The problem asks to calculate the probability of picking the first 7 digits of your phone number, assuming no digits are repeated and the digits are selected at random without replacement from the digits $0, 1, 2, ..., 9$.

Total number of possible outcomes

You are picking 7 digits out of the 10 digits available (since there are no repeated digits). The number of ways to select and arrange 7 digits from 10 digits is given by the number of permutations of 10 digits taken 7 at a time. This is calculated as:

$\text{Total number of arrangements} = P(10, 7) = \frac{10!}{(10-7)!} = \frac{10!}{3!}$

We can calculate this explicitly:

$10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3,628,800$ $3! = 3 \times 2 \times 1 = 6$ $P(10, 7) = \frac{3,628,800}{6} = 604,800$

Thus, the total number of possible outcomes is 604,800.

Favorable outcomes

There is only one favorable outcome, which is the exact sequence of the first 7 digits of your phone number.

Probability

The probability of picking the first 7 digits of your phone number in the correct order is the ratio of favorable outcomes to total outcomes:

$\text{Probability} = \frac{1}{604,800}$

Thus, the probability is:

$\boxed{\frac{1}{604,800}}$

Would you like further details on this calculation or have any questions?

Here are 5 related questions to explore:

1. How would the probability change if the digits could be repeated?
2. What is the probability of selecting any 7-digit number (without specific order) from the 10 digits?
3. How does this problem differ if we allowed digits in the phone number to repeat?
4. If you were to select 8 digits instead of 7, how would that affect the probability?
5. What is the probability of selecting the first 3 digits of your phone number instead of 7?

Tip: When calculating probabilities for ordered selections, always consider the number of permutations, which account for the specific sequence of choices.