You pick 6 digits \( (0-9) \) at random without replacement, and write them in the order picked. What is the probability that you have written the first 6 digits of your phone number? Assume there are no repeats of digits in your phone number. Express your answer in fraction form.

To find the probability that randomly picking 6 digits without replacement from the digits 0 to 9 will result in the first 6 digits of a specified phone number, we start by analyzing the total number of outcomes and the favorable outcomes. First, there are a total of 10 different digits (0 through 9). When selecting 6 digits without replacement, the total number of ways to arrange these 6 digits can be calculated as follows: \[ \text{Total ways to choose and arrange 6 digits} = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \] This expression arises because for the first digit we have 10 options, for the second digit 9 options (since we're not replacing), and so on until we have 5 options for the sixth digit. Calculating this gives: \[ 10 \times 9 \times 8 \times 7 \times 6 \times 5 = 30240 \] Now, let's consider the favorable outcomes, which in this case is only the specific sequence that matches the first 6 digits of the phone number. Since there is only one specific arrangement of those 6 digits, the number of favorable outcomes is \(1\). Now we can compute the probability \(P\): \[ P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{1}{30240} \] Thus, the probability of writing the first 6 digits of your phone number by picking 6 random digits without replacement is \[ \boxed{\frac{1}{30240}} \]