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Match List-I with List-II

Let A and B be any two events

List-IList-II
(A) P(A′)P(A')(I) P(A∩B)P(A);P(A)≠0\frac{P(A \cap B)}{P(A)}; P(A) \neq 0
(B) P(ϕ)P(\phi)(II) P(A∩B)P(B);P(B)≠0\frac{P(A \cap B)}{P(B)}; P(B) \neq 0
(C) P(A∣B)P(A\vert B)(III) 1−P(A)1 - P(A)
(D) P(B∣A)P(B\vert A)(IV) 0

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 3

P(A′)=1−P(A)P(A') = 1 - P(A)

This is the complement rule.

(A)→(III)(A) \to (III)


P(ϕ)=0P(\phi) = 0

ϕ\phi is the empty set (no outcome), so the probability is always 0.

(B)→(IV)(B) \to (IV)


P(A∣B)=P(A∩B)P(B); P(B)≠0P(A|B) = \dfrac{P(A \cap B)}{P(B)};\ P(B) \neq 0

In P(A∣B)P(A|B), the event after the bar (here BB) goes in the denominator.

(C)→(II)(C) \to (II)


P(B∣A)=P(A∩B)P(A); P(A)≠0P(B|A) = \dfrac{P(A \cap B)}{P(A)};\ P(A) \neq 0

In P(B∣A)P(B|A), the event after the bar (here AA) goes in the denominator.

(D)→(I)(D) \to (I)


(A)→(III),(B)→(IV),(C)→(II),(D)→(I)(A) \to (III),\quad (B) \to (IV),\quad (C) \to (II),\quad (D) \to (I)

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