Match List-I with List-II
Let A and B be any two events
|
List-I |
List-II |
|
(A) $P (A')$ |
(I) $\frac{P(A∩B)}{P(A)};P(A)≠ 0$ |
|
(B) $P (\phi)$ |
(II) $\frac{P(A∩B)}{P(B)};P(B)≠ 0$ |
|
(C) $P (A|B)$ |
(III) $1-P(A)$ |
|
(D) $P (B|A)$ |
(IV) 0 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
|
List-I |
List-II |
|
(A) $P (A')$ |
(III) $1-P(A)$ |
|
(B) $P (\phi)$ |
(IV) 0 |
|
(C) $P (A|B)$ |
(II) $\frac{P(A∩B)}{P(B)};P(B)≠ 0$ |
|
(D) $P (B|A)$ |
(I) $\frac{P(A∩B)}{P(A)};P(A)≠ 0$ |
(A)
Complement of an event.
$A' \text{ denotes the complement of } A.$
$A\cup A' = S,\quad A\cap A' = \emptyset$
By finite additivity,
$P(A)+P(A')=P(S)=1$
Therefore
$P(A')=1-P(A)$
Match: (A) → (III)
(B)
Null (empty) event.
By the probability axioms, the empty set has probability zero:
$P(\emptyset)=0$
Match: (B) → (IV)
(C)
Conditional probability of A given B (requirement: $P(B)\neq 0$).
Definition:
$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$
Equivalently $P(A\cap B)=P(A\mid B)\,P(B)$.
Match: (C) → (II)
(D)
Conditional probability of B given A (requirement: $P(A)\neq 0$).
Definition:
$P(B\mid A)=\frac{P(A\cap B)}{P(A)}$
Match: (D) → (I)