Question:
If \(E_1\) and \(E_2\) are equally likely, mutually exclusive and exhaustive events and \(P(A|E_1) = 0.2\), \(P(A|E_2) = 0.3\), then \(P(E_1|A)\) equal to...
If \(E_1\) and \(E_2\) are equally likely, mutually exclusive and exhaustive events and \(P(A|E_1) = 0.2\), \(P(A|E_2) = 0.3\), then \(P(E_1|A)\) equal to...
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Apply Bayes theorem with equal priors, so the priors cancel.
Updated On: Oct 1, 2026
- \(\frac{1}{5}\)
- \(\frac{4}{5}\)
- \(\frac{2}{3}\)
- \(\frac{2}{5}\)
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The Correct Option is D
Solution and Explanation
Step 1: Setup
\(P(E_1)=P(E_2)=\frac12\), \(P(A|E_1)=0.2\), \(P(A|E_2)=0.3\).
Step 2: Bayes theorem
\[ P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)} = \frac{\frac12(0.2)}{\frac12(0.2)+\frac12(0.3)} \]
Step 3: Simplify
\(\frac{0.2}{0.5} = \frac25\). Option (D).
Final Answer:
The probability is 2/5. \[ \boxed{\text{(D)}\ \frac25} \]
\(P(E_1)=P(E_2)=\frac12\), \(P(A|E_1)=0.2\), \(P(A|E_2)=0.3\).
Step 2: Bayes theorem
\[ P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)} = \frac{\frac12(0.2)}{\frac12(0.2)+\frac12(0.3)} \]
Step 3: Simplify
\(\frac{0.2}{0.5} = \frac25\). Option (D).
Final Answer:
The probability is 2/5. \[ \boxed{\text{(D)}\ \frac25} \]
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