# Quantitative Aptitude: Probability Questions – Set 7

1. Ticket numbered from 1 to 30 are mixed up and a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 4 or 6.
A) 6/13
B) 3/10
C) 5/9
D) 1/10
E) None
Option B
Solution:

n(s)=30
Multiple of 4 ={4,8,12,16,20,24,28}
Multiple of 6={6,12,18,24}
Then n(e)={4,6,8,12,16,18,20,24,28}|
P=n(e)/n(s)=9/30=3/10.
2. The probability of success of three students A, B and C in the one examination are 1/5, 1/4 and 1/3 respectively. Find the probability of success of at least two.
A) 2/3
B) 2/6
C) 1/6
D) 1/3
E) None
Option C
Solution:
P(A)=1/5, P(B)=1/4, P(C)=1/3
Required probability
[ P(A)P(B){1−P(C)} ] + [ {1−P(A)}P(B)P(C) ] + [ P(A)P(C){1−P(B)} ] + P(A)P(B)P(C)
= [1/4*1/3*4/5] + [3/4*1/3*1/5] + [2/3*1/4*1/5] + [1/4*1/3*1/5]
=4/60 + 3/60 + 2/60 + 1/60 ==>10/60=1/6.
3. A Bag contains 100 balloons, numbered from 1 to 100. If three balloons are selected at random and with replacement from the bag, what is the probability that the sum of the three numbers on the balloons selected from the bag will be odd?
A) 1/2
B) 1/4
C) 2/3
D) 1/5
E) None
Option A
Solution:

P(odd) =P (even) = 1/2 (because there are 50 odd and 50 even numbers)
Sum of the three numbers can be odd only under the following 4 scenarios:
Odd + Odd + Odd = 1/2*1/2*1/2 =1/8
Odd + Even + Even =1/2*1/2*1/2 =1/8
Even + Odd + Even = 1/2*1/2*1/2 =1/8
Even + Even + Odd =1/2*1/2*1/2 =1/8
4. The probability of occurrence of two events A and B are 1/5 and 1/3 resp. the probability of their simultaneous occurrence is 6/35. find the probability that either A or B must occur.
A) 44/110
B) 52/105
C) 49/120
D) 38/105
E) None
Option D
Solution:

P(A) = 1/5 , P(B) = 1/3, P(Both A and B)=6/35
Then P(A or B)= P(A)+ P(B)- P(Both A and B)
=1/5+1/3-6/35=38/105.
5. In a race, the odd favour of race bike A,B,C,D are 1:2, 1:3, 1:4 and 1:5 respectively. Find the probability that one of them wins the race.
A) 48/60
B) 57/60
C) 52/60
D) 59/60
E) None
Option B
Solution:

P(A)=1/3, P(B)=1/4, P(C)=1/5, P(D)=1/6
All the events are mutually exclusive hence,
Required probability = P(A)+P(B)+P(C)+P(D)
=1/3+1/4+1/5+1/6
=57/60.
6. In a party there are 6 couples. Out of them 5 people are chosen at random. Find the probability that there are at the least two couples ?
A) 4/33
B) 8/25
C) 5/33
D) 6/21
E) None
Option C
Solution:

Number of ways of (selecting at least two couples among five people selected) = 6C2*8C1
As remaining person can be any one among three couples left.
Required probability = 6C2*8C1/12C5
= 5/33.
7. Out of 25 applicants 13 boys and 12 girls. Two persons are to be selected for the job. Find the probability that at least one of the selected persons will be a boy.
A) 32/50
B) 42/50
C) 39/50
D) 28/50
E) None
Option C
Solution:

13C1*12C1/25C2 + 13C2 *12C0/25C2
=13*12/25*12 + 13*6*1/25*12
=13/25 + 13/50==> 39/50.
8. A coin is tossed twice if the coin shows head it is tossed again but if it shows a tail then a die is tossed. If 8 possible outcomes are equally likely. Find the probability that the die shows a number greater than 4, if it is known that the first throw of the coin results in a tail.
A) 2/3
B) 1/4
C) 2/5
D) 1/3
E) None
Option D
Solution:

S = { HH, HT, T1, T2, T 3, T 4, T 5, T6 }
Let A be the event that the die shows a number greater than 4 and B be the event that the first throw of the coin results in a tail then,
A = { T5, T6 }
B = { T1, T2, T 3, T 4, T 5, T6 }
Then P=P(A and B)/P(B)=(2/8)/(6/8)=1/3.

Directions (9-10): A, B and C try to hit a target. A can hit the target 1 times in 3 shorts, B 3 times in 5 shots and C 2 times in 5 shots. Find the probabilities of following events:

1. Two of them hit
A) 7/9
B) 2/3
C) 1/3
D) 1/5
E) None
Option C
Solution:
P(A) = 1/3 P(A’) = 2/3
P(B) = 3/5 P(B’) = 2/5
P(C) = 2/5 P(C’) = 3/5
Exactly two of them hit
= P( A & B hits & C misses ) + P( A & C hits & B misses ) + P( B & C hits & A misses )
= (1/3*3/5*3/5) + (1/3*2/5*2/5) + (3/5*2/5*2/3)
=9/75 + 12/75 + 4/75 ==>25/75=1/3.
2. Two at least hit
A) 12/17
B) 3/5
C) 5/9
D) 31/75
E) None
Option D
Solution:

Alteast two hit
= P(Exactly two hit) + P(all three hit)
=1/3(from problem a) + (1/3*3/5*2/5)
=1/3+6/75=31/75.

## 5 Thoughts to “Quantitative Aptitude: Probability Questions – Set 7”

1. NehA☼♥dear zindagi♥☼

T.Y.

2. Mohammad Samiuddin

n(s)=30
Multiple of 4 ={4,8,12,16,20,24,28}
Multiple of 6={6,12,18,24,30}

30 is also a multiple of 6 do 30 should be included.

Then n(e)={4,6,8,12,16,18,20,24,28,30}|
P=n(e)/n(s)=10/30=1/3.

1. Sathya Narayan

yes. why did the exclude the 30

3. Nishu

%.

4. kumkum ahuja

thanks:)