1. A and B are two events such that P(A) = 0.4 and P(A ∩ B) = 0.2 Then P(A ∩ B) is equal to ___________
a) 0.4
b) 0.2
c) 0.6
d) 0.8
Explanation: P(A ∩ B) = P(A – (A ∩ B))
= P(A) – P(A ∩ B)
= 0.6 – 0.2 Using P(A) = 1 – P(A)
= 0.4.
2. A problem in mathematics is given to three students A, B and C. If the probability of A solving the problem is 1⁄2 and B not solving it is 1⁄4. The whole probability of the problem being solved is 63⁄64 then what is the probability of solving it?
a) 1⁄8
b) 1⁄64
c) 7⁄8
d) 1⁄2
Explanation:
Let A be the event of A solving the problem
Let B be the event of B solving the problem
Let C be the event of C solving the problem
Given P(a) = 1⁄2, P(~B) = 1⁄4 and P(A ∪ B ∪ C) = 63/64
We know P(A ∪ B ∪ C) = 1 – P(A ∪ B ∪ C)
= 1 – P(A ∩ B ∩ C)
= 1 – P(A) P(B) P(C)
Let P(C) = p
ie 63⁄64 = 1 – (1⁄2)(1⁄4)(p)
= 1 – p⁄8
⇒ P =1/8 = P(C)
⇒P(C) = 1 – P = 1 – 1⁄8 = 7⁄8.
3. Let A and B be two events such that P(A) = 1⁄5 While P(A or B) = 1⁄2. Let P(B) = P. For what values of P are A and B independent?
a) 1⁄10 and 3⁄10
b) 3⁄10 and 4⁄5
c) 3⁄8 only
d) 3⁄10
Explanation: For independent events,
P(A ∩ B) = P(A) P(B)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= P(A) + P(B) – P(A) P(B)
= 1⁄5 + P (1⁄5)P
⇒ 1⁄2 = 1⁄5 + 4⁄5P
⇒ P= 3⁄8.
4. If A and B are two mutually exclusive events with P(~A) = 5⁄6 and P(b) = 1⁄3 then P(A /~B) is equal to ___________
a) 1⁄4
b) 1⁄2
c) 0, since mutually exclusive
d) 5⁄18
Explanation: As A and B are mutually exclusive we have
\(A\cap\bar{B}\)
And Hence
\(P(A/\bar{B})=\frac{P(A\cap\bar{B})}{P(\bar{B})}\)
\(\frac{1-P(\bar{A})}{1-P(\bar{B})}=\frac{1-\frac{5}{6}}{1-\frac{1}{3}}\)
\(P(A/\bar{B})=\frac{1}{4}\)
5. If A and B are two events such that P(a) = 0.2, P(b) = 0.6 and P(A /B) = 0.2 then the value of P(A /~B) is ___________
a) 0.2
b) 0.5
c) 0.8
d) 1⁄3
Explanation: For independent events,
P(A /~B) = P(a) = 0.2.
6. If A and B are two mutually exclusive events with P(a) > 0 and P(b) > 0 then it implies they are also independent.
a) True
b) False
Explanation: P(A ∩ B) = 0 as (A ∩ B) = ∅
But P(A ∩ B) ≠ 0 , as P(a) > 0 and P(b) > 0
P(A ∩ B) = P(A) P(B), for independent events.
7. Let A and B be two events such that the occurrence of A implies occurrence of B, But not vice-versa, then the correct relation between P(a) and P(b) is?
a) P(A) < P(B)
b) P(B) ≥ P(A)
c) P(A) = P(B)
d) P(A) ≥ P(B)
Explanation: Here, according to the given statement A ⊆ B
P(B) = P(A ∪ (A ∩ B)) (∵ A ∩ B = A)
= P(A) + P(A ∩ B)
Therefore, P(B) ≥ P(A)
8. In a sample space S, if P(a) = 0, then A is independent of any other event.
a) True
b) False
Explanation: P(a) = 0 (impossible event)
Hence, A is not dependent on any other event.
9. If A ⊂ B and B ⊂ A then,
a) P(A) > P(B)
b) P(A) < P(B)
c) P(A) = P(B)
d) P(A) < P(B)
Explanation: A ⊂ B and B ⊂ A => A = B
Hence P(a) = P(b).
10. If A ⊂ B then?
a) P(a) > P(b)
b) P(A) ≥ P(B)
c) P(B) = P(A)
d) P(B) = P(B)
Explanation: A ⊂ B => B ⊂ A
Therefore, P(A) ≥ P(B)