Animated Solution for Mathematics - Probability: Two dices are rolled. If both dices have six faces numbered 1,2,3,5,7 and 11, then the probability that the sum of the numbers on the top faces is less than or equal to 8 is :

Welcome, future engineer! Today, we are going to tackle a probability problem that might look simple at first glance but hides a beautiful, logical structure.

We are dealing with two dice, but these aren't your standard casino dice. These dice have faces numbered $\{1,2,3,5,7,11\}$. This small change forces us to abandon our intuition about standard dice and rely on the fundamental principles of probability.

Visualizing the Sample Space

When we roll two dice, the total number of outcomes is the product of the number of faces on each die. Since each die has 6 faces, our sample space $S$ contains $6\times 6=36$ possible outcomes.

Imagine a $6\times 6$ grid where the rows represent the first die and the columns represent the second. Every cell in this grid is a unique pair $(x,y)$.

Our goal is to find the probability of the event $E$, where the sum of the numbers on the top faces is less than or equal to 8, or mathematically, $x+y\le 8$.

The Systematic Walkthrough

Instead of guessing, let's be the architects of our own solution. We will fix the value of the first die, $x$, and see how many values of $y$ satisfy the condition $x+y\le 8$.

Case 1: $x=1$. The condition becomes $1+y\le 8$, so $y\le 7$. Looking at our set $\{1,2,3,5,7,11\}$, the values of $y$ that satisfy this are $\{1,2,3,5,7\}$. That gives us 5 favorable pairs.

Case 2: $x=2$. The condition becomes $2+y\le 8$, so $y\le 6$. The valid values for $y$ are $\{1,2,3,5\}$. That is 4 favorable pairs.

Case 3: $x=3$. The condition becomes $3+y\le 8$, so $y\le 5$. The valid values for $y$ are $\{1,2,3,5\}$. Again, we have 4 favorable pairs.

Case 4: $x=5$. The condition becomes $5+y\le 8$, so $y\le 3$. The valid values for $y$ are $\{1,2,3\}$. That gives us 3 favorable pairs.

Case 5: $x=7$. The condition becomes $7+y\le 8$, so $y\le 1$. The only valid value for $y$ is $\{1\}$. That is 1 favorable pair.

Case 6: $x=11$. The condition becomes $11+y\le 8$, which is impossible for any positive value of $y$. So, we have 0 favorable pairs.

The Final Calculation

Now, we simply sum the favorable outcomes:

The probability $P(E)$ is the ratio of favorable outcomes to the total sample space:

There you have it! By breaking the problem down into manageable cases, we turned a potentially confusing scenario into a clear, logical path.

Remember, in JEE, the most complex problems often yield to the most systematic approaches. Keep practicing, and keep falling in love with the logic behind the math! The final answer is $\frac{17}{36}$.