Animated Solution for Mathematics - Probability: A fair die is thrown until 2 appears. Then the probability, that 2 appears in even number of throws , is

Imagine you are standing in a quiet room, holding a single, fair six-sided die. You have one goal: roll a $2$. You are committed to a game that continues until that elusive $2$ appears.

The question is: what is the probability that this success occurs on an even-numbered throw? This is a classic JEE problem that tests your ability to translate a verbal description into a rigorous mathematical sequence.

Before we dive into the infinite, let us ground ourselves in the basics. We define our success, $p$, as rolling a $2$. Since the die is fair, we have:

Consequently, the probability of failure, $q$, which is rolling anything other than a $2$, is:

These two values, $p$ and $q$, are the DNA of our entire problem. Every scenario we construct will be built from these two fundamental probabilities.

We are interested in the success occurring on an even throw: $n=2,4,6,\dots$. Let us map this out.

For the success to happen on the $2^{nd}$ throw, the first throw must be a failure, and the second must be a success. The probability is $q\times p$.

If we want success on the $4^{th}$ throw, we must fail three times and then succeed. This is represented as $q\times q\times q\times p$, or $q^{3}p$.

For the $6^{th}$ throw, we need five failures followed by a success, which is $q^{5}p$. The pattern is undeniable: the probability of success on the $n$-th throw, where $n$ is even, is $q^{n-1}p$.

Now, we sum these probabilities to find the total probability $P$:

This is an infinite geometric progression. In any GP, the sum is determined by the first term $a$ and the common ratio $r$.

Here, the first term $a=qp$. To find the common ratio $r$, we divide the second term by the first:

Since $q=\frac{5}{6}$, our ratio $r=\frac{25}{36}$, which is less than $1$. This confirms that our series converges to a finite value.

We use the sum formula for an infinite GP:

Substituting our values, we get:

Now, let us plug in the numbers. The numerator is:

The denominator is:

Finally, we calculate $P=\frac{5/36}{11/36}$. The $36$ in the denominator of both fractions cancels out, leaving us with the beautiful, simple result:

Mathematics often feels like a series of hurdles, but when you break a problem down into its core components—defining your variables, visualizing the sequence, and identifying the underlying structure—the complexity dissolves. You have just navigated an infinite process and arrived at a precise, elegant solution.