Animated Solution for Mathematics - Probability: Let $A$ and $B$ be two events such that $P(B|A)=\frac{2}{5}$, $P(A|B)=\frac{1}{7}$ and $P(A\cap B)=\frac{1}{9}$. Consider (S1) $P(A^{\prime }\cup B)=\frac{5}{6}$, (S2) $P(A^{\prime }\cap B^{\prime })=\frac{1}{18}$. Then

The architecture of probability is best understood through the geometric landscape of events. We represent events $A$ and $B$ as circles on a Venn diagram, where the intersection is given as:

This intersection serves as the central anchor for all subsequent calculations.

We utilize the definition of conditional probability, which expresses the ratio of the intersection to the total area of the conditioning event. The formulas are:

Rearranging these to solve for the individual probabilities, we find:

To find the union $P(A\cup B)$, we must avoid the double-counting trap. We add the individual probabilities and subtract the intersection once:

Substituting our calculated values with a common denominator of $18$:

For Statement (S1), we evaluate $P(A^{\prime }\cup B)$. Using the complement rule, we note that $P(A^{\prime }\cup B)=1-P(A\cap B^{\prime })$. The term $P(A\cap B^{\prime })$ represents the region "only $A$":

Thus, $P(A^{\prime }\cup B)=1-\frac{1}{6}=\frac{5}{6}$. Statement (S1) is true.

For Statement (S2), we evaluate $P(A^{\prime }\cap B^{\prime })$. By De Morgan's Law, this is equivalent to the complement of the union:

Substituting our previous result:

Statement (S2) is also true. Both statements are correct.