1. If A is a perfect subset of B and P(a < Pb), then P(B – A) is equal to ____________
a) P(a) / P(b)
b) P(a)P(b)
c) P(a) + P(b)
d) P(b) – P(a)
Explanation:From Basic Theorem of probability,
P(B – A) = P(b) – P(a), this is true only if the condition given in the question is true.
2. What is the probability of an impossible event?
a) 0
b) 1
c) Not defined
d) Insufficient data
Explanation: If the probability of an event is 0, then it is called as an impossible event.
3. If A = A1 ∪ A2……..∪ An, where A1…An are mutually exclusive events then?
a) \(\sum_{i=0}^n P(A_i)\)
b) \(\sum_{i=1}^n P(A_i)\)
c) \(\prod_{i=0}^n P(A_i)\)
d) Not defined
Explanation: A = A1 ∪ A2……..∪ An, where A1…An
Since A1…An are mutually exclusive
P(a) = P(A1) + P(A2) + … + P(An)
Therefore p(a)=\(\sum_{i=1}^n P(A_i)\)
4. If P(B⁄A) = p(b), then P(A ∩ B) = ____________
a) p(b)
b) p(a)
c) p(b).p(a)
d) p(a) + p(b)
Explanation: P(B/A) = p(b) implies A and B are independent events
Therefore, P(A ∩ B) = p(a).p(b).
5. Two unbiased coins are tossed. What is the probability of getting at most one head?
a) 1⁄2
b) 1⁄3
c) 1⁄6
d) 3⁄4
Explanation: Total outcomes = (HH, HT, TH, TT)
Favorable outcomes = (TT, HT, TH)
At most one head refers to maximum one head,
Therefore, probability = 3⁄4.
6. If A and B are two events such that p(a) > 0 and p(b) is not a sure event, then P(A/B) = ?
a) 1 – P(A /B)
b) \(P\frac{(\bar{A})}{(\bar{B})}\)
c) Not Defined
d) \(\frac{1-P(A \cup B)}{P(\bar{B}}\)
Explanation:From definition of conditional probability we have
\(P(\bar{A}/\bar{B})=\frac{\bar{A}\cap\bar{B}}{P(\bar{B})}\)
Using De Morgan’s Law
=\(\frac{P(\bar{A \cup B})}{(\bar{B}}\)
=\(\frac{1-P(A \cup B)}{P(\bar{B}}\)
7. If A and B are two events, then the probability of exactly one of them occurs is given by ____________
a) P(A ∩ B) + P(A ∩ B)
b) P(A) + P(B) – 2P(A) P(B)
c) P(A) + P(B) – 2P(A) P(B)
d) P(A) + P(B) – P(A ∩ B)
Explanation: The set corresponding to the required outcome is
(A ∩ B) ∪ (A ∩ B)
Hence the required probability is
P(P(A ∩ B) ∪ (A ∩ B)) = (A ∩ B) + P(A ∩ B).
8. The probability that at least one of the events M and N occur is 0.6. If M and N have probability of occurring together as 0.2, then P(M) + P(N) is?
a) 0.4
b) 1.2
c) 0.8
d) Indeterminate
Explanation: Given : P(M ∪ N) = 0.6, P(M ∩ N) = 0.2
P(M ∪ N) + P(M ∩ N) = P(M) + P(N)
2 – (P(M ∪ N) + P(M ∩ N)) = 2 – (P(M) + P(N))
= (1 – P(M)) + (1 – P(N))
2 – (0.6 + 0.2) = P(M) + P(N)
P(M) + P(N) = 2 – 0.8
= 1.2
9. A jar contains ‘y’ blue colored balls and ‘x’ red colored balls. Two balls are pulled from the jar without replacing. What is the probability that the first ball Is blue and second one is red?
a) \(\frac{xy-y}{x^2+y^2+2xy-(x+y)}\)
b) \(\frac{xy}{x^2+y^2+2xy-(x+y)}\)
c) \(\frac{y^2-y}{x^2+y^2+2xy-(x+y)}\)
d) \(\frac{xy-y}{x^2+y^2+2xy-(x-y)}\)
Explanation: Number of blue balls = y
Number of Red balls = x
Total number of balls = x + y
Probability of Blue ball first = y / x + y
No. of balls remaining after removing one = x + y – 1
Probability of pulling secondball as Red=\(\frac{x}{x+y-1}\)
Required porbability=\(\frac{y}{(x+y)}\frac{x}{(x+y-1)}=\frac{xy}{x^2+y^2+2xy-(x+y)}\)
10. A survey determines that in a locality, 33% go to work by Bike, 42% go by Car, and 12% use both. The probability that a random person selected uses neither of them is?
a) 0.29
b) 0.37
c) 0.61
d) 0.75
Explanation: Given: p(b) = 0.33, P(c) = 0.42
P(B ∩ C) = 0.12
P(B ∩ C) = ?
P(B ∩ C) = 1 – P(B ∪ C)
= 1 – p(b) – P(c) + P(B ∩ C)
= 1 – 0.22 – 0.42 + 0.12
= 0.37.