Numbers of three digits with repetition using digits {1, 2, 3, …..9} are formed. The probability that a selected number is divisible by 5 is

Options

A. `2/9`
B. `3/9`
C. `1/9`
D. `1/8`

Answer

To solve the problem of finding the probability that a randomly selected three-digit number formed using the digits {1, 2, 3, ..., 9} is divisible by 5, we can follow these steps:

### Step 1: Determine the total number of three-digit numbers that can be formed.
Since we are allowed to use the digits 1 to 9 and repetition is allowed, we can choose any of the 9 digits for each of the three places (hundreds, tens, and units).

- Total choices for the first digit (hundreds place) = 9
- Total choices for the second digit (tens place) = 9
- Total choices for the third digit (units place) = 9

Thus, the total number of three-digit numbers is:
\[ \text{Total outcomes} = 9 \times 9 \times 9 = 9^3 = 729 \]

### Step 2: Determine the number of favorable outcomes (numbers divisible by 5).
A number is divisible by 5 if its last digit (units place) is either 0 or 5. However, since we are only using the digits from 1 to 9, the only digit we can use that makes the number divisible by 5 is 5.

- For the units place, we have only 1 choice: 5.
- For the hundreds place, we can still choose any of the 9 digits (1 to 9).
- For the tens place, we can again choose any of the 9 digits (1 to 9).

Thus, the number of favorable outcomes is:
\[ \text{Favorable outcomes} = 9 \times 9 \times 1 = 9^2 = 81 \]

### Step 3: Calculate the probability.
The probability that a selected number is divisible by 5 is given by the ratio of the number of favorable outcomes to the total number of outcomes:
\[ \text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{81}{729} \]

This simplifies to:
\[ \text{Probability} = \frac{81 \div 81}{729 \div 81} = \frac{1}{9} \]

### Conclusion:
The probability that a selected three-digit number formed using the digits {1, 2, 3, ..., 9} is divisible by 5 is \(\frac{1}{9}\).

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