Let's walk through the solution step-by-step.
1. Understand the problem: We need to determine the probability that a randomly chosen two-digit code, using numbers from 1 to 9 without repetition, has both digits as even numbers.
2. Identify the even numbers between 1 and 9: These numbers are 2, 4, 6, and 8. So, there are 4 even numbers in this range.
3. Identify the total number of two-digit combinations without repetition:
- The first digit can be any of the 9 digits (1-9).
- The second digit can be any of the remaining 8 digits (since repetition is not allowed).
- Therefore, the total number of possible combinations is: [tex]\( 9 \times 8 = 72 \)[/tex].
4. Calculate the number of favorable combinations where both digits are even:
- The first digit can be any of the 4 even numbers.
- The second digit can be any of the remaining 3 even numbers.
- Therefore, the number of favorable combinations is: [tex]\( 4 \times 3 = 12 \)[/tex].
5. Determine the probability that both digits are even:
- Probability is defined as the number of favorable outcomes divided by the total number of possible outcomes.
- Thus, the probability that both digits are even is: [tex]\( \frac{12}{72} = \frac{1}{6} \)[/tex].
6. Analyze the given expressions:
- Expression 1: [tex]\( P(\text{both even}) = \frac{(4 P_1)(3 P_1)}{9_2} \)[/tex]
- Expression 2: [tex]\( P(\text{both even}) = \frac{(6 C_1)(3 C_1)}{9 C_2} \)[/tex]
- Expression 3: [tex]\( P(\text{both even}) = \frac{(6 P_2)(6 P_1)}{9 P_2} \)[/tex]
- Expression 4: [tex]\( P(\text{both even}) = \frac{(5 C_1)(4 C_1)}{9 C_2} \)[/tex]
Let's check the correct one by matching our explanation to the formula structure:
- The total number of combinations is correctly [tex]\( 9 P_2 \)[/tex], which is [tex]\( 9 \times 8 = 72 \)[/tex]. Therefore, the denominator should be [tex]\( 9 P_2 \)[/tex].
- The number of favorable combinations (both even) is correctly calculated using combinations and permutations of the even numbers [tex]\( (4 \times 3 = 12) \)[/tex].
The correct expression for the probability that both digits are even is best captured by:
[tex]\[ P(\text{both even}) = \frac{(4P_1)(3P_1)}{9P_2} \][/tex]
However, interpreting the combinations and probabilistic notations correctly in terms of provided options:
[tex]\[ P(\text{both even}) = \frac{(4C_1)(3C_1)}{9C_2} \][/tex]
In summary, the accurate probability calculation in conventional combination or permutation terms best aligns with:
[tex]\[ P(\text{both even}) = \frac{(4P_1 \times 3P_1)}{9P_2} \][/tex]
Therefore, the final answer is indeed validated by the detailed explanation here.
