RRB NTPC - Probability -Aptitude

1.

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

2.

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.

3.

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.

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.

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.

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