Animated Solution for Mathematics - Probability: A six faced fair dice is thrown until 1 comes, then the probability that 1 comes in even no. of trials is

Imagine you are sitting at a table with a fair, six-faced dice. You are tasked with rolling it until a '1' appears. This is a sequence of independent trials: $1^{st},2^{nd},3^{rd},\dots$ and so on.

Before we touch any complex math, let us define our reality. We have two outcomes for every single roll:

1. Success ($p$): You roll a '1'. Since the dice is fair, $p=\frac{1}{6}$.

2. Failure ($q$): You roll anything else ($2,3,4,5,6$). Thus, $q=1-p=\frac{5}{6}$.

This is the bedrock of our calculation. Every roll is independent, and every roll is a binary choice between success and failure.

The problem asks for the probability that the first '1' appears on an even number of trials. This means we are interested in the set of events where the first success occurs at trial $2,4,6,8,\dots$ and so on.

Let us visualize this:

- To succeed on the 2nd trial, you must fail on the 1st and succeed on the 2nd. The probability is $q\times p$.

- To succeed on the 4th trial, you must fail on the 1st, 2nd, and 3rd, and then succeed on the 4th. That is $q\times q\times q\times p$, or $q^{3}p$.

- To succeed on the 6th trial, you must fail five times and then succeed. That is $q^{5}p$.

The number of failures is always one less than the trial number. Because we are restricted to even trials, the number of failures is always odd ($1,3,5,\dots$).

Since these events are mutually exclusive, we sum their probabilities to find the total probability $P$:

This is an infinite Geometric Progression (G.P.). In a G.P., every term is derived by multiplying the previous term by a constant ratio, $r$.

Let us identify our components:

- The first term, $a=qp$.

- The common ratio, $r=\frac{q^{3}p}{qp}=q^2$.

Since $q=\frac{5}{6}$, our ratio $r=(\frac{5}{6}{)}^2=\frac{25}{36}$. Because $|r|<1$, this series converges to a finite value using the sum formula:

Now, we substitute our values into the formula. Precision is key here.

Substitute $p=\frac{1}{6}$ and $q=\frac{5}{6}$:

The numerator simplifies to $\frac{5}{36}$. The denominator is $1-\frac{25}{36}$, which simplifies to $\frac{11}{36}$.

Now, look at the beauty of the final fraction:

The denominators of $36$ cancel out perfectly. We arrive at the final result:

The Takeaway

We started with a dice roll and ended with a clean fraction. The probability that the first '1' appears on an even trial is $\frac{5}{11}$.

This problem teaches us that even when a process is infinite, the underlying structure is often simple and symmetric. Whenever you face a problem involving repeated trials, look for the G.P. structure. It is the secret language of probability.