The smallest 4-digit number is 1000 and the largest is 9999. Therefore, there are 9,000 4-digit numbers. The position of 2 consecutive 5's can only be positions 1 and 2, 2 and 3, and 3 and 4. The remaining 2 digits can be any of the digits 0 to 9 when the 5's come at the start. So that's 5500 up to 5599, 100 numbers. When the 5's are not in the 1st and 2nd positions we have 1550 to 9559 or 1055 to 9955, that is, 90 for each pair of consecutive 5's. In total we have 100+90+90=280. Some of these will be numbers with 3 or 4 5's together. There are 10 from 5550 to 5559, 9 from 1555 to 9555, and 5555. That makes 20, but these will already be accounted for twice in the 280, and 5555 will be accounted for three times, therefore we need to remove the double and triple accounting by reducing 280 by 21=259. So out of 9,000 numbers, 259 will contain at least one pair of fives together. Take 259 away from 9,000 and we're left with 8,741, so there's a probability of 8741/9000=0.9712 or 97.12% that the 4-digit number will not contain two consecutive 5's.
