Question

The tree diagram shows the sample space of​ two-digit numbers that can be created using the digits 5​, 4​, 3​, and 7. What is the probability of choosing a number from the sample space that contains both 4 and 7​?
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Part 1
​P(4 and 7​)equals
  
enter your response here or
  
enter your response here​%
There are
  
enter your response here possible outcomes that include both 4 and 7. There are
  
enter your response here possible​ outcomes, which are equally likely.
​(Type integers,​ fractions, or mixed​ numbers.)   

Answer 1

To solve this problem involving the probability of choosing a two-digit number containing both 4 and 7 from a sample space created by the digits 5, 4, 3, and 7, we need to first determine the total number of two-digit outcomes possible and then identify how many of these contain both digits 4 and 7.

### Step 1: Determine Possible Outcomes

The two-digit numbers can be formed by using the digits 5, 4, 3, and 7, where each digit can be used in the tens place and the units place.

- For the **tens place**, you can choose any of the 4 digits (5, 4, 3, or 7).
- For the **units place**, you can also choose any of the 4 digits (5, 4, 3, or 7).

Thus, the total number of two-digit numbers is calculated as follows:

\[
\text{Total outcomes} = 4 \text{ (choices for tens)} \times 4 \text{ (choices for units)} = 16
\]

### Step 2: Identify Outcomes Containing Both 4 and 7

Next, we need to find the specific two-digit numbers that contain both 4 and 7. The combinations for our two-digit numbers that would meet this criteria are:

1. **Tens digit is 4** and **units digit is 7**: this gives us the number **47**.
2. **Tens digit is 7** and **units digit is 4**: this gives us the number **74**.

These are the only numbers that contain both 4 and 7.

### Counting Outcomes

Now we can count the outcomes that include both 4 and 7:

- There are **2 outcomes**: {47, 74}.

### Step 3: Calculate Probability

To find the probability \( P(4 \text{ and } 7) \):

\[
P(4 \text{ and } 7) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{2}{16} = \frac{1}{8}
\]

### Step 4: Convert to Percentage

To convert this probability into percentage form:

\[
P(4 \text{ and } 7) \approx 0.125 = 12.5\%
\]

### Final Responses

So we have:

- Number of possible outcomes containing both 4 and 7: **2**
- Total possible outcomes: **16**
- Probability \( P(4 \text{ and } 7) \): **\(\frac{1}{8}\)** or **12.5%**

Here are the final answers you could use:

- \( P(4 \text{ and } 7) = \frac{1}{8} \) or \( 12.5\% \)
- There are **2** possible outcomes that include both 4 and 7.
- There are **16** possible outcomes, which are equally likely.