explanation:
In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36
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
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 card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is?
1 / 13 | |
2 / 13 | |
1 / 26 | |
1 / 52 |
Question 16 Explanation:
Here, n(S) = 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
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 |
Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is?
3 / 20 | |
29 / 34 | |
47 100 | |
13 / 102 |
Question 17 Explanation:
Let S be the sample space
Then, n(S) = 52C2 = (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
=(13C1 x 13C1)
=> (13 x 13)= 169
P(E) = n(E) / n(S) = 169 / 1326 = 13 / 102
Then, n(S) = 52C2 = (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
=(13C1 x 13C1)
=> (13 x 13)= 169
P(E) = n(E) / n(S) = 169 / 1326 = 13 / 102
Question 18 |
One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)?
1 / 13 | |
3 / 13 | |
1 / 4 | |
9 / 52 |
Question 18 Explanation:
Clearly, there are 52 cards, out of which there are 12 face cards
Therefore, P(Getting a face card) = 12 / 52 = 3 / 13
Therefore, P(Getting a face card) = 12 / 52 = 3 / 13
Question 19 |
Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?
1 / 2 | |
2 / 5 | |
8 / 15 | |
9 / 20 |
Question 19 Explanation:
Here, S = {1, 2, 3, 4, .... , 19, 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
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 |
There are 76 people residing in a colony, 53 can read hindu, 46 can read times,39 can read deccan, 15 can read all.if 22 can read hindu and deccan, 23 can read deccan and times, then what is the number of persons who can read only times and hindu?
22 | |
32 | |
33 | |
44 |
Question 20 Explanation:
Let x = the number of persons read only times and Hindu
(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
(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 bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?
3 / 4 | |
4 / 7 | |
1 / 8 | |
3 / 7 |
Question 21 Explanation:
Let number of balls = (6 + 8) = 14
Number of white balls = 8
P(drawing a white ball) = 8 / 14 = 4 / 7
Number of white balls = 8
P(drawing a white ball) = 8 / 14 = 4 / 7
Question 22 |
10 friends meet foe movie, 20 for picnic and 5 for games, 4 for movies and picnic, 2 for movies and games, 0 for picnic and games 2 for all, how many students are there totally?
31 | |
15 | |
21 | |
25 |
Question 22 Explanation:
n(A U B U C) = n(A) + n(B) + n(C) – n(A n B) – n(A n C) – n(C n B) + n(A n B n C)
Total no.of students = 10 + 20 + 5 - 4 - 4 - 2 + 2 = 31
Total no.of students = 10 + 20 + 5 - 4 - 4 - 2 + 2 = 31
Question 23 |
There are two boxes namely A and B. A contains 20 green and 15 blue balls and B contains 7green and 8 blue balls. You can move the balls between the two boxes. If you are allowed tochoose a box at random then what will be the maximum probability of getting a green ball from the chosen box?
55/58 | |
40/58 | |
41/58 | |
39/58 |
Question 23 Explanation:
Total number of green balls = 20 + 7 = 27 and total number of blue balls = 15 + 8 = 23
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
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 |
Four boys and three girls stand in queue for an interview. The probability that they stand in alternate positions is?
1/35
| |
1/34 | |
1/17 | |
1/68 |
Question 24 Explanation:
Total number of possible arrangements for 4 boys and 3 girls in a queue = 7!
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
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 committee of 4 members is to be selected from a group of 4 women and 3 men. What is the probability that the committee has at least one man?
1/35 | |
3/7 | |
34/35 | |
4/7 |
Question 25 Explanation:
Probability of there being all women in the committee = 4C4 divided by 7C4 = 1/35
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
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 |
An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four marbles are picked at random, what is the probability that at least one is blue?
4/15 | |
69/91 | |
11/15 | |
22/91
|
Question 26 Explanation:
Total possible outcomes = n(S) <br>
=> Selection of 4 marbles out of 15 marbles <br>
=> 15C4 = (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>
=> 11C4 = 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 leather box contains 8 black balls and 6 white balls. Two draws of three balls each are made, the balls being replaced after the first draw. What is the probability that the balls were black in the first draw and white in the second draw?
70/8281 | |
140/20449 | |
25/5445 | |
35/5448 |
Question 27 Explanation:
Total number of balls = 8 + 6 = 14
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
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 |
Fourteen persons are sitting around a circular table facing the centre. What is the probability that three particular persons sit together?
2/9 | |
1/13 | |
2/13 | |
1/26 |
Question 28 Explanation:
n a circle of n different persons, the total number of arrangements possible = (n - 1)!
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
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 |
In how many different ways can the letter of the word “ECHRONICLEE” be arranged?
1663200 | |
8316000 | |
3326400 | |
4158000 |
Question 29 Explanation:
ECHRONICLEE has 11 letters
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
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 |
In a game there are 70 people in which 40 are boys and 30 are girls, out of which 10 people are selected at random. One from the total group, thus selected is selected as a leader at random. What is the probability that the person, chosen as the leader is a boy?
4/7 | |
4/9
| |
5/7
| |
2/7 |
Question 30 Explanation:
The total groups contains boys and girls in the ratio 4 : 3
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
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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