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This problem involves conditional probability.

### Let's break it down:

- **Probability of passing French 101 (P(A))** = 0.73
- **Probability of passing French 102 given that the student passed French 101 (P(B|A))** = 0.93

We are asked to find the probability that a student passes **both French 101 and French 102**. This is the joint probability, \( P(A \cap B) \), and we can calculate it using the formula:

\[
P(A \cap B) = P(A) \times P(B|A)
\]

### Substituting the values:

\[
P(A \cap B) = 0.73 \times 0.93 = 0.6789
\]

So, the probability that a student selected at random will pass both French 101 and French 102 is **0.6789** (or about 67.89%).

Would you like more details or clarifications on any part?

Here are 5 related questions to expand your understanding:

1. What is the probability of passing only French 101 but not French 102?
2. How would the probability change if the conditional probability \( P(B|A) \) were different?
3. How can we interpret joint probability in the context of multiple events?
4. What if the probability of passing French 101 were 0.9, how would that affect the result?
5. How does conditional probability differ from independent probability?

**Tip:** Always verify if the events are independent or dependent before calculating probabilities, as it affects the use of formulas like joint or conditional probability.

Class records at a particular college indicate that a student selected at random has probability 0.73 of passing French 101. For the student who passes French 101, the probability is 0.93 that he or she will pass French 102. What is the probability that a student selected at random will pass both French 101 and French 102?

Learn how to calculate the probability that a student will pass both French 101 and French 102. The probability of passing French 101 is 0.73, and the conditional probability of passing French 102 given passing French 101 is 0.93. The solution involves applying the conditional probability formula to determine the joint probability of passing both subjects.

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