# Quantitative Aptitude: Probability Questions – Set 4

Probability Questions for Bank PO Exams – SBI PO, NIACL, NICL, BoB PO, IBPS PO, and other exams

1. A bag contains 6 red, 2 blue and 4 green balls. 3 balls are chosen at random. What is the probability that at least 2 balls chosen will be red?
A) 2/7
B) 1/2
C) 1/3
D) 2/5
E) 3/7
Option B
Solution:
There will be 2 cases
Case 1: 2 red, 1 blue orgreen
Prob. = 6C2 × 6C1 / 12C3 = 9/22
Case 2: all 3 red
Prob. = 6C3 / 12C3 = 2/22
Add the cases, required prob. = 9/22 + 2/22 = 11/22 = 1/2
2. Tickets numbered 1 to 250 are in a bag. What is the probability that the ticket drawn has a number which is a multiple of 4 or 7?
A) 83/250
B) 89/250
C) 77/250
D) 93/250
E) 103/250
Option B
Solution:
Multiples of 4 up to 120 = 250/4 = 62
Multiples of 7 up to 120 = 250/7 = 35 (take only whole number before the decimal part)
Multiple of 28 (4×7) up to 250 = 250/28 = 8
So total such numbers are = 62 + 35 – 8 = 89
So required probability = 89/250
3. From a deck of 52 cards, 3 cards are chosen at random. What is the probability that all are face cards?
A) 14/1105
B) 19/1105
C) 23/1105
D) 11/1105
E) 26/1105
Option D
Solution:
There are 3*4 = 12 face cards in 52 cards
So required probability = 12C3 / 52C3 = 11/1105
4. One 5 letter word is to be formed taking all letters – S, A, P, T and E. What is the probability that this the word formed will contain all vowels together?
A) 2/5
B) 3/10
C) 7/12
D) 3/5
E) 5/12
Option A
Solution:
Total words that can be formed is 5! = 120
Now vowels together:
Take: S, P, T and AE
So their arrangement is 4! * 2! = 48
So required probability = 48/120 = 2/5
5. One 5-digit number is to be formed from numbers – 0, 1, 3, 5, and 6 (repetition not allowed). What is the probability that number formed will be even?
A) 8/15
B) 7/16
C) 7/15
D) 3/10
E) 13/21
Option B
Solution:
Two cases:
Case 1: 0 at last place
So 4 choices for 1st digit, 3 for 2nd, 2 for 3rd and 1 for 4th. So numbers = 4*3*2*1 = 24
Case 2: 6 at last place
For 5-digit number 0 cannot be placed at 1st place or cannot be 1st digit
So 3 choices (1, 3, 5) for 1st digit, 3 for 2nd, 2 for 3rd and 1 for 4th. So numbers = 3*3*2*1 = 18
So total choices = 24+18 = 42
Number total 5-digit numbers that can be formed from 0, 1, 3, 5, and 6
0 not allowed at 1st place, so 4 choices for 1st place, 4 for 2nd, 3 for 3rd, 2 for 4th and 1 for 5th. Sp total = 4*4*3*2*1 = 96
So required probability = 42/96 = 7/16

Directions (6-8): There are 3 bags containing 3 colored balls – Red, Green and Yellow.
Bag 1 contains:
24 green balls. Red balls are 4 more than blue balls. Probability of selecting 1 red ball is 4/13

Bag 2 contains:
Total balls are 8 more than 7/13 of balls in bag 1. Probability of selecting 1 red ball is 1/3. The ratio of green balls to blue balls is 1 : 2

Bag 3 contains:
Red balls equal total number of green and blue balls in bag 2. Green balls equal total number of green and red balls in bag 2. Probability of selecting 1 blue ball is 3/14.

1. 1 ball each is chosen from bag 1 and bag 2, What is the probability that 1 is red and other blue?
A) 15/128
B) 21/115
C) 17/135
D) 25/117
E) 16/109
Option D
Solution:
Let red = x, so blue = x-4
So
x/(24+x+(x-4)) = 4/13
Solve, x = 16
So bag 1: red = 16, green  = 24, blue = 12
NEXT:
bag 2: total = 8 + 7/13 * 52 = 36
green and blue = y and 2y. Let red balls = z
So z + y + 2y = 36…………………(1)
Now Prob. of red = 1/3
So z/36 = 1/3
Solve, z = 12
From (1), y = 8
So bag 2: red = 12, green  = 8, blue = 16
NEXT:
bag 3: red = 8+16 = 24, green = 12+8 = 20
Blue prob. = 3/14
So a/(24+20+a) = 3/14
Solve, a = 12
So bag 3: red = 24, green  = 20, blue = 12
Now probability that 1 is red and other blue::
16/52 * 16/36 + 12/52 * 12/36 = 25/117
2. Some green balls are transferred from bag 1 to bag 3. Now probability of choosing a blue ball from bag 3 becomes 3/16. Find the number of remaining balls in bag 1.
A) 60
B) 58
C) 52
D) 48
E) 44
Option E
Solution:
blue balls in bag 3 are 12
Let x green balls are transferred. So
12/(56+x) = 3/16   [56 was number of bags in bag 3 before transfer] Solve, x = 8
So remaining number of balls in bag 1 = 52-8 = 44
3. Green balls in ratio 4 : 1 from bags 1 and 3 respectively are transferred to bag 4. Also 4 and 8 red balls from bags 1 and 3 respectively . Now probability of choosing green ball from bag 4 is 5/11. Find the number of green balls in bag 4?
A) 12
B) 15
C) 10
D) 9
E) 11
Option C
Solution:
4x and x = 5x green balls
4+8 = 12 red balls
So 5x/(5x+12) = 5/11
Solve, x = 2
5*2 = 10 green balls

Directions (9-10): There are 3 people – A, B and C. Probability that A speaks truth is 3/10, probability that B speaks truth is 3/7 and probability that C speaks truth is 5/6. For a particular question asked, at most 2 people speak truth. All people answer to a particular question asked.

1. What is the probability that B will speak truth for a particular question asked?
A) 7/18
B) 14/33
C) 4/15
D) 9/28
E) 10/33
Option D
Solution:
In any case B speaks truth. Now at most 2 people speak truth for 1 question
So case 1: B and A speaks truth
Probability = 3/7 * 3/10 * (1-5/6) = 3/140
Case 2: B and C speaks truth
Probability = 3/7 * ( 1- 3/10) * 5/6 = 5/20
Case 3: Only B speaks truth
Probability = 3/7 * ( 1- 3/10) * (1-5/6) = 1/20
Add the three cases = 6/20 + 3/140 = 45/140 = 9/28
2. A speaks truth only when B does not speak truth, then what is the probability that C does not speak truth on a question?
A) 11/140
B) 21/180
C) 22/170
D) 13/140
E) None of these
Option A
Solution:
Case 1: B does not speak truth, A speaks truth
So A speaks truth here
Probability that C does not speak truth = 3/10 * (1 – 3/7) * ( 1- 5/6) = 1/35
Case 2: B speaks truth
So A does not speak truth here
Probability that C does not speak truth = ( 1- 3/10) * 3/7 * ( 1- 5/6) = 1/20
So total = 1/35 + 1/20 = 11/140

## 17 Thoughts to “Quantitative Aptitude: Probability Questions – Set 4”

1. ครђ

Nice questions …Thankyou Shubhra ma’am 🙂

2. gomathy priya

ty az:))

3. Debapriyo Mustafi

Mam/Sir,
I am having bit of a problem on understanding question no 9.
Why the value of C was included when we are just finding the probability of B and A speaking the truth ??

1. Shubhra

See this
All people answer to a particular question asked.

1. Player

(6-8)
Bag 3 contains:
Probability of selecting 1 blue ball is 1/3…..I think misprint in this value..

in sol : it is given
Blue prob. = 3/14……

1. Shubhra

yes 3/14

1. Player

A speaks truth only when B does not speak truth, then what is the probability that C does not speak truth on a question?
A) 11/140
B) 21/180
C) 22/170
D) 13/140
E) None of these

mam isme ek condition ye bhi to hogi when both A & B speak truth …because usne atmost 2 people speak truth bola h
than answer will be
1/10…….E hoga

2. Shubhra

But agar B truth bolega. To A kese truth bolega.
A speaks truth only when B lies

3. Player

ok mam …thanku 🙂

4. Alice

atmost 2 speak truth
it means none of them may nt speak truth,all three lie can be a condition?

2. EKTA

hello mam i am not getting question no 10 ..pls explain mam …thank u .. questions level is soo good

4. @chasmish_/_

Mam please explain ques 4
In solution why 4!×2!??

1. Shubhra

S , , P , T and AE together.
So 4 ! For this?

Next AE k liye 2!.
Got?

1. @chasmish_/_

Yes mam got???
Thankyou?

5. Ayushi

thanku mam plz upload more que on probability … 🙂

6. Ashin Wirathu

Thanks:)

7. kumkum ahuja

thanks:)