Animated Solution for Mathematics - Probability: Four fair dice $D_1,D_2,D_3$ and $D_4$; each having six faces numbered 1, 2, 3, 4, 5 and 6 are rolled simultaneously. The probability that $D_4$ shows a number appearing on one of $D_1,D_2$ and $D_3$ is

Why We Need a Better Way

Imagine you are standing in an exam hall, the clock is ticking, and you have four fair dice in your hands. You roll them, and the numbers tumble across the table. The problem asks for the probability that the fourth die, $D_4$, matches at least one of the first three dice, $D_1,D_2,$ or $D_3$.

Your first instinct might be to start counting. You might think, "$D_4$ could match $D_1$, or it could match $D_2$, or maybe it matches both $D_1$ and $D_2$, or perhaps all three!"

If you start down this path, you are entering a labyrinth. You will quickly find yourself drowning in a sea of overlapping cases, trying to account for every possible intersection. In the world of JEE Advanced, this is the trap; the examiners love to see if you can identify when a direct approach is a path to madness. We need a superpower, and that superpower is the Complement Rule.

The Complement Rule

The Complement Rule is one of the most beautiful concepts in probability. It tells us that if we want to find the probability of an event $E$ happening, we can simply calculate the probability of it NOT happening—the failure event $E^c$—and subtract that from 1.

Mathematically, we write this as:

Think about it: the total probability space is 1. If we remove the scenarios where absolutely nothing matches, what remains must be the scenarios where at least one match occurs. It is elegant, it is clean, and it is incredibly efficient.

Fixing the Outcome

Now, let us focus on the failure event $E^c$. For this to happen, $D_4$ must match none of the first three dice. Let us fix the outcome of $D_4$ to be some number $x$.

Since it is a standard die, $x$ can be any integer from the set $\{1,2,3,4,5,6\}$. For our failure condition to be satisfied, $D_1$ must not show $x$, $D_2$ must not show $x$, and $D_3$ must not show $x$.

Consider $D_1$. It has six faces, and only one of those faces is $x$. Therefore, there are exactly five faces that are not $x$. The probability that $D_1$ avoids $x$ is simply:

This logic applies perfectly to $D_2$ and $D_3$ as well.

Here is where the magic of independent events comes into play. The roll of $D_1$ has absolutely no physical or mathematical connection to the roll of $D_2$ or $D_3$.

Because they are independent, the probability of them all avoiding $x$ simultaneously is the product of their individual probabilities. We are looking for the probability that ($D_1 eq x$) AND ($D_2 eq x$) AND ($D_3 eq x$).

In the language of probability, the 'AND' operator for independent events translates directly to multiplication:

Substituting our values, we get:

Calculating this, we find:

Bringing It Home

We are almost there. We have the probability of failure, $\frac{125}{216}$. Now, we return to our Complement Rule equation:

Substituting our value, we get:

To perform this subtraction, we find a common denominator:

The numerator becomes 91. Thus, our final probability is:

Look at how far we have come! We didn't need to count hundreds of combinations or draw complex Venn diagrams. We used the symmetry of the dice, the power of the Complement Rule, and the logic of independent events to slice through the problem with surgical precision. This is the mindset of a topper.