In a survey, 60% of people like apples, 50% like bananas, and 20% like both. Given a randomly selected person likes at least one of the fruits, what is the probability they like ap... In a survey, 60% of people like apples, 50% like bananas, and 20% like both. Given a randomly selected person likes at least one of the fruits, what is the probability they like apples?
Understand the Problem
The question is asking for a conditional probability: the probability that a person likes apples given that they like at least one of the two fruits (apples or bananas). We are given the individual probabilities of liking apples and bananas, as well as the probability of liking both. We will use the formula for conditional probability: P(A|B) = P(A and B) / P(B), where A is liking apples and B is liking at least one of the fruits.
Answer
$0.75$
Answer for screen readers
$0.75$
Steps to Solve
- Find the probability of liking at least one fruit
We need to find $P(A \cup B)$, which is the probability of liking apples or bananas or both. We can use the formula: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ $P(A \cup B) = 0.6 + 0.4 - 0.2 = 0.8$
- Find the probability of liking apples and liking at least one fruit
We want to find $P(A \cap (A \cup B))$. Since $A$ is a subset of $A \cup B$, $A \cap (A \cup B) = A$. Therefore, $P(A \cap (A \cup B)) = P(A) = 0.6$.
- Apply the conditional probability formula
We want to find $P(A \mid A \cup B)$. Using the formula for conditional probability: $P(A \mid A \cup B) = \frac{P(A \cap (A \cup B))}{P(A \cup B)}$ $P(A \mid A \cup B) = \frac{P(A)}{P(A \cup B)} = \frac{0.6}{0.8} = \frac{3}{4} = 0.75$
$0.75$
More Information
The probability that a person likes apples given that they like at least one of the two fruits is 0.75 or 75%.
Tips
A common mistake is to incorrectly calculate $P(A \cup B)$ or to confuse $P(A \cap (A \cup B))$ with $P(A \cap B)$. Remember that $A \cap (A \cup B) = A$. Also, remember the conditional probability formula $P(A|B) = \frac{P(A \cap B)}{P(B)}$
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