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. PYQ June 19

ProbabilityPYQ June 19Question 3264 of 170

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

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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

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