Question

Events E₁ and E₂ form a partition of the sample space S. A is any event that P(E₁) = P(E₂) = \(\frac{1}{2}\), P(E₂/A) = \(\frac{1}{2}\) and P(A/E₂) = \(\frac{2}{3}\), then P(E₁/A) is

• A. \(\frac{1}{2}\)
• B. \(\frac{2}{3}\)
• C. 1
• D. \(\frac{1}{4}\)

Answer 1

The Correct Option is A

Given:

• \(E_1\) and \(E_2\) form a partition of the sample space \(S\). This means \(E_1\) and \(E_2\) are disjoint, and \(E_1 \cup E_2 = S\).
• \(P(E_1) = P(E_2) = \frac{1}{2}\)
• \(P(E_2|A) = \frac{1}{2}\)
• \(P(A|E_2) = \frac{2}{3}\)

We want to find \(P(E_1|A)\). We can use Bayes' theorem:

\(P(E_1|A) = \frac{P(A|E_1) \cdot P(E_1)}{P(A)}\)

We know \(P(E_1) = \frac{1}{2}\). We need to find \(P(A|E_1)\) and \(P(A)\).

Since \(E_1\) and \(E_2\) form a partition:

\(P(A) = P(A \cap E_1) + P(A \cap E_2)\)

\(P(A) = P(A|E_1)P(E_1) + P(A|E_2)P(E_2)\)

We know \(P(A|E_2) = \frac{2}{3}\) and \(P(E_2) = \frac{1}{2}\), so:

\(P(A \cap E_2) = (\frac{2}{3})(\frac{1}{2}) = \frac{1}{3}\)

Also, we are given \(P(E_2|A) = \frac{1}{2}\). Using the definition of conditional probability:

\(P(E_2|A) = \frac{P(E_2 \cap A)}{P(A)}\)

\(\frac{1}{2} = \frac{\frac{1}{3}}{P(A)}\)

\(P(A) = \frac{2}{3}\)

Now we have:

\(P(A) = P(A|E_1)P(E_1) + P(A|E_2)P(E_2)\)

\(\frac{2}{3} = P(A|E_1)(\frac{1}{2}) + (\frac{2}{3})(\frac{1}{2})\)

\(\frac{2}{3} = P(A|E_1)(\frac{1}{2}) + \frac{1}{3}\)

\(P(A|E_1)(\frac{1}{2}) = \frac{1}{3}\)

\(P(A|E_1) = \frac{2}{3}\)

Finally, we can calculate \(P(E_1|A)\):

\(P(E_1|A) = \frac{P(A|E_1) \cdot P(E_1)}{P(A)}\)

\(P(E_1|A) = \frac{(\frac{2}{3}) \cdot (\frac{1}{2})}{\frac{2}{3}}\)

\(P(E_1|A) = \frac{\frac{1}{3}}{\frac{2}{3}}\)

\(P(E_1|A) = \frac{1}{2}\)