Animated Solution for Mathematics - Probability: One ticket is selected at random from 50 tickets numbered 00, 01, 02, ..., 49. Then the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero, equals

A Probability Journey

Welcome, fellow traveler of the JEE landscape. Today, we are not just solving a probability problem; we are learning to navigate the shifting sands of a 'restricted sample space.'

Imagine you are standing in a room with $50$ tickets, labeled from $00$ to $49$. At first glance, the world seems vast.

But then, a condition is whispered into your ear: 'The product of the digits is zero.' Suddenly, the room shrinks. The tickets that don't have a zero on them vanish into thin air. This is the heart of conditional probability—the art of redefining your reality.

Defining the New Reality

When we are told that the product of the digits is zero, we are being told that at least one of the digits must be $0$. Let us be systematic, for in the JEE, a systematic mind is a winning mind. We split our search into two logical buckets.

First, consider the tickets where the first digit is $0$. These are the tickets $00,01,02,03,04,05,06,07,08,09$. There are exactly $10$ such tickets.

Second, consider the tickets where the second digit is $0$. These are $10,20,30,40$. Note that $00$ is also in this set, and $50$ is excluded because our range stops at $49$.

Now, here is where the trap lies: if we simply add $10+4$, we get $14$. We must ensure we have counted the ticket $00$ correctly. The set of tickets with at least one zero is $\{00,01,02,03,04,05,06,07,08,09,10,20,30,40\}$.

Counting these, we find the true size of our restricted sample space, $n(B)$:

We have successfully navigated the first hurdle. Our universe now consists of exactly $14$ tickets.

The Hunt for the Target

Now that we are living in this restricted world of $14$ tickets, we look for our target event $A$: the sum of the digits must be $8$. We don't need to look at all $50$ tickets anymore; we only look at our $14$ candidates.

Let us scan them: - Does $00$ sum to $8$? No. - Does $08$ sum to $8$? Yes, $0+8=8$. - Do $10,20,30,40$ sum to $8$? No, their sums are $1,2,3,4$ respectively.

It turns out that within our restricted set, there is only one ticket that satisfies the condition: the ticket $08$. Thus, the number of favorable outcomes $n(A\cap B)$ is exactly $1$.

The Elegant Conclusion

We have arrived at the final step. The probability of an event $A$ given $B$ is defined by the ratio of the favorable outcomes within the restricted space to the total number of outcomes in that restricted space:

Substituting our hard-earned values, we get:

Look at that result. It is clean, it is precise, and it is the product of your logical rigor. You didn't just guess; you built a framework, identified the constraints, and navigated the logic step-by-step.

This is the mindset that conquers the JEE Advanced. Remember, whenever you see a 'given that' in a probability problem, stop, take a breath, and redefine your universe. The final answer is $1/14$.