Question: Discuss the validity of the following statement. If the statement is always true, explain why. If…

Discuss the validity of the following statement. If the statement is always true, explain why. If not, give a counterexample. If the odds for E equal the odds against E’, then P(E) = \frac{1}{2}.

Choose the correct answer below. A. The statement is false because the odds for E is equal to the odds against E’ only when P(E) is less than 0.5. B. The statement is true because because when P(E) is equal to 0.5, the odds for E is 50% chance and the odds against E’ is also 50% chance. C. The statement is false because the odds for E is always equal to the odds against E’ regardless the value of P(E). D. The statement is true because the odds for E is equal to the odds against E’ when P(E) is equal to 0.5.

Solution

To determine the validity of the statement, we need to understand the relationship between the odds for an event \(E\) and its probability \(P(E)\). The odds for an event \(E\) are defined as the ratio of the probability of \(E\) to the probability of its complement \(E’\): \[ \text{Odds for } E = \frac{P(E)}{P(E’)} \] Since \(P(E) + P(E’) = 1\), we have \(P(E’) = 1 - P(E)\). So, \[ \text{Odds for } E = \frac{P(E)}{1 - P(E)} \] The statement says that the odds for \(E\) equal the odds against \(E’\). The odds against \(E’\) means the odds against the complement of \(E\). Thus, the odds against \(E’\) is: \[ \frac{P(E’)}{P(E)} = \frac{1 - P(E)}{P(E)} \] If these odds are equal, we have: \[ \frac{P(E)}{1 - P(E)} = \frac{1 - P(E)}{P(E)} \] Cross-multiplying gives: \[ P(E)^2 = (1 - P(E))^2 \] Taking the square root of both sides: \[ P(E) = 1 - P(E) \] Solving for \(P(E)\): \[ 2P(E) = 1 \] \[ P(E) = \frac{1}{2} \] Thus, the statement is true only when \(P(E) = \frac{1}{2}\). Therefore, the correct answer is: D. The statement is true because the odds for \(E\) is equal to the odds against \(E’\) when \(P(E)\) is equal to 0.5.