Let A and B be two events associated with an experiment such that P(A B)=P(A) P(B). Assertion (A) : P(A B)=P(A) and P(B A)= P(B) Reason (R): P(A B)=P(A)+P(B).

Let A and B be two events associated with an experiment such that…

MCQ

Let $A$ and $B$ be two events associated with an experiment such that $P(A \cap B)=P(A) P(B)$.
Assertion (A) : $P(A \mid B)=P(A)$ and $P(B \mid A)=$ $P(B)$
Reason (R): $P(A \cup B)=P(A)+P(B)$.

A
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
✓
(A) is true but (R) is false.
D
(A) is false but (R) is true.

✓

Answer

Correct option: C.

(A) is true but (R) is false.

(c) : Since, $P(A \cap B)=P(A) P(B)$, therefore, $A$ and $B$ are independent events.
$
\therefore \quad P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{P(A) P(B)}{P(B)}=P(A)
$
Similarly, $P(B \mid A)=P(B)$.
Thus, assertion is true.
However, Reason is not correct for independent events. For example, when a dice is rolled once, then the events $A$ : 'an even number' shows up and $B$ : 'a multiple of 3 ' show up are independent as
$
\begin{aligned}
P(A) P(B)=\frac{3}{6} \times \frac{2}{6}=\frac{1}{6} & =P(A \cap B) \\
& (\because A=\{2,4,6\} \text { and } B=\{3,6\})
\end{aligned}
$
But $P(A \cup B)=P(\{2,3,4,6\})$
$
=\frac{4}{6} \neq P(A)+P(B) \quad\left(\because P(A)+P(B)=\frac{3}{6}+\frac{2}{6}=\frac{5}{6} \neq \frac{4}{6}\right)
$

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