1.
6 Coins are tossed simultaneously. find the probability to get 2 hands
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.
2.
If P (A) = 0.18, P (B) = 0.5 and P (B|A) = 0.2, find P(A n B)?
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
3.
If P(A) = 5/13, P(B) = 7/13, and P(A ∩ B) = 8/13, Find P(A ∪ B)?
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.
4.
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?
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
5.
Let A and B be independent events with P (A) = 0.7 and P(B) = 0.7. Find P(A n B)?
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.
6.
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
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.
7.
What is the probability that when a hand of 5 cards is drawn from a well shuffled deck of 52 cards,
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.
8.
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?
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
9.
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.
Answer: Option 'C'
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
10.
A family has two children. find the probability that both the children are girls given that at least one of them is a girl?
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
11.
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 ?
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.
12.
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).
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.
13.
If P(A) = 2/15, P(B) = 4/15, and P(A ∪ B) = 6/15 Find P(A|B)
Answer: Option 'C'
P(A|B) = P(A ∪ B)/P(B)
P(A|B) = (6/15)/(4/15) = 3/2.
14.
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?
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
15.
The probability of obtaining an even prime number on each die, when a pair of dice is rolled
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.