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:
- Tens digit is 4 and units digit is 7: this gives us the number 47.
- 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.