Question

A, B, C are mutually exclusive and exhaustive events of a random experiment and E is an event that occurs in conjunction with one of the events A, B, C. The conditional probabilities of E given the happening of A, B, C are respectively 0.6, 0.3 and 0.1. If \( P(A) = 0.30 \) and \( P(B) = 0.50 \), then \( P(C \mid E) = \):

• A. \( \frac{2}{35} \)
• B. \( \frac{15}{35} \)
• C. \( \frac{18}{35} \)
• D. \( \frac{17}{35} \)

Hint:

In problems involving conditional probability with mutually exclusive and exhaustive events, use the Law of Total Probability to calculate \( P(E) \), and then apply Bayes' Theorem to find the conditional probability.

Answer 1

The Correct Option is A

To determine \( P(C \mid E) \), use Bayes' theorem:
\( P(C \mid E) = \frac{P(E \mid C)P(C)}{P(E)} \).
Step 1: Calculate \( P(E) \)
Since A, B, and C are mutually exclusive and exhaustive, \( P(E) \) is:
\( P(E) = P(E \mid A)P(A) + P(E \mid B)P(B) + P(E \mid C)P(C) \)
Given:
\( P(A) = 0.30 \), \( P(B) = 0.50 \), \( P(E \mid A) = 0.6 \), \( P(E \mid B) = 0.3 \), \( P(E \mid C) = 0.1 \).
Therefore,
\( P(E) = 0.6 \times 0.30 + 0.3 \times 0.50 + 0.1 \times (1 - 0.30 - 0.50) \).
\( P(E) = 0.18 + 0.15 + 0.1 \times 0.2 \).
\( P(E) = 0.18 + 0.15 + 0.02 \).
\( P(E) = 0.35 \).
Step 2: Calculate \( P(C) \)
\( P(C) = 1 - P(A) - P(B) = 1 - 0.30 - 0.50 = 0.20 \).
Step 3: Calculate \( P(C \mid E) \)
\( P(C \mid E) = \frac{P(E \mid C)P(C)}{P(E)} = \frac{0.1 \times 0.20}{0.35} \).
\( P(C \mid E) = \frac{0.02}{0.35} = \frac{2}{35} \).
The correct answer is \( \frac{2}{35} \).

Answer 2

We are given that: - A, B, C are mutually exclusive and exhaustive events. - The conditional probabilities are: \[ P(E \mid A) = 0.6, \quad P(E \mid B) = 0.3, \quad P(E \mid C) = 0.1. \] - The probabilities of \( A \) and \( B \) are: \[ P(A) = 0.30, \quad P(B) = 0.50. \] - We need to find \( P(C \mid E) \). 
Step 1: Use the Total Probability Theorem. The total probability of event \( E \) is given by the Law of Total Probability: \[ P(E) = P(E \mid A)P(A) + P(E \mid B)P(B) + P(E \mid C)P(C). \] Substitute the given values: \[ P(E) = (0.6 \times 0.30) + (0.3 \times 0.50) + (0.1 \times P(C)). \] This simplifies to: \[ P(E) = 0.18 + 0.15 + 0.1P(C). \] Thus, we have: \[ P(E) = 0.33 + 0.1P(C). \] 
Step 2: Use Bayes' Theorem to find \( P(C \mid E) \). Bayes' Theorem tells us that: \[ P(C \mid E) = \frac{P(E \mid C)P(C)}{P(E)}. \] Substitute the known values: \[ P(C \mid E) = \frac{0.1 \times P(C)}{0.33 + 0.1P(C)}. \] 
Step 3: Find \( P(C) \). Since \( A, B, C \) are mutually exclusive and exhaustive events, we have: \[ P(A) + P(B) + P(C) = 1. \] Substitute the known values: \[ 0.30 + 0.50 + P(C) = 1 \quad \Rightarrow \quad P(C) = 1 - 0.80 = 0.20. \] 
Step 4: Substitute \( P(C) \) into the Bayes' Theorem equation. Now, substitute \( P(C) = 0.20 \) into the equation for \( P(C \mid E) \): \[ P(C \mid E) = \frac{0.1 \times 0.20}{0.33 + 0.1 \times 0.20} = \frac{0.02}{0.33 + 0.02} = \frac{0.02}{0.35} = \frac{2}{35}. \] Thus, the correct answer is: \[ \boxed{\frac{2}{35}}. \]