If E and F are two events such that P(E F)=P(E), then Option 1: P(E|F)=0 Option 2: P(F|E)=0 Option 3: P(E|F)=1 Option 4: P(F|E)=1

Answer: P(E|F)=1. Solution: P(E F) = P(E) forces P(F) = P(E F), i.e. F is essentially contained in E. Therefore P(E, F) = (P(E F))/(P(F)) = 1. P(F, E) need not be 1, since E may be strictly larger than F.

Algebra

Probability

Medium

core

If E and F are two events such that $P(E \cup F)=P(E)$ , then

$P(E|F)=0$

$P(F|E)=0$

$P(E|F)=1$

$P(F|E)=1$

✅ Correct Option: 3

$P(E\cup F) = P(E)$ forces $P(F) = P(E\cap F)$ , i.e. $F$ is essentially contained in $E$ .

Therefore $P(E|F) = \dfrac{P(E\cap F)}{P(F)} = 1$ .

$P(F|E)$ need not be $1$ , since $E$ may be strictly larger than $F$ .

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If E and F are two events such that P(E∪F)=P(E)P(E \cup F)=P(E)P(E∪F)=P(E), then

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If E and F are two events such that $P(E \cup F)=P(E)$ , then