Q.A bag contains 3 yellow and 5 brown balls. Another bag contains 4 yellow and 6 brown balls. If one ball is drawn from each bag, what is the probability that, a) both the balls are of the same colour? b) the balls are of different colour?

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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