Question

1. given: p(e) = 0.27, p(f) = 0.46, and p(e ∩ f) = 0.11 find p(e|f). a) .76 b) .12 c) .88 d) .24 e) .41

Explanation:

Step1: Recall Conditional Probability Formula

The formula for conditional probability is \( P(E|F) = \frac{P(E \cap F)}{P(F)} \), where \( P(E \cap F) \) is the probability of both \( E \) and \( F \) occurring, and \( P(F) \) is the probability of \( F \) occurring.

Step2: Substitute Given Values

We are given \( P(E \cap F) = 0.111 \) and \( P(F) = 0.46 \). Substitute these values into the formula:
\( P(E|F) = \frac{0.111}{0.46} \)

Step3: Calculate the Result

Perform the division: \( \frac{0.111}{0.46} \approx 0.2413 \), which is approximately \( 0.24 \) (matching option d) or more precisely, let's check the calculation again. Wait, \( 0.111\div0.46 = 0.2413\approx0.24 \)? Wait, no, wait: \( 0.46\times0.24 = 0.1104 \), which is close to \( 0.111 \). Wait, maybe I miscalculated. Wait, \( 0.111\div0.46 = 0.2413\approx0.24 \) (option d) or wait, the options are a) 0.76, b) 0.12, c) 0.88, d) 0.24, e) 0.41. Wait, let's do the division properly: \( 0.111 \div 0.46 \). Let's multiply numerator and denominator by 1000 to eliminate decimals: \( 111 \div 460 \approx 0.2413 \), which is approximately 0.24, so option d.

Answer:

d. 0.24