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 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)
Instructions to take Aptitude Test
- Click on start to start taking the test.
- Click on the option (A, B, C or D) to figure out the right answer.
- You can answer multiple times till you get the right answer.
- Once you get the right answer, explanations (if any) for the same will be showcased down.
- On click of list, you get to see total no of questions, no of questions you answered and no of questions pending to answer.
- On click of question number you will go to that particular question.
- On click of END your test will end.
- On click of ‘Get Results’ you will get to see correct answer for each questions.
Probability - Test
Probability - Question and Answers
You scored %%SCORE%% out of %%TOTAL%%.
Your performance has been rated as %%RATING%%
Question 1
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18/145
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18/29
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36/135
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36/145
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= 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
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1 / 13
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3 / 13
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1 / 4
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9 / 52
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Total no .of cards = 52
So probability = 12 / 52 = 3 / 13
Question 3
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10 / 21
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11 / 21
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2 / 7
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5 / 7
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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
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1 / 6
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1 / 8
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1 / 9
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1 / 12
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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
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1 / 15
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25 / 57
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35 / 256
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1 / 221
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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
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1 / 216
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1 / 36
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4 / 216
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3 / 216
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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
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1 / 10
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2 / 5
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2 / 7
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5 / 7
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Question 8
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1 / 3
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3 / 4
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7 / 19
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8 / 21
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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
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4
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5
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9
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10
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No. of students those did not complete any task: 50 - 45 = 5
Question 10
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30
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35
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40
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60
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200 = 125 + 115 – Both
Both = 240 - 200 = 40
Question 11
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3 / 4
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1 / 4
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3 / 8
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7 / 8
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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
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1 / 22
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3 / 22
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2 / 91
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2 / 77
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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
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21 / 46
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25 / 117
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1 / 50
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3 / 25
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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
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6 / 13
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5 / 26
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5 / 13
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7 / 26
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=> 6C2 / 13C2
=> 5 / 26
Question 15
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1 / 2
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3 / 4
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3 / 8
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5 / 16
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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
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1 / 13
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2 / 13
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1 / 26
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1 / 52
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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
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3 / 20
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29 / 34
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47 100
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13 / 102
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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
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1 / 13
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3 / 13
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1 / 4
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9 / 52
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Therefore, P(Getting a face card) = 12 / 52 = 3 / 13
Question 19
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1 / 2
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2 / 5
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8 / 15
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9 / 20
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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
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22
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32
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33
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44
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(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
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3 / 4
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4 / 7
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1 / 8
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3 / 7
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Number of white balls = 8
P(drawing a white ball) = 8 / 14 = 4 / 7
Question 22
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31
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15
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21
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25
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Total no.of students = 10 + 20 + 5 - 4 - 4 - 2 + 2 = 31
Question 23
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55/58
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40/58
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41/58
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39/58
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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
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1/35
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1/34
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1/17
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1/68
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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
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1/35
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3/7
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34/35
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4/7
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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
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4/15
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69/91
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11/15
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22/91
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=> 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
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70/8281
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140/20449
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25/5445
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35/5448
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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
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2/9
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1/13
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2/13
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1/26
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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
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1663200
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8316000
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3326400
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4158000
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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
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4/7
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4/9
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5/7
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2/7
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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
Done
Hope it has benefitted you!!