How would you have solved this question without having to use the probability of not picking three and instead use the probability of picking 3 

Sure! Let's solve the problem by directly calculating the probability of picking the card numbered three at least once.

We start by considering the two scenarios where the card numbered three can be picked at least once in two draws:

1. Picking the card numbered three on the first draw.
2. Picking the card numbered three on the second draw.

However, we must also consider the scenario where the card numbered three is picked in both draws. This will ensure that we do not double-count scenarios.

Let's break it down:

### Scenario Analysis

1. **Picking the card numbered three on the first draw:**
   - The probability of picking three on the first draw is \(\frac{1}{10}\).
   - The second draw can be any card (10 possibilities).
   - Total probability for this scenario is \(\frac{1}{10} \times 1 = \frac{1}{10}\).

2. **Picking the card numbered three on the second draw:**
   - The probability of picking three on the second draw is \(\frac{1}{10}\).
   - The first draw can be any card (10 possibilities).
   - Total probability for this scenario is \(\frac{1}{10} \times 1 = \frac{1}{10}\).

3. **Picking the card numbered three on both draws:**
   - The probability of picking three on both draws is \(\frac{1}{10} \times \frac{1}{10} = \frac{1}{100}\).

### Combining Probabilities

To find the overall probability of picking the card numbered three at least once, we need to combine the probabilities of the first two scenarios and subtract the probability of double-counting the scenario where the card is picked both times.

The combined probability of picking three at least once is:
\[ P(\text{at least one three}) = P(\text{three on first}) + P(\text{three on second}) - P(\text{three on both}) \]

Substituting the calculated probabilities:
\[ P(\text{at least one three}) = \frac{1}{10} + \frac{1}{10} - \frac{1}{100} \]

Simplifying:
\[ P(\text{at least one three}) = \frac{10}{100} + \frac{10}{100} - \frac{1}{100} \]
\[ P(\text{at least one three}) = \frac{20}{100} - \frac{1}{100} \]
\[ P(\text{at least one three}) = \frac{19}{100} \]

Thus, the probability of picking the card numbered three at least once in two draws is:
\[ \frac{19}{100} = 0.19 \text{ or } 19\% \]

Would you like more details or have any questions?

Here are some related questions for further practice:
1. What is the probability of picking the card numbered three exactly once in two draws?
2. How does the probability change if the cards are picked without replacement?
3. What is the probability of picking two cards, both numbered three, in two draws with replacement?
4. If three cards are picked with replacement, what is the probability of picking the card numbered three at least once?
5. What is the probability of picking the card numbered three at least once in four draws with replacement?
6. If the box contains cards numbered 1 to 20, how does the probability change for two draws with replacement?
7. How would you calculate the probability if the cards were drawn without replacement?
8. What is the expected number of times the card numbered three will be picked in two draws with replacement?

**Tip**: When solving probability problems, it is often helpful to consider complementary events as an alternative method to reach the same solution.

Learn how to calculate the probability of picking the card numbered three at least once in two draws, with step-by-step solutions and detailed explanations. This problem involves basic probability concepts, including calculating probabilities of complementary events and combining probabilities for multiple scenarios. Suitable for high school students and those preparing for probability problems in exams.

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