If any four numbers are slected and they are multiplied, then the probability that the last digit will be `1,3,5 or 7` is

Options

A. `4//625`
B. `18//625`
C. `16//625`
D. None of these

Explanation

To solve the problem of finding the probability that the last digit of the product of four selected numbers will be 1, 3, 5, or 7, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Possible Last Digits**:
The last digits of any number can be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. We need to focus on the last digits that can yield a product ending in 1, 3, 5, or 7.

2. **Determine Valid Last Digits**:
The last digits that can contribute to a product ending in 1, 3, 5, or 7 are:
- 1
- 3
- 5
- 7

3. **Count the Total Last Digits**:
There are a total of 10 possible last digits (0 through 9).

4. **Count the Favorable Last Digits**:
Out of these 10 digits, 4 digits (1, 3, 5, 7) are favorable for our condition.

5. **Calculate the Probability of Selecting a Favorable Last Digit**:
The probability of selecting a last digit that is either 1, 3, 5, or 7 is:
\[ P(\text{favorable last digit}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{4}{10} = \frac{2}{5} \]

6. **Calculate the Probability for Four Selections**:
Since we are selecting 4 numbers, and we want all of them to have a last digit of 1, 3, 5, or 7, we raise the probability of selecting a favorable last digit to the power of 4:
\[ P(\text{all four digits are favorable}) = \left(\frac{2}{5}\right)^4 \]

7. **Calculate the Final Probability**:
\[ P(\text{all four digits are favorable}) = \frac{2^4}{5^4} = \frac{16}{625} \]

### Final Answer:
The probability that the last digit of the product of four selected numbers will be 1, 3, 5, or 7 is \(\frac{16}{625}\).

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