Animated Solution for Mathematics - Probability: A random variable $X$ has the following probability distribution: $$\begin{array}{cccccc}X& 0& 1& 2& 3& 4\\ P(X)& k& 2k& 4k& 6k& 8k\end{array}$$ The value of $P(1<X<4|X\le 2)$ is equal to:

Imagine you are standing in front of a probability distribution table. It looks simple, almost innocent, with its rows of $X$ and $P(X)$. But beneath that simplicity lies a beautiful, logical structure waiting to be unraveled.

Today, we are going to master the art of conditional probability, a concept that often trips up even the brightest students because they overcomplicate the arithmetic. Let's dive in.

When you see a distribution table where probabilities are defined in terms of a constant $k$, your first instinct might be to sum them up, equate them to $1$, and solve for $k$. Stop!

Before you reach for your pen to calculate $k=\frac{1}{21}$, look at the question. We are asked for $P(1<X<4|X\le 2)$. This is a conditional probability.

The formula is:

Notice that both the numerator and the denominator are sums of probabilities involving $k$. If you write out the ratio, you will see $k$ in every term. It is a beautiful, silent cancellation waiting to happen.

By not solving for $k$, you save time and avoid the risk of a silly arithmetic error.

Let's define our events. The condition, which we call event $B$, is $X\le 2$. This is our new universe.

We don't care about $X=3$ or $X=4$ anymore; they are outside our current reality. The outcomes satisfying $B$ are $X=0$, $X=1$, and $X=2$.

The total probability of this condition is:

This $7k$ is the denominator of our final answer.

Where Paths Cross

Now, we look at the target event $A$, which is $1<X<4$. Since $X$ is a discrete random variable taking integer values, this means $X$ can be $2$ or $3$.

But we need the intersection $A\cap B$. This is the crucial step. We need the outcomes that are in both $A$ and $B$.

Event $A$ gives us $\{2,3\}$, and event $B$ gives us $\{0,1,2\}$. The only overlap is $X=2$. Therefore:

We have everything we need. The numerator is $4k$, and the denominator is $7k$. When we put them together, we get:

Since $k$ is a probability constant, it cannot be zero, so we can safely cancel it out. We are left with $\frac{4}{7}$.

It is clean, it is elegant, and it is correct. Remember, in JEE Advanced, the most complex-looking problems often have the most beautiful, simple solutions if you just take a moment to look at the structure before you start calculating.