According to bayee's theorem, P(Ei / A) = P(Ei)P(A / Ei)/sumj=1n P(Ej)P(A / Ej) here

Option a: E1, E2, ..., En are mutually exclusive. Solution: Bayes' Theorem Analysis Bayes' theorem states: P(Ei | A) = P(Ei) × P(A | Ei)/sumj=1n P(Ej) × P(A | Ej) For this theorem to be applicable, certain conditions must hold: 1. The events E1, E2, ..., En must form a partition of the sample space. 2. A partition requires that the events are mutually exclusive (no two events can occur simultaneously: Ei Ej = for i ≠ j ) and exhaustive (their union covers the entire sample space). The denominator sumj=1n P(Ej) P(A|Ej) = P(A) (by the Total Probability Theorem), which requires the Ej 's to be mutually exclusive and exhaustive. Option A states that E1, E2, ..., En are mutually exclusive — this is the correct fundamental requirement. Options B, C, D are incorrect conditions — the posterior and likelihood probabilities do not need to be 1, and disjoint sets is a loose restatement that does not capture the full condition accurately in the context given. Hence, Option A is the correct answer.

According to bayee's theorem, $P(E_i / A) = \frac{P(E_i)P(A / E_i)}{\sum_{j=1}^{n} P(E_j)P(A / E_j)}$ here

$P(E_i / A), P(E_j / A), \dots$ are equal to $1$

$P(A / E_i), P(A_2 / E_i), \dots$ Are equal to $1$

$A$ & $E_i$'s are disjoint sets

According to bayee's theorem, $P(E_i / A) = \frac{P(E_i)P(A / E_i)}{\sum_{j=1}^{n} P(E_j)P(A / E_j)}$ here

The correct answer is Option a: $E_1, E_2, \dots, E_n$ are mutually exclusive. **Bayes' Theorem Analysis** Bayes' theorem states: $$P(E_i | A) = \frac{P(E_i) \cdot P(A | E_i)}{\sum_{j=1}^{n} P(E_j) \cdot P(A | E_j)}$$ For this theorem to be applicable, certain conditions must hold: 1. The events $E_1, E_2, \dots, E_n$ must form a **partition** of the sample space. 2. A partition requires that the events are **mutually exclusive** (no two events can occur simultaneously: $E_i \cap E_j = \emptyset$ for $i \neq j$) and **exhaustive** (their union covers the entire sample space). The denominator $\sum_{j=1}^{n} P(E_j) P(A|E_j) = P(A)$ (by the Total Probability Theorem), which requires the $E_j$'s to be mutually exclusive and exhaustive. **Option A** states that $E_1, E_2, \dots, E_n$ are mutually exclusive — this is the correct fundamental requirement. **Options B, C, D** are incorrect conditions — the posterior and likelihood probabilities do not need to be 1, and disjoint sets is a loose restatement that does not capture the full condition accurately in the context given. Hence, **Option A** is the correct answer.

ProbabilityMCQPYQ June 19Question 3264 of 187

All Questions

According to bayee's theorem, $\displaystyle P(E_i / A) = \frac{P(E_i)P(A / E_i)}{\sum_{j=1}^{n} P(E_j)P(A / E_j)}$ here

Options

\displaystyle E_1, E_2, \dots, E_n

are mutually exclusive

\displaystyle P(E_i / A), P(E_j / A), \dots

are equal to

\displaystyle 1

\displaystyle P(A / E_i), P(A_2 / E_i), \dots

Are equal to

\displaystyle 1

\displaystyle A

\displaystyle E_i

's are disjoint sets

For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option a — $\displaystyle E_1, E_2, \dots, E_n$ are mutually exclusive

All Options:

A $\displaystyle E_1, E_2, \dots, E_n$ are mutually exclusive
B $\displaystyle P(E_i / A), P(E_j / A), \dots$ are equal to $\displaystyle 1$
C $\displaystyle P(A / E_i), P(A_2 / E_i), \dots$ Are equal to $\displaystyle 1$
D $\displaystyle A$ & $\displaystyle E_i$ 's are disjoint sets

Detailed Solution & Explanation

**Bayes' Theorem Analysis** Bayes' theorem states:

P(E_i | A) = \frac{P(E_i) \cdot P(A | E_i)}{\sum_{j=1}^{n} P(E_j) \cdot P(A | E_j)}

For this theorem to be applicable, certain conditions must hold: 1. The events

\displaystyle E_1, E_2, \dots, E_n

must form a **partition** of the sample space. 2. A partition requires that the events are **mutually exclusive** (no two events can occur simultaneously:

\displaystyle E_i \cap E_j = \emptyset

for

\displaystyle i \neq j

) and **exhaustive** (their union covers the entire sample space). The denominator

\displaystyle \sum_{j=1}^{n} P(E_j) P(A|E_j) = P(A)

(by the Total Probability Theorem), which requires the

\displaystyle E_j

's to be mutually exclusive and exhaustive. **Option A** states that

\displaystyle E_1, E_2, \dots, E_n

are mutually exclusive — this is the correct fundamental requirement. **Options B, C, D** are incorrect conditions — the posterior and likelihood probabilities do not need to be 1, and disjoint sets is a loose restatement that does not capture the full condition accurately in the context given. Hence, **Option A** is the correct answer.

About This Chapter: Probability

Paper

Paper 3: Quantitative Aptitude

Weightage

5-7 Marks

Key Topics

Probability Operations, Expected Value

A logic-heavy chapter dealing with random experiments, events (mutually exclusive, exhaustive), set theory probability, conditional probability, and Bayes' Theorem. It forms the basis for Theoretical Distributions.

View Official ICAI Syllabus

Exam Strategy Tip

Always draw a quick Venn Diagram or tree when faced with 'At least one' or 'Only A but not B' wording. It saves you from double-counting.

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According to bayee's theorem, P(Ei/A)=P(Ei)P(A/Ei)∑j=1nP(Ej)P(A/Ej)\displaystyle P(E_i / A) = \frac{P(E_i)P(A / E_i)}{\sum_{j=1}^{n} P(E_j)P(A / E_j)}P(Ei​/A)=∑j=1n​P(Ej​)P(A/Ej​)P(Ei​)P(A/Ei​)​ here