Question

If A is any event associated with sample space and if E₁, E₂, E₃ are mutually exclusive and exhaustive events. Then which of the following are true? 

(A) \(P(A) = P(E_1)P(E_1|A) + P(E_2)P(E_2|A) + P(E_3)P(E_3|A)\) 
(B) \(P(A) = P(A|E_1)P(E_1) + P(A|E_2)P(E_2) + P(A|E_3)P(E_3)\) 
(C) \(P(E_i|A) = \frac{P(A|E_i)P(E_i)}{\sum_{j=1}^{3} P(A|E_j)P(E_j)}, \; i=1,2,3\) 
(D) \(P(A|E_i) = \frac{P(E_i|A)P(E_i)}{\sum_{j=1}^{3} P(E_i|A)P(E_j)}, \; i=1,2,3\) 

Choose the correct answer from the options given below:

• (A) and (C) only
• (A) and (D) only
• (B) and (D) only
• (B) and (C) only

Hint:

Memorize the statements of the Law of Total Probability and Bayes' Theorem.

\textbf{Total Probability}: Used to find the probability of an event A by considering all possible scenarios (the partition $E_i$). $P(A) = \sum P(A|E_i)P(E_i)$.
\textbf{Bayes' Theorem}: Used to "reverse" the conditioning. If you know $P(A|E_i)$, you can find $P(E_i|A)$. $P(E_i|A) = \frac{P(A|E_i)P(E_i)}{P(A)}$.

Answer 1

The Correct Option is D

Step 1: Conceptual Foundation:
This problem addresses the Law of Total Probability and Bayes' Theorem. Events E₁, E₂, and E₃ constitute a partition of the sample space, meaning they are mutually exclusive and exhaustive.

Step 3: Detailed Analysis:
Evaluation of each statement follows.

(A) $P(A) = P(E_1)P(E_1|A) + P(E_2)P(E_2|A) + P(E_3)P(E_3|A)$
This formula deviates from the standard Law of Total Probability. It incorrectly relates P(A) to conditional probabilities of E_i given A. The expression $P(E_i)P(E_i|A)$ does not simplify to P(A). Therefore, statement (A) is false.

(B) $P(A) = P(A|E_1)P(E_1) + P(A|E_2)P(E_2) + P(A|E_3)P(E_3)$
This statement correctly represents the Law of Total Probability. It calculates the total probability of event A by summing the probabilities of A conditioned on each event in the partition. Given that $P(A|E_i)P(E_i) = P(A \cap E_i)$, the formula simplifies to $P(A) = P(A \cap E_1) + P(A \cap E_2) + P(A \cap E_3)$, which is valid for mutually exclusive and exhaustive events. Thus, statement (B) is true.

(C) $P(E_i|A) = \frac{P(A|E_i)P(E_i)}{\sum_{j=1}^{3} P(A|E_j)P(E_j)}, i=1,2,3$
This statement accurately formulates Bayes' Theorem. The numerator, $P(A|E_i)P(E_i)$, is equivalent to $P(A \cap E_i)$ by the multiplication rule. The denominator, $\sum_{j=1}^{3} P(A|E_j)P(E_j)$, represents P(A) according to the Law of Total Probability (statement B). Consequently, the formula reduces to $P(E_i|A) = \frac{P(A \cap E_i)}{P(A)}$, which is the definition of conditional probability. Therefore, statement (C) is true.

(D) $P(A|E_i) = \frac{P(E_i|A)P(E_i)}{\sum_{j=1}^{3} P(E_i|A)P(E_j)}, i=1,2,3$
This formulation is incorrect, appearing to be a misapplication of Bayes' Theorem where the roles of A and E_i are confused. Statement (D) is thus false.

Step 4: Conclusion:
The accurate statements are (B) and (C). This corresponds to option (4).