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Probability - Quantitative Aptitude Problems and Solutions


  • 1. A family has two children. find the probability that both the children are girls given that at least one of them is a girl? 
A.) 1/4
B.) 2/3
C.) 1/3
D.) 1/4

Answer: Option 'C'

Let b stand for boy and g for girl. The sample space of the experiment is 
S = {(g, g), (g, b), (b, g), (b, b)} 
Let E and F denote the following events : 
E : ‘both the children are girls’ 
F : ‘at least one of the child is a girl’ 
Then E = {(g,g)} and F = {(g,g), (g,b), (b,g)} 
Now E n F = {(g,g)} 
Thus P(F) = 3/4
and P (E n F )= 1/4 
Therefore P(E|F) = P(E ∩ F)/P(F) = (1/4)/(3/4) = 1/3

  • 2. Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number? 
A.) 2/7
B.) 6/7
C.) 7/2
D.) 4/7

Answer: Option 'D'

Let A be the event ‘the number on the card drawn is even’ and B be the 
event ‘the number on the card drawn is greater than 3’. We have to find P(A|B). 
Now, the sample space of the experiment is S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} 
Then A = {2, 4, 6, 8, 10}, B = {4, 5, 6, 7, 8, 9, 10} 
and A n B = {4, 6, 8, 10} 
Also P(A) = 5/2, P(B) = 7/10 and P(A n B) = 4/10 
Then P(A|B) = P(A n B)/P(B) = (4/10)/(7/10) = 4/7

  • 3. A die is thrown three times. Events X and Y are defined as below:
    X : 4 on the third throw 
    Y : 6 on the first and 5 on the second throw 
    What is the probability of X given that Y has already occurred. 
A.) 2/6
B.) 1/6
C.) 1/5
D.) None of these

Answer: Option 'B'

The sample space has 216 outcomes.
Now X = (1,1,4) (1,2,4) ... (1,6,4) (2,1,4) (2,2,4) ... (2,6,4)
(3,1,4) (3,2,4) ... (3,6,4) (4,1,4) (4,2,4) ...(4,6,4)
(5,1,4) (5,2,4) ... (5,6,4) (6,1,4) (6,2,4) ...(6,5,4) (6,6,4)
Y = {(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6)}
and X n Y = {(6,5,4)}.
Now P(Y) = 6/216 
and P (X n Y) = 1/216 

  • 4. A die is thrown twice and the sum of the numbers appearing is observed to be 6. find the conditional probability that the number 4 has appeared at least once? 
A.) 1/5
B.) 3/5
C.) 2/5
D.) None of these

Answer: Option 'C'

Let E be the event that ‘number 4 appears at least once’ and F be the event that ‘the sum of the numbers appearing is 6’. 
Then, E = {(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4), (5,4), (6,4)} 
and F = {(1,5), (2,4), (3,3), (4,2), (5,1)} 
We have P(E) = 11/36 
and P (F) = 5/36 
Also EnF = {(2,4), (4,2)} 
Therefore P(EnF) = 2/36 
the required probability
P(E|F) = P(EnF)/P(F) = (2/36)/(5/36) = 2/5.

  • 5. Given that E and F are events such that P(E) = 0.16, P(F) = 0.4 and P(E n F) = 0.4,
    find P (E|F) and P(F|E) 
A.) 3/4,2
B.) 1/4,1
C.) 1,1/4
D.) None of these

Answer: Option 'C'

Here, E and F are events 
P(E|F) = P(EnF)/P(F) = 0.4/0.4 = 1 
P(F|E) = P(EnF)/P(E) = 0.4/0.16 = 1/4.

  • 6. If P (A) = 0.18, P (B) = 0.5 and P (B|A) = 0.2, find P(A n B)? 
A.) 0.32
B.) 0.36
C.) 0.16
D.) 0.64

Answer: Option 'B'

P(B|A) = P(A n B)/P(A) 
P(A n B) = P(B|A) × P(A) 
P(A n B) = 0.2 × 0.18 
P(A n B) = 0.36

  • 7. If P(A) = 5/13, P(B) = 7/13, and P(A ∩ B) = 8/13, Find P(A ∪ B)? 
A.) 4/13
B.) 5/13
C.) 6/13
D.) None of these

Answer: Option 'A'

P(A ∩ B) = P(A) + P(B) - P(A ∪ B) 
= 8/13 = 5/13 + 7/13 - P(A ∪ B) 
= P(A ∪ B) = 5/13 + 7/13 - 8/13 
= P(A ∪ B) = 4/13.

  • 8. If P(A) = 2/15, P(B) = 4/15, and P(A ∪ B) = 6/15 Find P(A|B) 
A.) 6/15
B.) 3/4
C.) 3/2
D.) None of these

Answer: Option 'C'

P(A|B) = P(A ∪ B)/P(B) 
P(A|B) = (6/15)/(4/15) = 3/2.

  • 9. If P(A) = 6/17, P(B) = 5/17, and P(A ∪ B) = 4/17 Find P(B|A)? 
A.) 6/3
B.) 2/5
C.) 2/7
D.) 2/3

Answer: Option 'D'

P(B|A) = P(A ∪ B)/P(A) 
P(B|A) = (4/17)/(6/17) = 4/6 = 2/3.

  • 10. An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black? 
A.) 1/7
B.) 2/7
C.) 7/3
D.) 3/7

Answer: Option 'D'

Let E and F denote respectively the events that first and second ball drawn are black. We have to find P(E n F) or P (EF).
Now P(E) = P (black ball in first draw) = 10/15 
Also given that the first ball drawn is black, i.e., event E has occurred, now there are 9 black balls and five white balls left in the urn. Therefore, the probability that the second ball drawn is black, given that the ball in the first draw is black, is nothing but the conditional probability of F given that E has occurred.
That is P(F|E) = 9/14 
By multiplication rule of probability, we have 
P (E n F) = P(E) P(F|E) 
= 10/15 × 9/14 = 3/7



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