Probability Rules for Mutually Exclusive and Exhaustive Events

This question delves into fundamental probability concepts, specifically focusing on events that are both mutually exclusive and exhaustive within a given sample space. Let A be an event, and let $E_1, E_2, E_3$ be three events that partition the sample space. This means they do not overlap ($E_i \cap E_j = \emptyset$ for $i \neq j$) and together they cover the entire sample space ($E_1 \cup E_2 \cup E_3 = S$). A key consequence of these properties is that the sum of their probabilities equals 1:

$P(E_1) + P(E_2) + P(E_3) = 1$

Understanding the Law of Total Probability (Option B)

Option (B) states:

$P(A) = P(A|E_1)P(E_1) + P(A|E_2)P(E_2)+P(A|E_3)P(E_3)$

This formula is a direct representation of the Law of Total Probability. It calculates the probability of event A by considering the conditional probabilities of A occurring given each of the partitioning events ($E_1, E_2, E_3$) and weighting them by the probabilities of those partitioning events. Since $E_1, E_2, E_3$ are mutually exclusive and exhaustive, this law applies perfectly. Therefore, Option (B) is true.

Applying Bayes' Theorem (Option C)

Option (C) states:

$P(E_i|A) = \frac{P(A|E_i)P(E_i)}{\sum_{j=1}^{3}P(A|E_j)P(E_j)}, i = 1,2,3$

Let's analyze this formula. The numerator, $P(A|E_i)P(E_i)$, relates to the joint probability $P(A \cap E_i)$. The denominator is $\sum_{j=1}^{3}P(A|E_j)P(E_j)$. From our analysis of Option (B), we know this sum is equal to $P(A)$ according to the Law of Total Probability.

Substituting $P(A)$ into the denominator, the formula becomes:

$P(E_i|A) = \frac{P(A|E_i)P(E_i)}{P(A)}$

This is the standard form of Bayes' Theorem. It allows us to reverse the conditional probability, finding the probability of $E_i$ given A, using the probability of A given $E_i$. Since the conditions (mutually exclusive and exhaustive events) are met, Bayes' Theorem holds true. Thus, Option (C) is also true.

Evaluating Incorrect Options (A and D)

• Option (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 incorrectly uses $P(E_i|A)$ instead of $P(A|E_i)$ and multiplies by $P(E_i)$ instead of $P(A)$. It does not align with standard probability theorems like the Law of Total Probability or Bayes' Theorem.
• Option (D): $P(A|E_i) = \frac{P(E_i|A)P(E_i)}{\sum_{j=1}^{3}P(E_j|A)P(E_j)}$
  This expression incorrectly formulates Bayes' Theorem. While the numerator resembles part of the calculation, the denominator is not structured correctly to represent $P(A)$ or any standard normalization constant in this context. The correct denominator should involve terms like $P(A|E_j)P(E_j)$, as seen in Option (C).

Conclusion

Based on the analysis, both Option (B) (Law of Total Probability) and Option (C) (Bayes' Theorem) are correct statements given that $E_1, E_2, E_3$ are mutually exclusive and exhaustive events.