Let's break down the problem step by step to find the probability of forming a 5-digit number with specific digits in certain places.

Step 1: Understand the Problem

• Digits Available: 0, 1, 2, 3, 4
• Form a 5-digit number: We need to make a 5-digit number using these digits.
• Specific Requirement: The digit '2' must be in the tens place, and the digit '0' must be in the units place.

Step 2: Determine Total Possible 5-Digit Numbers

For a 5-digit number:

• The first digit (thousands place) cannot be '0' (otherwise it wouldn't be a 5-digit number).
• The remaining digits can be any of the five digits available (0, 1, 2, 3, 4).

So, for the total number of 5-digit combinations:

• First digit: Can be 1, 2, 3, or 4 (4 options).
• Second digit: Can be any of the 5 digits (0, 1, 2, 3, 4).
• Third digit: Can be any of the 5 digits (0, 1, 2, 3, 4).
• Fourth digit: Can be any of the 5 digits (0, 1, 2, 3, 4).
• Fifth digit (units place): Can be any of the 5 digits (0, 1, 2, 3, 4).

Thus, the total number of possible 5-digit numbers is: $4 \times 5 \times 5 \times 5 \times 5 = 2500$

Step 3: Determine Favorable Outcomes

Now, let's consider the specific conditions:

• 2 must be in the tens place: This fixes the fourth digit as '2'.
• 0 must be in the units place: This fixes the fifth digit as '0'.

For the remaining three digits (first, second, and third places):

• First digit: Can still be 1, 3, or 4 (not 0 or 2 to ensure it's a 5-digit number) → 3 options.
• Second digit: Can be any of the 5 digits (0, 1, 2, 3, 4).
• Third digit: Can be any of the 5 digits (0, 1, 2, 3, 4).

Thus, the number of favorable outcomes is: $3 \times 5 \times 5 = 75$

Step 4: Calculate the Probability

The probability of getting a 5-digit number with 2 in the tens place and 0 in the units place is: $\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{75}{2500} = \frac{3}{100} = 0.03$

Final Answer

The probability is 0.03 or 3%.

Would you like any further details on this? Here are some related questions you might find interesting:

1. What if we had to place a different digit in the tens place?
2. How would the probability change if we allowed repeating digits?
3. What happens if we use a different set of digits, like 1, 2, 3, 4, 5?
4. How many 5-digit numbers can be formed if no digits repeat?
5. What is the probability of forming a number where '0' is in the hundreds place?

Tip: When calculating probabilities, always ensure to count both the total possible outcomes and the favorable outcomes accurately!