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# Probability Questions For IBPS RRB PO

Download Top-20 IBPS RRB PO Probability Questions PDF. Probability questions based on asked questions in previous year exam papers very important for the IBPS RRB PO (Officer Scale-I, II & III) exam

Question 1:Â In a bag there are 7 red balls, 5 blue balls and 4 green balls and if 3 balls are picked from the bag then what is the probability that all the balls are of the same colour ?

a)Â 7/80

b)Â 9/80

c)Â 11/80

d)Â 13/80

e)Â 17/80

Question 2:Â What is the probability of selecting 2 red balls and 1 blue ball in the particular given order one at a time from a bag consisting of 6 red balls and 3 blue balls ?

a)Â 5/26

b)Â 5/28

c)Â 5/21

d)Â 15/28

e)Â 5/7

Question 3:Â 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)Â 0.33

b)Â 0.40

c)Â 0.50

d)Â 0.25

e)Â 0.55

Question 4:Â Ravi rolled a dice twice. What is the probability that sum of both the outcome is less than 5?

a)Â 1/6

b)Â 1/4

c)Â 1/3

d)Â 1/2

e)Â None of the above.

Question 5:Â 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)Â 21/64

b)Â 3/32

c)Â 41/64

d)Â 21/32

e)Â 1/2

Question 6:Â A fair dice is rolled twice. Find the probability that he gets at least one composite number.

a)Â $\frac{5}{36}$

b)Â $\frac{3}{4}$

c)Â $\frac{5}{9}$

d)Â $\frac{4}{9}$

e)Â $\frac{5}{12}$

Question 7:Â 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)Â 93.75%

b)Â 92.25%

c)Â 95.75%

d)Â 96.25%

e)Â 94.75%

Question 8:Â Amit rolls 2 dies. What is the probability that the number that turns up on both the dies is a prime number?

a)Â $\frac{3}{8}$

b)Â $\frac{4}{9}$

c)Â $\frac{1}{2}$

d)Â $\frac{1}{3}$

e)Â $\frac{1}{4}$

Question 9:Â 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)Â 10%

b)Â 25%

c)Â 20%

d)Â 50%

e)Â 35%

Question 10:Â 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)Â 45/153

b)Â 55/153

c)Â 65/153

d)Â 4/9

e)Â 58/153

Question 11:Â 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)Â 1/3

b)Â 1/9

c)Â 2/9

d)Â 1/6

e)Â 2/3

Question 12:Â 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)Â 1/55

b)Â 54/55

c)Â 45/55

d)Â 3/55

e)Â 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 13:Â If 3 marbles are drawn at random, what is the probability that none is red ?

a)Â ${3 \over 8}$

b)Â ${1 \over {16}}$

c)Â ${1 \over {10}}$

d)Â ${3 \over {16}}$

e)Â None of these

Question 14:Â If 2 marbles are drawn at random, what is the probability that both are green?

a)Â ${1 \over 8}$

b)Â ${5 \over {16}}$

c)Â ${2 \over 7}$

d)Â ${3 \over 8}$

e)Â None of these

Question 15:Â If 4 marbles are drawn at random, what is the probability that 2 are red and 2 are blue ?

a)Â ${{11} \over {16}}$

b)Â ${3 \over {16}}$

c)Â ${11 \over {72}}$

d)Â ${3 \over {65}}$

e)Â None of these

Instructions

Study the following information carefully to answer the questions that follow :
A box contains 2 blue caps, 4 red caps, 5 green caps and 1 yellow cap.

Question 16:Â If three caps are picked at random, what is the probability that two are red and one is green ?

a)Â latek1

b)Â latek2

c)Â latek3

d)Â latek4

e)Â None of these

Question 17:Â If two caps are picked at random, what is the probability that at least one is red ?

a)Â latek1

b)Â latek2

c)Â latek3

d)Â latek4

e)Â None of these

Question 18:Â If one cap is picked at random, what is the probability that it is either blue or yellow ?

a)Â latek1

b)Â latek2

c)Â latek3

d)Â latek4

e)Â None of these

Question 19:Â If two caps are picked at random, what is the probability that both are blue ?

a)Â latek1

b)Â latek2

c)Â latek3

d)Â latek4

e)Â None of these

Question 20:Â If four caps are picked at random, what is the probability that none is green ?

a)Â latek1

b)Â latek2

c)Â latek3

d)Â latek4

e)Â None of these

Total number of balls=7+5+4=16
Probability of having three balls as red=7C3/16C3
=1/16
Probability of having three balls as blue=5C3/16C3
=1/56
Probability of having three balls as green=4C3/16C3
=1/140
So total probability=(1/16)+(1/56)+(1/140)
= (49/560) = (7/80)

Given we have 6 red balls and 3 blue balls
So the required probability is $\dfrac{6_{C_{1}}}{9_{C_1}} \times \dfrac{5_{C_{1}}}{8_{C_1}} \times \dfrac{3_{C_{1}}}{7_{C_1}}$
=(6*5*3)/(9*8*7)
=5/28

All the multiples of 4 till 50 are 12 i.e from 4,8â€¦â€¦.48
All the multiples of 5 till 50 are 10 i.e from 5,10â€¦.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

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 $= \frac{ \text{Favourable cases}}{\text{Total cases}}$
$= \frac{6}{36} =\frac{1}{6}$

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.

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 = $1 – \frac{4}{9}$ = $\frac{5}{9}$

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.

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 = Â½
Now the outcomes of both the dies are independent of each other. Hence, the probability that both dies will have a prime outcome = $\frac{1}{2}*\frac{1}{2}$ = $\frac{1}{4}$

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.

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.

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).

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)

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

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

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

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

Number of ways of drawing 4 marbles out of 16

=> $n(S) = C^{16}_4 = \frac{16 \times 15 \times 14 \times 13}{1 \times 2 \times 3 \times 4}$

= $1820$

Out of the four drawn marbles, 2 are red and 2 are blue.

=> $n(E) = C^8_2 \times C^3_2 = \frac{8 \times 7}{1 \times 2} \times \frac{3 \times 2}{1 \times 2}$

= $28 \times 3 = 84$

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

= $\frac{84}{1820} = \frac{3}{65}$