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: Understanding the Concept:
This question is about the theorem of total probability and Bayes' theorem. The events E\textsubscript{1}, E\textsubscript{2}, E\textsubscript{3} form a partition of the sample space, as they are mutually exclusive (they cannot happen at the same time) and exhaustive (one of them must happen).
Step 3: Detailed Explanation:
Let's analyze each statement.
(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 is not the standard formula for the law of total probability. The formula relates P(A) to conditional probabilities of A given E\textsubscript{i}, not the other way around. From the definition of conditional probability, $P(E_i)P(E_i|A) = P(E_i \cap A)/P(A) * P(E_i)$, which doesn't simplify nicely to P(A). So, (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 is the correct statement of the Law of Total Probability. It expresses the total probability of an event A in terms of the probabilities of A conditional on each event in a partition of the sample space. Since $P(A|E_i)P(E_i) = P(A \cap E_i)$, this formula is equivalent to $P(A) = P(A \cap E_1) + P(A \cap E_2) + P(A \cap E_3)$, which is true because E\textsubscript{i} are mutually exclusive and exhaustive. So, (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 is the correct statement of Bayes' Theorem. The numerator, by the multiplication rule, is $P(A \cap E_i)$. The denominator, by the law of total probability (from statement B), is P(A). So, the formula becomes $P(E_i|A) = \frac{P(A \cap E_i)}{P(A)}$, which is the definition of conditional probability. So, (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 is an incorrect formulation. It seems to be a misstatement of Bayes' theorem, with the roles of A and E\textsubscript{i} confused. So, (D) is false.
Step 4: Final Answer:
The true statements are (B) and (C). This corresponds to option (4).