Question

If A is any event associated with sample space and if E\textsubscript{1}, E\textsubscript{2}, E\textsubscript{3} 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 Overview:
This problem involves the theorem of total probability and Bayes' theorem. Events E\textsubscript{1}, E\textsubscript{2}, and E\textsubscript{3} constitute a partition of the sample space because they are mutually exclusive (cannot occur simultaneously) and exhaustive (one must occur).
Step 3: Detailed Analysis:
We will evaluate 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 statement does not represent the standard law of total probability. The correct formula links P(A) to conditional probabilities of A given E\textsubscript{i}, not the reverse. Using the definition of conditional probability, $P(E_i)P(E_i|A) = P(E_i \cap A)/P(A) * P(E_i)$, which does not simplify to P(A). Thus, (A) is incorrect.
(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 articulates the Law of Total Probability. It calculates the total probability of event A based on its probabilities conditional on each event within a sample space partition. As $P(A|E_i)P(E_i) = P(A \cap E_i)$, this formula is equivalent to $P(A) = P(A \cap E_1) + P(A \cap E_2) + P(A \cap E_3)$, which is valid because the E\textsubscript{i} events are mutually exclusive and exhaustive. Therefore, (B) is correct.
(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 accurate statement of Bayes' Theorem. The numerator is $P(A \cap E_i)$ by the multiplication rule. The denominator, by the law of total probability (as shown in statement B), is P(A). Consequently, the formula simplifies to $P(E_i|A) = \frac{P(A \cap E_i)}{P(A)}$, which is the definition of conditional probability. Thus, (C) is correct.
(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. It appears to be a misstatement of Bayes' theorem, where the roles of A and E\textsubscript{i} have been interchanged. Therefore, (D) is incorrect.
Step 4: Conclusion:
The correct statements are (B) and (C). This aligns with option (4).