 70 cards are marked from 30 to 99 and 2 are drawn at random. Find the probability that one card drawn is
a multiple of 3 and another one is a multiple of 13

To solve this problem, we will approach it in steps:

1. **Total number of cards**:
   - The cards are numbered from 30 to 99, inclusive.
   - So, the total number of cards = \( 99 - 30 + 1 = 70 \).

2. **Cards that are multiples of 3**:
   - The smallest multiple of 3 in this range is \( 30 \) and the largest is \( 99 \).
   - The multiples of 3 in this range are \( 30, 33, 36, \dots, 99 \).
   - This forms an arithmetic sequence where \( a = 30 \), \( d = 3 \), and the last term is \( 99 \).
   - To find the number of terms, use the formula for the \( n \)-th term of an arithmetic sequence:
     \[
     a_n = a + (n - 1) \cdot d
     \]
     Set \( a_n = 99 \):
     \[
     99 = 30 + (n - 1) \cdot 3
     \]
     Solving for \( n \):
     \[
     99 - 30 = (n - 1) \cdot 3
     \]
     \[
     69 = (n - 1) \cdot 3
     \]
     \[
     n - 1 = 23 \quad \Rightarrow \quad n = 24
     \]
     Therefore, there are 24 cards that are multiples of 3.

3. **Cards that are multiples of 13**:
   - The smallest multiple of 13 in this range is \( 39 \) and the largest is \( 91 \).
   - The multiples of 13 in this range are \( 39, 52, 65, 78, 91 \).
   - This forms an arithmetic sequence where \( a = 39 \), \( d = 13 \), and the last term is \( 91 \).
   - Using the same approach, find the number of terms:
     \[
     a_n = a + (n - 1) \cdot d
     \]
     Set \( a_n = 91 \):
     \[
     91 = 39 + (n - 1) \cdot 13
     \]
     Solving for \( n \):
     \[
     91 - 39 = (n - 1) \cdot 13
     \]
     \[
     52 = (n - 1) \cdot 13
     \]
     \[
     n - 1 = 4 \quad \Rightarrow \quad n = 5
     \]
     Therefore, there are 5 cards that are multiples of 13.

4. **Cards that are multiples of both 3 and 13**:
   - These would be multiples of \( 3 \times 13 = 39 \).
   - The multiples of 39 in this range are \( 39, 78 \), so there are 2 cards that are multiples of both 3 and 13.

5. **Number of favorable outcomes**:
   - We want one card to be a multiple of 3 and the other to be a multiple of 13.
   - First, subtract the multiples of both from the sets of multiples of 3 and 13.
     - Cards that are only multiples of 3: \( 24 - 2 = 22 \).
     - Cards that are only multiples of 13: \( 5 - 2 = 3 \).
   - Now, we compute the number of favorable outcomes:
     - Choose one card from the multiples of 3 (22 cards) and one card from the multiples of 13 (3 cards).
     - So, the number of favorable outcomes = \( 22 \times 3 = 66 \).

6. **Total number of ways to choose 2 cards from 70**:
   - The total number of ways to choose 2 cards from 70 is:
     \[
     \binom{70}{2} = \frac{70 \times 69}{2} = 2415
     \]

7. **Probability**:
   - The probability is the ratio of favorable outcomes to total outcomes:
     \[
     P = \frac{66}{2415}
     \]
     Simplifying this fraction:
     \[
     P = \frac{22}{805}
     \]

Thus, the probability that one card is a multiple of 3 and the other is a multiple of 13 is \( \frac{22}{805} \).

Would you like more details or have any questions?

Here are 5 related questions to expand on this:
1. How would the probability change if 3 cards were drawn instead of 2?
2. What is the probability of drawing two cards that are both multiples of 3?
3. Can you find the probability of drawing two cards, one a multiple of 5 and another a multiple of 13?
4. How does the presence of cards that are multiples of both 3 and 13 affect the probability?
5. What is the probability that both cards drawn are multiples of 13?

**Tip:** When working with probability, always ensure you account for overlaps in sets by removing common elements, as we did with multiples of both 3 and 13.

70 cards are marked from 30 to 99 and 2 are drawn at random. Find the probability that one card drawn is a multiple of 3 and another one is a multiple of 13.

Learn how to calculate the probability of drawing two cards at random from a set numbered 30 to 99, where one card is a multiple of 3 and the other is a multiple of 13. This problem involves arithmetic sequences and combinatorics, suitable for students in Grades 9-12.

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