Question

if 56% of the people at a certain conference are doctors, 50% are men, and 66% are doctors or men (or both), what is the probability that a person selected at random at this conference is a male doctor?
write your answer as a decimal (not as a percentage).

Explanation:

Step1: Recall the principle of inclusion - exclusion for probabilities.

The formula for the probability of the union of two events \( A \) and \( B \) is \( P(A\cup B)=P(A)+P(B)-P(A\cap B) \), where \( A \) is the event of being a doctor and \( B \) is the event of being a man. We know \( P(A) = 0.56\), \( P(B)=0.50\), and \( P(A\cup B)=0.66 \). We need to find \( P(A\cap B) \) (the probability of being a male doctor).

Step2: Rearrange the inclusion - exclusion formula.

From \( P(A\cup B)=P(A)+P(B)-P(A\cap B) \), we can solve for \( P(A\cap B) \) by rearranging the formula: \( P(A\cap B)=P(A)+P(B)-P(A\cup B) \).

Step3: Substitute the given values into the formula.

Substitute \( P(A) = 0.56\), \( P(B) = 0.50\), and \( P(A\cup B)=0.66 \) into the formula: \( P(A\cap B)=0.56 + 0.50-0.66 \).
First, calculate \( 0.56+0.50 = 1.06 \). Then, subtract \( 0.66 \) from \( 1.06 \): \( 1.06 - 0.66=0.4 \).

Answer:

\( 0.4 \)