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Question 6
There are three boxes on the table: The first box contains 2 quarters, the second box contains 2 nickels, and the last box contains 1 quarter and 1 nickel. You choose a box at random, then you pick a coin at random from the chosen box.

If the coin you picked is a quarter, what's the chance that the other coin in the box is also a quarter?
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To solve this problem, we need to use conditional probability. Specifically, we want to find the probability that the other coin in the box is also a quarter given that we have picked a quarter. We will apply Bayes' theorem for this purpose.

Let's denote:
- \( A \) as the event that the box chosen is the first box (which contains 2 quarters).
- \( B \) as the event that the coin picked is a quarter.

We need to find \( P(A|B) \), the probability that we chose the first box given that we picked a quarter.

By Bayes' theorem:
\[ P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)} \]

First, let's calculate each component:

1. **\( P(A) \):** The probability of choosing the first box.
\[ P(A) = \frac{1}{3} \]

2. **\( P(B|A) \):** The probability of picking a quarter given that we chose the first box.
\[ P(B|A) = 1 \] (since the first box contains only quarters).

3. **\( P(B) \):** The total probability of picking a quarter. This is calculated by considering all possible ways to pick a quarter:
\[ P(B) = P(B|A)P(A) + P(B|B')P(B') + P(B|C)P(C) \]
where
- \( B' \) is the event of choosing the second box (which contains 2 nickels),
- \( C \) is the event of choosing the third box (which contains 1 quarter and 1 nickel).

We already know:
\[ P(A) = \frac{1}{3}, \quad P(B'|B') = \frac{1}{3}, \quad P(C) = \frac{1}{3} \]

Now, let's find:
\[ P(B|B') = 0 \] (since the second box contains no quarters),
\[ P(B|C) = \frac{1}{2} \] (since the third box contains 1 quarter and 1 nickel).

So,
\[ P(B) = P(B|A)P(A) + P(B|B')P(B') + P(B|C)P(C) \]
\[ P(B) = 1 \cdot \frac{1}{3} + 0 \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{1}{3} \]
\[ P(B) = \frac{1}{3} + 0 + \frac{1}{6} \]
\[ P(B) = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \]

Now, applying Bayes' theorem:
\[ P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)} = \frac{1 \cdot \frac{1}{3}}{\frac{1}{2}} = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{1}{3} \cdot \frac{2}{1} = \frac{2}{3} \]

Thus, the probability that the other coin in the box is also a quarter, given that you picked a quarter, is \( \frac{2}{3} \).

Do you want more details or have any questions?

Here are 5 more questions you might find interesting:
1. How do you calculate the probability of an event happening given multiple conditions?
2. Can you explain the concept of conditional probability in simple terms?
3. How does Bayes' theorem apply to real-life situations?
4. What is the difference between independent and dependent events in probability?
5. How can you calculate the probability of multiple events happening in sequence?

**Tip:** Always list out all possible outcomes and their probabilities when dealing with probability problems to ensure you consider all scenarios.

Explore a probability problem involving three boxes with different coin compositions. Learn how to apply Bayes' theorem to find the probability that the other coin in the box is also a quarter, given that a quarter was picked. Suitable for advanced high school students.

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