We are given the digits 0, 6, 0, 6 and must form four-digit numbers, ensuring the first two digits do not form "00". Let's determine how many valid numbers exist and then find how many are greater than 5000.

Step 1: Find Total Valid Numbers

Since we have two occurrences each of the digits 0 and 6, the total distinct four-digit numbers that can be formed is calculated by: $\frac{4!}{2!2!} = \frac{24}{4} = 6.$ Listing all possible numbers:

• 0066 (invalid, as it starts with 00)
• 0606
• 0660
• 6006
• 6060
• 6600

Removing 0066, we are left with 5 valid numbers.

Step 2: Count Numbers Greater Than 5000

To be greater than 5000, the first digit must be 6. The valid numbers starting with 6 are:

• 6006
• 6060
• 6600

There are 3 such numbers.

Step 3: Compute Probability

The probability is: $\frac{\text{favorable cases}}{\text{total valid cases}} = \frac{3}{5}.$

Thus, the final answer is $\frac{3}{5}$ or 0.6.

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Do you want details on another approach? Here are 5 related questions:

1. How many four-digit numbers can be formed using the digits {1, 1, 2, 2}?
2. What is the probability of picking a number greater than 4000 from the set {2, 2, 4, 4}?
3. How would the probability change if we added the digit 9 to our set?
4. How do we calculate permutations of numbers with repeating digits?
5. How can we generalize this method for a different set of digits?

Tip: When dealing with permutations of digits with repetition, always use the formula $\frac{n!}{r_1!r_2!...}$ to account for duplicates.