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

The correct answer is Option (3) → \( P(E|F)=1 \)

By the addition rule:

    \( \displaystyle P(E\cup F)=P(E)+P(F)-P(E\cap F) \)

Since \( P(E\cup F)=P(E) \):

    \( \displaystyle P(F)-P(E\cap F)=0 \)

    \( \displaystyle P(E\cap F)=P(F) \)

So, for P(F) ≠ 0 (needed for \( P(E|F) \) to be defined),

    \( \displaystyle P(E|F)=\frac{P(E\cap F)}{P(F)} \)

    \( \displaystyle =\frac{P(F)}{P(F)}=1 \)

Why the other options are incorrect

• Option (1) \( P(E|F)=0 \) — Incorrect. \( P(E|F)=1 \), not 0.

• Option (2) \( P(F|E)=0 \) — Incorrect. \( P(F|E)=\frac{P(F)}{P(E)} \), which is 0 only if P(F) = 0.

• Option (4) \( P(F|E)=1 \) — Incorrect. \( P(F|E)=\frac{P(F)}{P(E)} \), which is 1 only if P(F) = P(E); the given condition does not ensure this.