Let A and B be two events such that \( \mathrm{P(A\cap B)}=0 \) then \( \mathrm{P\big(A|(A\cup B)\big)} \) is equal to:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{\mathrm{P(A)}}{\mathrm{P(A)+P(B)}} \)
By the definition of conditional probability:
\( \displaystyle P\big(A|(A\cup B)\big)=\frac{P\big(A\cap(A\cup B)\big)}{P(A\cup B)} \)
Since A is a subset of \( A\cup B \), \( A\cap(A\cup B)=A \).
Since \( P(A\cap B)=0 \):
\( \displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B) \)
\( \displaystyle =P(A)+P(B) \)
Hence
\( \displaystyle P\big(A|(A\cup B)\big)=\frac{P(A)}{P(A)+P(B)} \)