Suppose I give you a list of 20 problems to study, from which I will randomly pick 12 questions for an exam.

For whatever reason, you prepare for this exam by completing and understanding how to solve 10 questions of the 20, so there are 10 questions you do not know how to solve.To pass the midterm, you must correctly solve at least half of the 12 questions. Compute the probability that you will pass the exam

To solve this problem, we'll use combinatorics and the hypergeometric distribution. Here's a breakdown of the situation:

- **Total questions available**: 20
- **Questions you know how to solve**: 10
- **Questions you don't know how to solve**: 10
- **Total questions picked for the exam**: 12
- **You need to correctly solve at least half of the questions**: At least \( 6 \) correct answers out of \( 12 \).

Let’s calculate the probability of passing, which means solving 6 or more of the 12 exam questions correctly.

### Step 1: Total Number of Ways to Select 12 Questions from 20
The total number of ways to select 12 questions out of 20 is given by the combination formula:

\[
\binom{20}{12} = \frac{20!}{12!(20-12)!} = \binom{20}{12}
\]

### Step 2: Probability Distribution (Hypergeometric)

This is a hypergeometric problem, where you are drawing 12 questions without replacement from a population of 20 questions, and you want to count how many of the drawn questions are from the subset of 10 you know how to solve.

The probability of getting exactly \( k \) questions correct (where \( k \) is at least 6) is given by the hypergeometric formula:

\[
P(\text{correct} = k) = \frac{\binom{10}{k} \binom{10}{12-k}}{\binom{20}{12}}
\]

- \( \binom{10}{k} \) is the number of ways to choose \( k \) questions that you know how to solve from the 10 you know.
- \( \binom{10}{12-k} \) is the number of ways to choose \( 12-k \) questions from the 10 you don’t know.
- \( \binom{20}{12} \) is the total number of ways to select 12 questions from 20.

### Step 3: Calculate the Probability of Getting at Least 6 Correct
You need to sum the probabilities for \( k = 6, 7, 8, 9, 10 \) correct answers. Therefore, the probability of passing is:

\[
P(\text{pass}) = \sum_{k=6}^{10} \frac{\binom{10}{k} \binom{10}{12-k}}{\binom{20}{12}}
\]

Let me calculate this for you.The probability that you will pass the exam by correctly solving at least 6 out of the 12 questions is approximately **0.675** or **67.5%**.

Would you like more details on how this was calculated or have any further questions? Here are 5 related questions to consider:

1. What is the probability of solving exactly 6 questions correctly?
2. How does the probability change if you prepare for 12 questions instead of 10?
3. What if the exam has only 10 questions but you still need to solve 6 or more to pass?
4. How would the probability change if you needed 8 correct answers to pass instead of 6?
5. How can this concept be generalized to other types of exams with different numbers of questions?

**Tip**: Understanding the hypergeometric distribution is useful for problems involving sampling without replacement, common in exams and selection processes.

Suppose I give you a list of 20 problems to study, from which I will randomly pick 12 questions for an exam. For whatever reason, you prepare for this exam by completing and understanding how to solve 10 questions of the 20, so there are 10 questions you do not know how to solve. To pass the midterm, you must correctly solve at least half of the 12 questions. Compute the probability that you will pass the exam.

Learn how to calculate the probability of passing an exam by solving at least 6 out of 12 questions, using combinatorics and the hypergeometric distribution. This detailed example covers probability theory concepts such as combination formulas and hypergeometric probability distribution, ideal for students studying undergraduate statistics or probability.

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