RRB NTPC - Probability -Aptitude

1.

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

2.

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.

3.

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

4.

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.

5.

If P(A) = 4/5 and P (B) = 2/5, find P (A n B) if A and B are independent events. 

   A.) 8/23
   B.) 8/25
   C.) 3/25
   D.) None of these

Answer: Option 'B'

P (A n B) = P(A) . P(B) 
P (A n B) = 4/5 . 2/5 
P (A n B) = 8/25.

6.

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.

7.

6 Coins are tossed simultaneously. find the probability to get 2 hands 

   A.) 15/32
   B.) 5/64
   C.) 15/64
   D.) None of these

Answer: Option 'C'

26 = 64, ( 1 = 2, 2 = 4, 4 = 16, 16 = 32, 32=64 times ) 
6c2 = 6!/2! = (6 × 5 × 4 × 3 × 2 × 1)/ (6 - 2)! × (2 × 1) = 15
Probability = 15/64.

8.

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

9.

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

10.

What is the probability that when a hand of 6 cards is drawn from a well shuffled deck of 52 cards, it contains 2 Queen 

   A.) 291/1017926
   B.) 29187/1017926
   C.) 29187/101792
   D.) 2987/101926

Answer: Option 'B'

nCr = n!/(n-r)!r!
Total number of possible hands = 52C6
52C6 = (52!)/((52-6)! × 6!)
52C6 = 61075560.
Number of hands with 2 Queen and 4 non-Queen cards = 4C2 × 48C4
4C2 = (4!)/(2! × 2!) = 6.
48C4 = (48!)/(44! × 4!) = 3 × 47 × 46 × 45 = 291870
(other 2 cards must be chosen from the rest 48 cards)
P (2 Queen) = (4C2 × 48C4)/52C6 = 29187/1017926.

11.

Let A and B be independent events with P (A) = 0.7 and P(B) = 0.7. Find P(A n B)? 

   A.) 4.9
   B.) 0.049
   C.) 0.49
   D.) None of these

Answer: Option 'C'

P (A n B) = P(A) . P(B) 
P (A n B) = 0.7 . 0.7 
P (A n B) = 0.47.

12.

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.

13.

What is the probability that when a hand of 5 cards is drawn from a well shuffled deck of 52 cards,

   A.) 192/37015
   B.) 182/379015
   C.) 192/37901
   D.) 192/379015

Answer: Option 'D'

nCr = n!/(n-r)!r! 
Total number of possible hands = 52C5
52C5 = 2274090 
Number of hands with 4 Queens = 4C4 × 48C1 
4C4 = 24 
4C1 = 48 
(other 1 card must be chosen from the rest 48 cards) 
Hence P (a hand will have 4 Queens) = (4C4 × 48C1)/52C5 = 192/379015.

14.

The probability of obtaining an even prime number on each die, when a pair of dice is rolled 

   A.) 0
   B.) 1/3
   C.) 1/26
   D.) 1/36

Answer: Option 'D'

The probability of getting an even prime (2) for a pair of dice is only once.
That is 1 on each die.
probability of getting 1 on a die is 1/6.
probability of getting 1 on 2 dice is 1/6 × 1/6 = 1/36.

15.

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 

16.

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.

17.

If A and B are two events such that P (A) = 3/4, P (B) = 1/2 and P (A n B) = 3/8, find P (not A and not B). 

   A.) 3/8
   B.) 1/8
   C.) 1/4
   D.) None of these

Answer: Option 'B'

P(not A and not B) = 1 - (P(A) + P(B) - P(AB)) 
which you might find somewhere in your text.
Substituting in our probabilities we get: 
P(not A and not B) = 1 - (3/4 + 1/2 - 3/8) 
P(not A and not B) = 1 - (7/8) 
P(not A and not B) = 1/8.

18.

Given three identical boxes I, II and III, each containing two coins. In box I, both coins are 1 ruppe coins, in box II, both are 2 rupee coins and in the box III, there is one 1 rupee and one 2 rupee coin.  A person chooses a box at random and takes out a coin. If the coin is of 1 rupee, what is the probability that the other coin in the box is also of 1 rupee? 

   A.) 2/9
   B.) 2/7
   C.) 2/3
   D.) 2/5

Answer: Option 'C'

Let E1, E2 and E3 be the events that boxes I, II and III are chosen, respectively. 
Then P(E1) = P(E2) = P(E3) = 1/3 
Also, let A be the event that ‘the coin drawn is of 1 rupee' 
Then P(A|E1) = P(a 1 rupee coin from bag I) = 2/2 = 1 
P(A|E2) = P(a 1 rupee coin from bag II) = 0 
P(A|E3) = P(a 1 rupee coin from bag III) = 1/2 
the probability that the other coin in the box is of 1 rupee 
= the probability that 1 rupee coin is drawn from the box I. 
= P(E1|A) 
from Bayes' theorem, we know that
P(E1|A) = [P(E1 )P(A|E1 )]/ [P(E1 )P(A|E1 )+P(E2 )P(A|E2 )+P(E3 )P(A|E3 )] 
= (1/3 × 1)/(1/3 × 1 + 1/3 × 0 + 1/3 × 1/2)
= (1/3)/(1/3 × 1 + 1/3 × 0 + 1/3 × 1/2) 
= (1/3)/(1/3 × 1/6) = (1/3)/(3/6) = (1/3)/2 = 2/3

19.

Let A and B be independent events with P (A) = 0.2 and P(B) = 0.8. Find P(A/B)? 

   A.) 0.2
   B.) 0.3
   C.) 1.2
   D.) None of these

Answer: Option 'A'

P(A/B) = P (A n B)/P(B) 
Here, P (A n B) = 0.16 
P(A/B) = 0.16/0.8 = 0.2

20.

A die is thrown. If G is the event 'the number appearing is a multiple of 3' and H be the event 'the number appearing is even' then find whether G and H are independent ? 

   A.) G and H are not independent events.
   B.) G and H are independent events.
   C.) Only G independent event
   D.) None of these

Answer: Option 'B'

We know that the sample space is S = {1, 2, 3, 4, 5, 6}
Now G = { 3, 6}, F = { 2, 4, 6} and E n F = {6}
Then P(G) = 2/6 = 1/3 
P(H) = 3/6 = 1/2 and P(G ∩ H) = 1/6 
P(G n H) = P(G). P (H) 
G and H are independent events.


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