For a biased die the probabilities for the different faces to turn up are given below : Face: 1, 2,

Q: Why do we just ignore faces 3, 4, 5, and 6? Shouldn't they still be part of the calculation?

That is a fantastic question! In the world of conditional probability, we are essentially 'zooming in' on a specific subset of outcomes. Once we are given the information that the outcome is either Face 1 or Face 2, the other faces become impossible in this specific scenario. They have zero probability in this new, restricted universe. We are not ignoring them; we are acknowledging that the condition has rendered them irrelevant to the current state of knowledge.

Q: Wait, why is the intersection $P(E_1 \cap E_2)$ just $P(E_1)$?

Think about the set theory behind it. $E_1$ is the event 'Face 1'. $E_2$ is the event 'Face 1 or Face 2'. Since $E_1$ is a subset of $E_2$ (if Face 1 happens, then 'Face 1 or Face 2' has automatically happened), the intersection of the two is simply the smaller set, which is $E_1$. It is like saying, 'What is the intersection of the set of all students and the set of all students who passed?' The intersection is just the students who passed!

Q: Is there a shortcut? Can I just take the ratio of the probabilities directly?

Yes, you absolutely can! Since the total probability space is now just the sum of the probabilities of the allowed outcomes, you can think of it as 'normalizing' the probabilities. You are essentially calculating $\frac{P(E_1)}{P(E_1) + P(E_2)}$. It is a valid and efficient shortcut, but always remember the formula behind it so you do not get confused when the problems become more complex!

Question

For a biased die the probabilities for the different faces to turn up are given below : $\begin{array}{|l|c|c|c|c|c|c|} \hline 	ext{Face} & 1 & 2 & 3 & 4 & 5 & 6 \ \hline 	ext{Pr ob.} & 0.1 & 0.32 & 0.21 & 0.15 & 0.05 & 0.17 \ \hline \end{array}$ This die is tossed and you are told that either face 1 or face 2 has turned up. Then the probability that it is face 1 is .........

Sigma Percentile · Accepted Answer

### The Art of the Reduced Universe: Mastering Conditional Probability

Welcome, future engineer. Today, we are going to dismantle a classic problem that appears simple on the surface but holds the key to understanding one of the most powerful concepts in probability theory: the conditional sample space.

When you look at a biased die, you are not just looking at numbers; you are looking at a system where reality is skewed. In a fair die, every face has a probability of $\frac{1}{6}$. But here, the universe is uneven, with probabilities ranging from $0.1$ to $0.32$. This is the reality of the problem, and we must respect it.

### Phase 1: The Shrinking Universe

Imagine you are holding this die and you toss it. You do not see the result yet, but a voice tells you, "The result is either Face 1 or Face 2." In that single moment, your entire world changes.

Before that voice spoke, you were considering six possibilities. Now, you are only considering two. This is the essence of conditional probability. We are not dealing with the original sample space of $\{1, 2, 3, 4, 5, 6\}$ anymore; our new, restricted universe is just $\{1, 2\}$.

In JEE Advanced, the trap is often not the calculation; it is the failure to redefine the sample space. When we are told that the outcome is restricted to a subset, we must treat that subset as our new "total" probability. We are effectively "zooming in" on the events that are actually possible.

### Phase 2: The Mathematical Toolkit

To solve this, we rely on the fundamental definition of conditional probability. Let $E_1$ be the event that "Face 1 turns up" and $E_2$ be the event that "Face 1 or Face 2 turns up." We are looking for the probability of $E_1$ given that $E_2$ has occurred, denoted as $P(E_1 | E_2)$.

The formula is our North Star:

$$P(E_1 | E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)}$$

Let us pause and look at the numerator, $P(E_1 \cap E_2)$. This represents the probability that both $E_1$ and $E_2$ happen. Since $E_1$ is a subset of $E_2$, the intersection is simply $E_1$ itself, which is $0.1$.

### Phase 3: The Calculation

Now, let us look at the denominator, $P(E_2)$. This is the probability of our new, restricted universe. Since the events "Face 1" and "Face 2" are mutually exclusive, we simply sum their individual probabilities:

$$P(E_2) = P(	ext{Face } 1) + P(	ext{Face } 2)$$

Substituting the values from our data table:

$$P(E_2) = 0.1 + 0.32 = 0.42$$

This $0.42$ is the total weight of our new universe. We are now ready to perform the final calculation by taking our numerator and dividing it by our new total:

$$P(E_1 | E_2) = \frac{0.1}{0.42}$$

### Phase 4: The Final Polish

In an exam setting, you want to be clean and precise. To clear the decimals, multiply both the numerator and the denominator by $100$:

$$P(E_1 | E_2) = \frac{10}{42}$$

Now, we simplify by dividing both terms by $2$:

$$P(E_1 | E_2) = \frac{5}{21}$$

The probability that the face is 1, given the condition, is **$\frac{5}{21}$**.

### The Takeaway

Why did we go through the formal definition? Because in harder problems, the intuition of "just dividing" might fail you if the events are not mutually exclusive or if the intersection is not a simple subset.

By mastering the formula $P(A|B) = \frac{P(A \cap B)}{P(B)}$, you are building a robust mental framework. You have learned to shrink the universe, identify the intersection, and normalize the probability. Keep this mindset, and you will find that even the most "biased" problems become fair game for your intellect.

Visualized Solution

Analyzing the Biased Die Data

The Given Condition

Defining the Events

Conditional Probability Formula

Analyzing the Intersection

Calculating $P (E_{2})$

Evaluating $P (E_{2})$

Substituting into the Formula

Simplifying the Fraction (Part 1)

Simplifying the Fraction (Part 2)

Final Conclusion

The Sigma Insight: Conditional Probability

The Art of the Reduced Universe

Phase 1

Phase 2

Phase 3

Phase 4

The Takeaway

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