Question

a high school offers both spanish and french classes. the probability that a student takes both spanish and french is 0.24. the probability that a student takes spanish given that the student takes french is 0.32. what is the probability that a student takes french? a. 0.01 b. 0.08 c. 0.56 d. 0.75

Explanation:

Step1: Recall Conditional Probability Formula

The formula for conditional probability is \( P(A|B) = \frac{P(A \cap B)}{P(B)} \), where \( P(A|B) \) is the probability of \( A \) given \( B \), \( P(A \cap B) \) is the probability of both \( A \) and \( B \) occurring, and \( P(B) \) is the probability of \( B \) occurring. Here, let \( A \) be "taking Spanish" and \( B \) be "taking French". We know \( P(A \cap B) = 0.24 \) and \( P(A|B) = 0.32 \). We need to find \( P(B) \)? Wait, no, wait. Wait, the question is "What is the probability that a student takes French?" Wait, no, re-reading: Wait, the problem says: "A high school offers both Spanish and French classes. The probability that a student takes both Spanish and French is 0.24. The probability that a student takes Spanish given that the student takes French is 0.32. What is the probability that a student takes French?" Wait, no, wait, the conditional probability formula: \( P(Spanish | French) = \frac{P(Spanish \cap French)}{P(French)} \). So we have \( 0.32 = \frac{0.24}{P(French)} \). Wait, no, wait, solving for \( P(French) \): \( P(French) = \frac{P(Spanish \cap French)}{P(Spanish | French)} \)? Wait, no, wait, the formula is \( P(A|B) = \frac{P(A \cap B)}{P(B)} \), so rearranged, \( P(B) = \frac{P(A \cap B)}{P(A|B)} \). Wait, no, \( A \) is Spanish, \( B \) is French. So \( P(Spanish | French) = \frac{P(Spanish \cap French)}{P(French)} \), so \( P(French) = \frac{P(Spanish \cap French)}{P(Spanish | French)} \)? Wait, no, that would be if we know \( P(Spanish | French) \) and \( P(Spanish \cap French) \), but wait, no, the question is "What is the probability that a student takes French?" Wait, maybe I misread. Wait, the options are 0.01, 0.08, 0.56, 0.75. Wait, maybe the problem is: Wait, maybe the given is \( P(Spanish \cap French) = 0.24 \), \( P(Spanish | French) = 0.32 \), find \( P(French) \). Then using \( P(Spanish | French) = \frac{P(Spanish \cap French)}{P(French)} \), so \( 0.32 = \frac{0.24}{P(French)} \), so \( P(French) = \frac{0.24}{0.32} = 0.75 \)? Wait, no, 0.24 divided by 0.32 is 0.75? Wait, 0.24 / 0.32 = (24/100)/(32/100) = 24/32 = 3/4 = 0.75. Wait, but that would be D. But wait, maybe the problem is the other way? Wait, maybe the question is "What is the probability that a student takes French given that they take Spanish?" No, the question says "What is the probability that a student takes French?" Wait, no, let's re-express:

Wait, the problem statement: "A high school offers both Spanish and French classes. The probability that a student takes both Spanish and French is 0.24. The probability that a student takes Spanish given that the student takes French is 0.32. What is the probability that a student takes French?"

So let \( F \) be the event that a student takes French, \( S \) be the event that a student takes Spanish.

We know:

- \( P(S \cap F) = 0.24 \)
- \( P(S | F) = 0.32 \)

We need to find \( P(F) \).

Using the conditional probability formula:

\( P(S | F) = \frac{P(S \cap F)}{P(F)} \)

Rearranging to solve for \( P(F) \):

\( P(F) = \frac{P(S \cap F)}{P(S | F)} \)

Substitute the known values:

\( P(F) = \frac{0.24}{0.32} = 0.75 \)

Wait, but 0.24 divided by 0.32 is 0.75? Let's check: 0.32 times 0.75 is 0.24. Yes, because 0.32 * 0.75 = 0.24. So that works. So the probability that a student takes French is 0.75, which is option D.

Wait, but maybe I misread the question. Let me check again. The question says: "What is the probability that a student takes French?" Given that \( P(S \cap F) = 0.24 \) and \( P(S | F) = 0.32 \). So using the formula, we get \( P(F) = 0.75 \), which is option D.

Step1: Identify the formula for conditional probability

The formula for conditional probability is \( P(A|B) = \frac{P(A \cap B)}{P(B)} \), where \( A \) is the event "takes Spanish" and \( B \) is the event "takes French".

Step2: Rearrange the formula to solve for \( P(B) \)

From \( P(S|F) = \frac{P(S \cap F)}{P(F)} \), we rearrange to get \( P(F) = \frac{P(S \cap F)}{P(S|F)} \).

Step3: Substitute the given values

We know \( P(S \cap F) = 0.24 \) and \( P(S|F) = 0.32 \). Substituting these values into the formula:
\( P(F) = \frac{0.24}{0.32} = 0.75 \).

Answer:

D. 0.75