Phone Numbers as Possible Combinations of 7 digits: Probability

1.) If the first digit isn't allowed to be a 0 or a 1, how many 7-digit phone numbers are possible?

Solution: For each digit in the phone number, there are 10 possibilities (0 - 9) if there are no restrictions. In this case the first number isn't allowed to be 0 or 1. For phone-number digits 2 - 7, there are $10^{6}$ possibilities: 10 * 10 * 10 * 10 * 10 * 10. Multiply this by the number of possibilities for the first digit (only 8 since two digits, 1 and 0, are excluded), and you get your answer:

No. of possibilities for the first digit * No. of possibilities for the remaining 6 digits

= $8*10^{6}$ = 8,000,000 possible phone numbers.

2.) If the above condition stays in place, along with the added condition that the first three digits of the phone number cannot be 911, how many possible phone numbers are there?

Solution: If the probability of an event A occurring is P(A), then the probability of event A not occurring is 1 - P(A). Consider the case where the event A is the choosing of the first three digits of the phone number to be 911:

The probability space of the first three digits is $8*10*10$ if we consider only the condition of part 1. The probability of the first three digits being 911 is $\frac{1}{8}*\frac{1}{10}*{\frac{1}{10}} = \frac{1}{800}$

So, the probability of the first three digits not being 911 is $ 1 -\frac{1}{800} = \frac{799}{800}$

Finally, the number of possible phone numbers with the first digit not being a 0 or a 1, and the first 3 digits not being 911, is
$799 * 10^{4} $ = 7,990,000.

(799 is the number of possibilities for the first 3 digits, and $10^{4}$ is the number of possibilities for the last 4 digits.)