Probability Questions For IBPS RRB Clerk

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probability questions for ibps rrb clerk
probability questions for ibps rrb clerk

Probability Questions For IBPS RRB Clerk

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Question 1: If a number is randomly selected from the first 50 natural numbers then what is the probability that it is divisible by either 4 or 5 ?

a) 0.33

b) 0.40

c) 0.50

d) 0.25

e) 0.55

Question 2: If all the words that are formed by using the letters in the word “ALIEN” are arranged as listed in the dictionary then how many words are listed before “ILANE” ?

a) 64

b) 60

c) 61

d) 62

e) 63

Question 3: There are 15 chairs in a row and 5 people are to be seated in them such that odd number of chairs should be present between any two people.In how many can it be done ?

a) 70

b) 77

c) 74

d) 72

e) 75

Question 4: There are 4 male and 4 female badminton players and if 2 teams comprising of 1 male and 1 female has to be made then in how many ways can it be done ?

a) 18

b) 80

c) 144

d) 36

e) 72

Question 5: A sum of Rs 80 has to be made by using Rs 1 or Rs 5 coins.In how many different ways can it be done ?

a) 10

b) 17

c) 14

d) 15

e) 16

Question 6: Ravi rolled a dice twice. What is the probability that sum of both the outcome is less than 5?

a) 1/6

b) 1/4

c) 1/3

d) 1/2

e) None of the above.

Question 7: Raghu tosses a coin 6 times. The probability that the number of times he gets heads will not be greater than the number of times he gets tails is

a) 21/64

b) 3/32

c) 41/64

d) 21/32

e) 1/2

Question 8: A fair dice is rolled twice. Find the probability that he gets at least one composite number.

a) $\frac{5}{36}$

b) $\frac{3}{4}$

c) $\frac{5}{9}$

d) $\frac{4}{9}$

e) $\frac{5}{12}$

Question 9: A box contains 5 blue pens, 6 green pens and 10 black pens. The number of ways in which 3 pens can be selected such that at least 2 pens are of the same colour is

a) 1000

b) 1300

c) 1330

d) 1030

e) 1310

Question 10: Raghu has gotten calls for interviews from top 4 management institutes of a country. The probability that he converts an interview of an institute is 50%. What is the probability of Raghu converting the interview of at least 1 management institute?

a) 93.75%

b) 92.25%

c) 95.75%

d) 96.25%

e) 94.75%

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Question 11: Find the number of rectangles in the following diagram:

a) 225

b) 315

c) 360

d) 441

e) 482

Question 12: Amit rolls 2 dies. What is the probability that the number that turns up on both the dies is a prime number?

a) $\frac{3}{8}$

b) $\frac{4}{9}$

c) $\frac{1}{2}$

d) $\frac{1}{3}$

e) $\frac{1}{4}$

Question 13: A circular disc of diameter 3.5 cm is thrown inside a circle of radius 7 cm. What is the probability that the disc will remain completely within the circle?

a) 10%

b) 25%

c) 20%

d) 50%

e) 35%

Question 14: A committee of 5 members is to be formed from 4 men and 4 women. If the committee must have at least 3 men, the number of ways in which the committee can be formed is

a) 24

b) 32

c) 36

d) 40

e) 28

Question 15: A bag contains 10 black balls, 5 red balls and 3 green balls. 2 balls are selected from this bag. What is the probability of that both the balls are of the same colour?

a) 45/153

b) 55/153

c) 65/153

d) 4/9

e) 58/153

Question 16: A dice has its faces numbered from 1 to 6. The dice is biased such that the faces with even numbers are twice as likely as the faces with the odd numbers to show up on the top. If the die is rolled randomly, what is the probability of the die showing number 3?

a) 1/3

b) 1/9

c) 2/9

d) 1/6

e) 2/3

Question 17: Letters of the word DIRECTOR are arranged in such a way that all the vowel come together .Find the No of ways making such arrangement?

a) 4320

b) 720

c) 2160

d) 120

e) None of these

Question 18: In a box carrying one dozen of oranges one third have become bad.If 3 oranges taken out from the box random ,what is the probability that at least one orange out of the 3 oranges picked up is good ?

a) 1/55

b) 54/55

c) 45/55

d) 3/55

e) None of these

Instructions

Study the information carefully to answer the following questions:

A bucket contains 8 red, 3 blue and 5 green marbles.

Question 19: If 3 marbles are drawn at random, what is the probability that none is red ?

a) ${3 \over 8}$

b) ${1 \over {16}}$

c) ${1 \over {10}}$

d) ${3 \over {16}}$

e) None of these

Question 20: If 2 marbles are drawn at random, what is the probability that both are green?

a) ${1 \over 8}$

b) ${5 \over {16}}$

c) ${2 \over 7}$

d) ${3 \over 8}$

e) None of these

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Answers & Solutions:

1) Answer (B)

All the multiples of 4 till 50 are 12 i.e from 4,8…….48
All the multiples of 5 till 50 are 10 i.e from 5,10….50
In the above multiples there are few numbers which are common multiples of 4 and 5 and so they are repeated twice so they are to be removed
LCM of 4 and 5 is 20
Multiples of 20 are 20 and 40 which lie below 50
Therefore total numbers are 12+10-2
=22-2
=20
Probability of selecting a number which is either divisible by 4 or 5 from first 50 natural numbers is 20/50
=2/5
=0.4

2) Answer (C)

Total 5! words can be formed using the 5 letters and so by taking A as the first letter is 4!=24
By taking E as first letter we have 4! words=24
By taking I as first letter and A as the second letter we have 3! Words
By taking I as first letter and E as the second letter we have 3! Words
By taking I as first letter and L as second letter and A as third letter we have 2 words
And the second word is ILANE
So 24+24+6+6+1=61 words come before the word ILANE

3) Answer (B)

In the given condition if all the 5 people are seated in either odd places or even places then the number of chairs between them will be odd and so we have odd places 8 starting from 1,3….15 and even places 7 starting from 2,4…14 and so filing 5 persons in 8 chairs=8C5 =(8*7*6)/(3*2)=56
5 persons sitting in 7 places i.e i 7C5 ways=(7*6)/(1*2)=21
So total number of ways=56+21=77 ways

4) Answer (D)

Two teams are to be selected so 4 players i.e 2 males and 2 females are to be selected.
First select 2 males out of four i.e 4C2=(4*3)/2 =6 ways
then select 2 females out of four i.e 4C2=(4*3)/2 =6 ways
Then first male can be paired with two females and second male can be paired with only one female.
Therefore required ways=6*6*2*1
=72 ways

5) Answer (B)

let the number of Rs 1 coins be x
And the number of Rs 2 coins be y
Therefore x+5y=80
So for every x there should be one y and vice versa
Y take values from y=0 to y=16
I.e for every y from 0 to 16,we will have x
So for y=0 we have x=80
For y=1 we have x=75
So on
Therefore 16+1=17 ways are possible

6) Answer (A)

When a dice is rolled we get outcome from (1,2,3,4,5,6) so favourable cases (Sum = 2,3,4)
Favourable cases = (1,1) (1,2) (2,1) (1,3) (3,1) (2,2)
Total cases = 6*6 = 36
So probability $= \frac{ \text{Favourable cases}}{\text{Total cases}}$
$= \frac{6}{36} =\frac{1}{6}$

7) Answer (D)

Probability of getting a head = 0.5
Probability of getting a tail = 0.5
Number of heads $\leq$ Number of tails.
Therefore, the number of heads can be 0,1, 2 or 3.
Sample space = $2^6$ = 64.
0 heads can be obtained in 6C0 = 1 way.
1 head can be obtained in 6C1 = 6 ways.
2 heads can be obtained in 6C2 = 30/2 = 15 ways.
3 heads can be obtained in 6C3 = 20 ways.

Required probability = (20+15+6+1)/64 = 42/64 = 21/32.
Therefore, option D is the right answer.

8) Answer (C)

The probability that he gets at least one composite number = 1 – p(no composite number)

The numbers which are not composite = 1, 2, 3, 5

Hence, the probability that he gets them both the times = $\frac{4}{6} * \frac{4}{6} = \frac{4}{9}$

Required probability = $1 – \frac{4}{9}$ = $\frac{5}{9}$

9) Answer (D)

Number of ways in which at least 2 pens of the same colour can be selected = Total number of ways in which 3 pens can be selected – Number of ways in which 3 pens of different colour can be selected.

There are 5+6+10 = 21 pens in total.
Total number of ways in which 3 pens can be selected = 21C3 = (21*20*19)/(1*2*3) = 1330 ways.
Number of ways in which 3 pens of different colour can be selected = 5*6*10 = 300 ways

Therefore, Number of ways in which at least 2 pens of the same colour can be selected = 1330 – 300 = 1030. Therefore, option D is the right answer.

10) Answer (A)

Raghu can get selected in 1 college, 2 colleges, 3 colleges or all 4 colleges. Let us subtract the probability of Raghu not getting selected in any of the colleges from the total probability to get the required probability.

Probability of converting the interview of a college = 0.5
Therefore, probability of not converting the interview of a college = 1-0.5 = 0.5
Probability of not converting any of the 4 colleges = 0.5*0.5*0.5*0.5 = 0.0625 or 6.25%
Therefore, probability of converting at least 1 college = 100 – 6.25 = 93.75%.
Therefore, option A is the right answer.

11) Answer (B)

There are 7 horizontal lines and 6 vertical lines in the diagram.

For a rectangle to be formed, 2 out of these 7 horizontal lines and 2 out of these 6 vertical lines must be selected.

Therefore, total number of rectangles = 7C2*6C2 = 21*15 = 315.
Therefore, option B is the right answer.

12) Answer (E)

Total possible outcomes for each dice = 6
Numbers on a dice which are prime = 2, 3, 5
Hence, the probability of getting a prime number on any dice = ½
Now the outcomes of both the dies are independent of each other. Hence, the probability that both dies will have a prime outcome = $\frac{1}{2}*\frac{1}{2}$ = $\frac{1}{4}$

13) Answer (B)

For the disc to remain completely remain within the circle, the centre of the disc must at-least be 3.5cm away from the circumference of the circle.

Radius of the outer circle = $7$ cm
Radius of the inner circle = $7 – 3.5$ = $3.5$ cm.

The centre of the circle must fall within the inner circle. Of the total area of the outer circle, only the inner circle is the area that the centre of the coin can fall within without violating the conditions.

Probability = Ratio of the areas of the $2$ circles.

The radius of the outer circle is twice the radius of the inner circle.
Ratio of areas = $\frac{\pi r^2}{\pi (2r)^2}$
Therefore, the areas will be in the ratio $1:4$.
Therefore, the required probability is $25$%. Hence, option B is the right answer.

14) Answer (E)

The committee can have either 3 men or 4 men.
If 3 men are to be selected, the number of ways in which the committee can be formed is 4C3 * 4C2
= 4*6 = 24 ways.

If 4 men are to be selected, the committee can be formed in 4C4*4C1 = 1*4 = 4 ways.
Therefore, the total number of ways in which the committee can be formed is 24+4 = 28 ways. Hence, option E is the right answer.

15) Answer (E)

The number of ways in which 2 black balls can be selected is 10C2 = 45.
The number of ways in which 2 red balls can be selected is 5C2 = 10.
The number of ways in which 2 green balls can be selected is 3C2 = 3.

Therefore, the total number of ways in which 2 balls can be selected is 45 + 10 + 3 = 58.
There are 18 balls in total (10+5+3)
Number of ways in which 2 balls can be selected = 18C2 = 153.
Therefore, the probability is 58/153.

Hence, option E is the right answer.

16) Answer (B)

Let the probability of an odd face appearing be 1/x and an even face appearing be 2/x.
3*(1/x) + 3*(2/x) = 1 (Since the total probability is 1).
9/x = 1.
=> x= 9.
The probability of the dice showing number 3 is 1/9 (Since it is an odd number).

17) Answer (A)

Word – DIRECTOR

So “I,E,O” are there are 3! ways to arrange the vowels

Now “D,R,C,T,R” are the remaining alphabets ,

Condition is that the vowels should always be together so we can assume the vowels as a single alphabet/unit say “X” (‘X’=’I,E,O’) so now we have a new word – “D,R,C,T,R,X”

Possible arrangements for this word = 6!

Thus total number of ways to rearrange DIRECTOR with vowels grouped together = (Possible arrangements of ‘DRCTRX’) $\times$ (Possible arrangements of vowels)

= 6! $\times$ 3! = $720 \times 6 = 4320$

=> Ans – (A)

18) Answer (B)

Total number of oranges in the box = 12

Number of ways of selecting 3 oranges out of 12 oranges, n(S) = $C^{12}_3$

= $\frac{12 \times 11 \times 10}{1 \times 2 \times 3} = 220$

Number of oranges which became bad = $\frac{12}{3}=4$

Number of ways of selecting 3 oranges out of 4 bad oranges = $C^4_3 = C^4_1 = 4$

Number of desired selection of oranges, n(E) = 220 – 4 = 216

$\therefore$ $P(E) = \frac{n(E)}{n(S)}$

= $\frac{216}{220}= \frac{54}{55}$

=> Ans – (B)

19) Answer (C)

Number of ways of drawing 3 marbles out of 16

$n(S) = C^{16}_3 = \frac{16 \times 15 \times 14}{1 \times 2 \times 3}$

= $560$

Out of the three drawn marbles, none is red, i.e., they will be either blue or green.

=> $n(E) = C^8_3 = \frac{8 \times 7 \times 6}{1 \times 2 \times 3}$

= $56$

$\therefore$ Required probability = $\frac{n(E)}{n(S)}$

= $\frac{56}{560} = \frac{1}{10}$

20) Answer (E)

Number of ways of drawing 2 marbles out of 16

$n(S) = C^{16}_2 = \frac{16 \times 15}{1 \times 2}$

= $120$

Out of the two drawn marbles, both are green

=> $n(E) = C^5_2 = \frac{5 \times 4}{1 \times 2}$

= $10$

$\therefore$ Required probability = $\frac{n(E)}{n(S)}$

= $\frac{10}{120} = \frac{1}{12}$

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