Animated Solution for Mathematics - Probability: Let $X$ and $Y$ be two events such that $P(X)=\frac{1}{3}$, $P(X|Y)=\frac{1}{2}$ and $P(Y|X)=\frac{2}{5}$. Then

Connecting the Dots

Welcome, future engineer. Today, we are not just solving a probability problem; we are learning to navigate the relationships between events.

In the world of JEE Advanced, probability is rarely about simple coin tosses. It is about understanding how one event influences another. When you see conditional probabilities like $P(X|Y)$ and $P(Y|X)$, do not panic. Instead, visualize them as bridges connecting two islands: Event $X$ and Event $Y$.

Finding the Intersection

We are given $P(X)=\frac{1}{3}$, $P(X|Y)=\frac{1}{2}$, and $P(Y|X)=\frac{2}{5}$. The most important step in any conditional probability problem is to find the intersection, $P(X\cap Y)$.

Think of this as the shared territory. Without it, we are lost. We know the definition of conditional probability is:

By rearranging this, we get the intersection: $P(X\cap Y)=P(Y|X)\cdot P(X)$. Substituting our values, we have:

This is our anchor. We have successfully mapped the shared region between $X$ and $Y$. Notice that option [B] suggests $P(X\cap Y)=\frac{1}{5}$, which is $\frac{3}{15}$. Since our calculation gives $\frac{2}{15}$, we can confidently discard option [B].

Unlocking the Total Probability

Now that we have the intersection, the rest of the puzzle falls into place. We need to find $P(Y)$.

We use the other conditional probability provided: $P(X|Y)=\frac{P(X\cap Y)}{P(Y)}$. We know $P(X|Y)=\frac{1}{2}$ and we just found $P(X\cap Y)=\frac{2}{15}$.

Substituting these into our equation:

Solving for $P(Y)$, we get $P(Y)=2\cdot \frac{2}{15}=\frac{4}{15}$. This confirms that option [D] is correct. We are building a complete picture of our sample space, piece by piece.

Evaluating the Options

Let us look at option [A]: $P(X^{\prime }|Y)$. This asks for the probability of $X$ not happening, given that $Y$ has occurred.

Using the complement rule for conditional probability, we know that $P(X^{\prime }|Y)=1-P(X|Y)$. Since $P(X|Y)=\frac{1}{2}$, we have $1-\frac{1}{2}=\frac{1}{2}$. Thus, option [A] is also correct.

Finally, let us check option [C]: $P(X\cup Y)$. The addition theorem tells us that:

Plugging in our values:

Converting to a common denominator of 15, we get:

Since $\frac{7}{15} eq \frac{2}{5}$, option [C] is incorrect.

We have navigated the logic, verified the options, and arrived at the truth. The key takeaway here is simple: whenever you are faced with conditional probabilities, do not try to guess the relationships.

Calculate the intersection $P(X\cap Y)$ first. It is the core link that holds the entire problem together. Keep practicing this systematic approach, and you will find that even the most complex probability problems become clear and manageable. You have got this!