Animated Solution for Mathematics - Probability: In a bag there are 20 cards 10 names $A$ and another 10 names $B$. Cards are drawn randomly one by one with replacement then find probability that second $A$ comes before third $B$.

Imagine you are standing in front of a bag containing $20$ cards. Half are labeled $A$, and half are labeled $B$. You are drawing them one by one, with replacement.

The question asks for the probability that the second $A$ appears before the third $B$. At first glance, this feels like an infinite, daunting sequence of possibilities.

You might be tempted to start writing out infinite series, calculating the probability of $A$ winning on the second draw, the third, the fourth, and so on, into the abyss. But stop. Take a breath.

In JEE Advanced, when you see a problem that looks like it requires an infinite series, there is almost always a hidden geometric or combinatorial symmetry waiting to be unlocked.

Think of this problem not as a sequence of draws, but as a race. $A$ is a runner who needs $2$ steps to reach the finish line. $B$ is a runner who needs $3$ steps to reach the finish line.

We are drawing cards, and every time we draw an $A$, runner $A$ takes a step. Every time we draw a $B$, runner $B$ takes a step.

The game ends the moment one of them crosses their finish line. The question is simply: what is the probability that $A$ wins this race?

Here is the "Aha!" moment that separates the masters from the novices. If $A$ needs $m=2$ successes and $B$ needs $n=3$ successes, what is the maximum number of trials this race can possibly last?

If we play for $m+n-1$ trials, which is $2+3-1=4$ trials, the game must be over.

Why? Because if the game hasn't ended by the $4$th trial, it would mean $A$ has fewer than $2$ successes (so $A\le 1$) AND $B$ has fewer than $3$ successes (so $B\le 2$).

But if $A\le 1$ and $B\le 2$, the total number of trials would be $A+B\le 3$. This contradicts the fact that we have conducted $4$ trials! Therefore, in any sequence of $4$ trials, it is mathematically impossible for the game to still be ongoing.

Since the game must end by the $4$th trial, we can rephrase our condition: $A$ wins if and only if $A$ gets at least $2$ successes within those $4$ trials. If $A$ gets $2$ or more successes in $4$ trials, $A$ has crossed the finish line before $B$ could possibly reach $3$.

This transforms our terrifying infinite problem into a simple Binomial Distribution problem. We have $n=4$ trials, and the probability of success (drawing an $A$) is $p=\frac{1}{2}$. The probability of failure (drawing a $B$) is $q=\frac{1}{2}$.

We need to find the probability $P(X\ge 2)$, where $X$ is the number of $A$'s. This is simply the sum of the probabilities of getting exactly $2$, $3$, or $4$ $A$'s:

Using the binomial formula $P(X=k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k{q}^{n-k}$, we calculate each term:

For $X=2$:

For $X=3$:

For $X=4$:

Adding these together, we get:

Look at that! We didn't need to sum an infinite series. We didn't need to get lost in the weeds of conditional probability.

By understanding the constraints of the race, we reduced a complex problem to a simple, elegant calculation. This is the power of conceptual clarity in mathematics.

Keep this "Maximum Trials" trick in your toolkit—it will save you precious minutes and prevent countless errors in the exam hall. The final answer is $\frac{11}{16}$.