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This problem involves a sequence where the probability of making the \( n \)-th free throw depends on the proportion of successful throws in the first \( n-1 \) attempts.

Let’s break it down step by step:

### Known:
1. Frank makes his first free throw and misses his second.
   - So, for \( n = 1 \), \( P(\text{make 1st}) = 1 \).
   - For \( n = 2 \), \( P(\text{make 2nd}) = 0 \).

2. For \( n \geq 3 \), the probability of making the \( n \)-th free throw is the proportion of successful free throws in the first \( n - 1 \) attempts. This means:
   \[
   P(\text{make } n) = \frac{\text{number of successful throws in the first } n-1}{n-1}
   \]
   
### Approach:
To compute the expected number of successful free throws after 100 attempts, we must sum the expected success probabilities for all attempts from 1 to 100.

We know the first attempt is a success and the second is a miss, so we start from \( n = 3 \). Each probability from \( n = 3 \) to \( n = 100 \) will depend on the number of successful throws in the previous \( n-1 \) throws.

Let’s calculate the expected value. I will work through this computation step by step.Frank can expect to make 50 successful free throws out of 100 attempts based on the given probability structure.

Would you like more details on the calculation process or have any questions? Here are 5 related questions that could deepen understanding:

1. How would the expected value change if Frank had a different initial success/failure pattern?
2. How does the dependency of each attempt's probability on previous outcomes affect the overall expectation?
3. What would happen to the expected number of free throws if the formula for probability changed slightly?
4. How can this problem be generalized to any number of total attempts?
5. What happens to the expected number of successes if Frank misses both of his first two throws?

**Tip:** In problems with dependent probabilities, simulating or iterating through the sequence is a useful technique to avoid complex algebraic expressions.

Frank is shooting free throws. He makes his first free throw and misses his second free throw. For n ≥ 3, the probability of making the nth free throw is equal to the proportion of free throws he made during his first n − 1 attempts. How many free throws can Frank expect to make in 100 attempts?

Explore how to calculate the expected number of successful free throws Frank can make in 100 attempts, where each free throw's probability depends on the proportion of successful prior attempts. Includes step-by-step reasoning and probability calculations suitable for Grades 10-12.

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