Question

In a certain town, 40% of people have brown hair, 30% of people have brown eyes and 12% have both brown hair and brown eyes.
How many people in town have neither brown hair nor brown eyes?

• A. 0.41
• B. 0.42
• C. 0.43
• D. 0.44

Answer 1

The Correct Option is B

To determine the percentage of people in the town who have neither brown hair nor brown eyes, we can use the principle of Inclusion-Exclusion. Let's solve the problem step by step: 

1. Let \(P(A)\) be the probability (or percentage) of people having brown hair, which is 40%. Thus, \(P(A) = 0.40\).
2. Let \(P(B)\) be the probability (or percentage) of people having brown eyes, which is 30%. Thus, \(P(B) = 0.30\).
3. Let \(P(A \cap B)\) be the probability (or percentage) of people having both brown hair and brown eyes, which is 12%. Thus, \(P(A \cap B) = 0.12\).
4. Using the principle of Inclusion-Exclusion, the probability of people having either brown hair or brown eyes or both can be calculated as follows:

\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)

Substituting the values, we have:

\(P(A \cup B) = 0.40 + 0.30 - 0.12 = 0.58\)

1. This means 58% of the people have either brown hair, brown eyes, or both.
2. Therefore, the percentage of people who have neither brown hair nor brown eyes is the complement of \(P(A \cup B)\):

\(1 - P(A \cup B) = 1 - 0.58 = 0.42\)

1. Thus, 42% of the people in the town have neither brown hair nor brown eyes.

Hence, the correct answer is 0.42.