Animated Solution for Mathematics - Probability: Let $X$ and $Y$ be two events such that $P(X|Y)=1/2,P(Y|X)=1/3$ and $P(X\cap Y)=1/6$. Which of the following is (are) correct?

We start with the raw data: $P(X|Y)=\frac{1}{2}$, $P(Y|X)=\frac{1}{3}$, and $P(X\cap Y)=\frac{1}{6}$.

The definition of conditional probability is our most powerful tool. It serves as the bridge between the intersection and the individual event.

We know that:

By rearranging this, we find that $P(Y)=\frac{P(X\cap Y)}{P(X|Y)}$. Substituting our values, we get:

We apply the same logic to $P(X)$ using $P(Y|X)=\frac{P(X\cap Y)}{P(X)}$, which gives us:

We have successfully mapped the individual probabilities of our two events.

Now that we know $P(X)=\frac{1}{2}$ and $P(Y)=\frac{1}{3}$, we can tackle the union. The Addition Theorem of Probability is a beautiful piece of geometry:

We subtract the intersection because when we add the two circles $X$ and $Y$, the overlapping region $X\cap Y$ is counted twice. Subtracting it once restores the balance.

Plugging in our values:

This confirms that Option A is correct.

Next, we face the ultimate question: Are $X$ and $Y$ independent? Independence is a profound concept—it means the occurrence of one event provides absolutely no information about the occurrence of the other.

Mathematically, this is defined by the condition:

Let us test it:

Since this matches our given $P(X\cap Y)=\frac{1}{6}$, we have proven that $X$ and $Y$ are indeed independent. Consequently, Option B is correct, and Option C is logically discarded.

Finally, let us look at $P(X^c\cap Y)$. This represents the probability that $Y$ occurs but $X$ does not.

Geometrically, this is the part of $Y$ that lies outside of $X$. We calculate this as:

Since $\frac{1}{6} eq \frac{1}{3}$, we see that Option D is incorrect. Through this process, we have understood the mechanics of how events interact.