# 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

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

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

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

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