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Let's analyze the problem step by step.

We are given:
- 70 cards marked from 30 to 99.
- Two cards are drawn at random.
- We are asked to find the probability that one card is a multiple of 3 and the other is a multiple of 13.

### Step 1: Find the total number of cards
The cards range from 30 to 99, inclusive. So, the total number of cards is:
\[ 99 - 30 + 1 = 70 \]

### Step 2: Count the multiples of 3 between 30 and 99
To find how many cards are multiples of 3, we check the range from 30 to 99:
- The first multiple of 3 is \( 30 \) (since \( 30 \div 3 = 10 \)).
- The last multiple of 3 is \( 99 \) (since \( 99 \div 3 = 33 \)).

So, the multiples of 3 are \( 30, 33, 36, ..., 99 \), which is an arithmetic sequence. The number of terms in this sequence is:
\[
\frac{99 - 30}{3} + 1 = 24 \text{ multiples of 3}.
\]

### Step 3: Count the multiples of 13 between 30 and 99
Next, find the multiples of 13 in this range:
- The first multiple of 13 is \( 39 \) (since \( 39 \div 13 = 3 \)).
- The last multiple of 13 is \( 91 \) (since \( 91 \div 13 = 7 \)).

The multiples of 13 are \( 39, 52, 65, 78, 91 \), so there are 5 multiples of 13.

### Step 4: Probability calculation
We now calculate the probability of drawing one card that is a multiple of 3 and one card that is a multiple of 13.

The total number of ways to draw 2 cards from 70 cards is:
\[
\binom{70}{2} = \frac{70 \times 69}{2} = 2415
\]

The number of ways to select one multiple of 3 (from 24 multiples) and one multiple of 13 (from 5 multiples) is:
\[
24 \times 5 = 120
\]

Thus, the probability of drawing one card that is a multiple of 3 and one card that is a multiple of 13 is:
\[
P = \frac{120}{2415}
\]
Simplifying this fraction:
\[
P = \frac{40}{805}
\]

Therefore, the probability is \( \frac{40}{805} \).

Would you like more details or have any questions?

Here are 5 related questions:
1. How would the probability change if there were more cards marked from 20 to 99?
2. What is the probability of drawing two multiples of 3?
3. How would you calculate the probability if the cards had no multiples of 13?
4. Can you find the probability of drawing two cards that are multiples of 5?
5. How do you calculate the probability if the range of cards was extended to 1–100?

**Tip**: When solving probability problems, always break the process into logical steps like counting favorable outcomes and the total number of outcomes.

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.

Find the probability that one card drawn is a multiple of 3 and another is a multiple of 13 from a set of 70 cards numbered 30 to 99. This problem explores key probability and combinatorics concepts suitable for Grades 9-12.

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