## RRB NTPC - Probability -Aptitude

• 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

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

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

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

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

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

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

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

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

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

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

• 11. Three cards are drawn successively, without replacement from a pack of 52 well shuffled cards. What is the probability that first two cards are queens and the third card drawn is an ace?
A.) 2/5530
B.) 3/5525
C.) 2/5525
D.) 4/5525

Let Q denote the event that the card drawn is queen and A be the event that
the card drawn is an ace. Clearly, we have to find P (QQA)
Now P(Q) = 4/52
Also, P (Q|Q) is the probability of second queen with the condition that one queen has
already been drawn. Now there are three queen in (52 - 1) = 51 cards.
Therefore P(Q|Q) = 3/51
P(A|QQ) is the probability of third drawn card to be an ace, with the condition
that two queens have already been drawn. Now there are four aces in left 50 cards.
Therefore P(A|QQ) = 4/50
By multiplication law of probability, we have
P(QQA) = P(Q) P(Q|Q) P(A|QQ)
= 4/52 × 3/51 × 4/50
= 2/5525.

• 12. 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

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.

• 13. An unbiased die is thrown twice. Let the event A be ‘odd number on the first throw’ and B the event  ‘odd number on the second throw’. Check the independence of the events A and B.
A.) A or B are independent events
B.) A and B are not independent events
C.) A and B are independent events
D.) None of these

If all the 36 elementary events of the experiment are considered to be equally likely, we have
P(A) = 18/36 = 1/2
= and P(B) = 18/36 = 1/2
Also P(A n B) = P (odd number on both throws)
= 9/36 = 1/4
Now P(A) P(B) = 1/2 × 1/2 = 1/4
P(A n B) = P(A) × P(B)
A and B are independent events

• 14. 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

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.

• 15. 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

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

• 16. 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

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

• 17. 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

• 18. Let A and B be independent events with P (A) = 0.13 and P(B) = 0.3. Find P(B/A)?
A.) 2.34
B.) 1.3
C.) 0.3
D.) None of these

P(B/A) = P (A n B)/P(A)
Here, P (A n B) = P(A) × P(B) = 0.13 × 0.3 = 0.39
P(B/A) = 0.39/0.13 = 0.3

• 19. 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

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.

• 20. Events A and B are such that P (A) = 1/3, P(B) = 7/6, and P(not A or not B) = 1/4. State whether A and B are independent?
A.) A and B are independent
B.) A and B are not independent
C.) A and B are neither or not independent
D.) None of these

P(A) = 1/3, P(B) = 7/6
Also P(A n B) = P
P(A n B) = 1/4
Now P(A) P(B) = 1/3 × 7/6 = 7/18
But P(A n B) = 1/4
Clearly P(A n B) ≠ P(A) × P(B)
Thus, A and B are not independent events

• 21. 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

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.

• 22. 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

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

• 23. 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.

• 24. 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

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.