Q.$A$, $B$, and $C$ try to hit a target simultaneously but independently. Their respective probabilities of hitting the target are $\frac{3}{4}$, $\frac{1}{2}$ and $\frac{5}{8}$. Find the probability that the target a) is hit exactly by one of them b) is not hit by any one of them c) is hit d) is exactly hit by two of them.

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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