Question

Let E and F be two events, if $P(E|F)=0.5$, $P(E|\overline{F})=0.6$ and $P(F)=0.6$ then $P(E)$ equals

• A. 0.56
• B. 0.44
• C. 0.54
• D. 0.46

Hint:

The Theorem of Total Probability is like a weighted average. You are essentially finding the "average" chance of E happening, weighted by how likely F and not-F are to occur.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:

Given \( P(E|F) = 0.5 \), \( P(E|\overline{F}) = 0.6 \) and \( P(F) = 0.6 \); find \( P(E) \).

Step 2: Detailed Explanation:

• Use the total probability theorem: \( P(E) = P(E|F)P(F) + P(E|\overline{F})P(\overline{F}) \).

• \( P(\overline{F}) = 1 - 0.6 = 0.4 \).

• \( P(E) = 0.5 \times 0.6 + 0.6 \times 0.4 = 0.30 + 0.24 = 0.54 \).

Step 3: Final Answer:
The correct option is (C) 0.54.