Q.If $P(A) = \dfrac{1}{4}$, $P(B) = \dfrac{1}{3}$ and $P(A - B) = \dfrac{1}{6}$, then the events $A$ and $B$ are mutually

Find $P(A\cap B)$ from $P(A-B)=P(A)-P(A\cap B)$, then compare with $P(A)P(B)$.

Testing independence via $P(A\cap B)=P(A)P(B)$ is a core CBSE/NCERT Class 12 probability idea.

By definition $P(A-B)=P(A)-P(A\cap B)$, so

$P(A\cap B) = P(A) - P(A-B) = \frac14 - \frac16 = \frac{3-2}{12} = \frac{1}{12}.$

Now $P(A)\,P(B) = \frac14 \cdot \frac13 = \frac{1}{12}$.

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Exactly one of $A,B$ = (only $A$) + (only $B$) = $P(A)+P(B)-2P(A\cap B)$.

The probability of "exactly one" of two events is a CBSE/NCERT Class 12 probability result.

The event "exactly one of $A$ and $B$" is $(A \cap B') \cup (A' \cap B)$, whose probability is

$P(A\cap B') + P(A'\cap B) = [P(A)-P(A\cap B)] + [P(B)-P(A\cap B)].$

Combining:

$= P(A) + P(B) - 2P(A\cap B).$

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"Exactly one" $=P(A)+P(B)-2\,P(A\cap B)=x$ and "both" $=P(A\cap B)=y$; solve for $P(A)+P(B)$.

This reuses the "exactly one" identity from the CBSE/NCERT Class 12 probability chapter.

We are told:

$x = P(\text{exactly one}) = P(A)+P(B)-2P(A\cap B),$

$y = P(A\cap B).$

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$P(A/B)=P(A)$ and $P(B/A)=P(B)$ mean neither event's occurrence changes the other's probability — so $A$ and $B$ are independent events.

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