Let S = {E1, E2, ..., E8} be a sample space of random experiment such that P(En) = n/36 for every n

Q: Why do we use the complement? Can't we just count the sets $A$ directly?

Direct counting is a trap. You would need to count sets with 5, 6, 7, or 8 elements, which is combinatorially explosive and prone to error. The complement rule $P(A) = 1 - P(A^c)$ reduces the problem to finding partitions of integers up to 7, which is much cleaner and faster.

Q: Why is the sum $\le 7$ and not $7.2$?

The indices $i$ are discrete integers from the set $\{1, 2, ..., 8\}$. The sum of integers must be an integer. Since $7.2$ is the threshold, the largest possible integer sum satisfying the inequality is $7$.

Q: Is there a formula for the number of partitions, or do I have to list them manually?

While there are partition functions in number theory, for small numbers like 7, manual systematic listing is safer and faster. Just be careful to be methodical—start from sum 0 and work your way up to 7, ensuring you don't miss any combinations, especially when dealing with triplets or quadruplets.

Question

Let $S = \{E_1, E_2, ..., E_8\}$ be a sample space of random experiment such that $P(E_n) = \frac{n}{36}$ for every $n = 1, 2, ..., 8$. Then the number of elements in the set $\{A \subset S : P(A) \ge \frac{4}{5}\}$ is ____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given a sample space $S = \{E_1, E_2, ..., E_8\}$ where the probability of each event is weighted by its index: $P(E_n) = \frac{n}{36}$. Immediately, we observe this is not a uniform distribution.

The higher the index, the more weight the event carries. We are tasked with finding the number of subsets $A \subseteq S$ such that $P(A) \ge \frac{4}{5}$.

Many students attempt to construct sets $A$ directly, leading to a combinatorial explosion. In JEE Advanced, when faced with a condition like $P(A) \ge \frac{4}{5}$, the strategic approach is to utilize the **Complement**.

### The Master Equation

The condition $P(A) \ge \frac{4}{5}$ implies $1 - P(A^c) \ge \frac{4}{5}$, which simplifies to $P(A^c) \le \frac{1}{5}$. We have transformed a "greater than" problem into a "less than" problem, focusing on small, manageable sets $A^c$.

Since $P(E_n) = \frac{n}{36}$, the probability of the complement set $A^c$ is the sum of the probabilities of its individual elements:

$$P(A^c) = \sum_{i \in A^c} \frac{i}{36}$$

Setting this to be less than or equal to $\frac{1}{5}$, we obtain:

$$\frac{\sum_{i \in A^c} i}{36} \le \frac{1}{5}$$

Multiplying by $36$, we find that the sum of the indices in $A^c$ must satisfy:

$$\sum_{i \in A^c} i \le \frac{36}{5} = 7.2$$

Since the indices $i$ are integers, the sum of the indices must be an integer $S \le 7$. This reduces the problem to partitioning integers up to $7$ using the set $\{1, 2, ..., 8\}$.

### The Systematic Grind

We now list the subsets $A^c$ systematically based on the sum of their elements:

- **Sum = 0:** The empty set $\emptyset$. (**1 way**)
- **Sum = 1:** The set $\{1\}$. (**1 way**)
- **Sum = 2:** The set $\{2\}$. (**1 way**)
- **Sum = 3:** The sets $\{3\}$ and $\{1, 2\}$. (**2 ways**)
- **Sum = 4:** The sets $\{4\}$ and $\{1, 3\}$. (**2 ways**)
- **Sum = 5:** The sets $\{5\}$, $\{1, 4\}$, and $\{2, 3\}$. (**3 ways**)
- **Sum = 6:** The sets $\{6\}$, $\{1, 5\}$, $\{2, 4\}$, and $\{1, 2, 3\}$. (**4 ways**)
- **Sum = 7:** The sets $\{7\}$, $\{1, 6\}$, $\{2, 5\}$, $\{3, 4\}$, and $\{1, 2, 4\}$. (**5 ways**)

### Final Calculation

Summing the number of ways found for each possible total:

$$1 + 1 + 1 + 2 + 2 + 3 + 4 + 5 = 19$$

By shifting our perspective to the complement, we turned a daunting probability problem into a simple counting exercise. The total number of subsets $A$ satisfying the condition is **19**.

Animated Solution for Mathematics - Probability: Let $S = {E_{1}, E_{2}, ..., E_{8}}$ be a sample space of random experiment such that $P (E_{n}) = \frac{n}{36}$ for every $n = 1, 2, ..., 8$ . Then the number of elements in the set ${A \subset S : P (A) \geq \frac{4}{5}}$ is ____.

Enter Numerical Value:

Visualized Solution

Understanding the Sample Space $S$

Defining the Condition $P (A) \geq \frac{4}{5}$

Using the Complement Rule

Translating to Index Sums

Case 1: Sum is $0, 1, 2$

Case 2: Sum is $3, 4$

Case 3: Sum is $5$

Case 4: Sum is $6$

Case 5: Sum is $7$

Final Summation

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

The Master Equation

The Systematic Grind

Final Calculation

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Enter Numerical Value:

Visualized Solution

Understanding the Sample Space .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Defining the Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(A)≥54​

Using the Complement Rule

Translating to Index Sums

Case 1: Sum is .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 0,1,2

Case 2: Sum is .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3,4

Case 3: Sum is .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 5

Case 4: Sum is .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 6

Case 5: Sum is .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 7