A seven digit telephone number has its first four digits known, but the last three digits are forgotten. If the caller randomly dials the final three digits, after correctly dialing the first four, what is the probability of dialing the correct full number on that attempt?

Introduction / Context:
This question focuses on basic counting and probability. We know the first four digits of a seven digit phone number but not the last three. The last three digits are dialed at random, and we want the probability that the guess is exactly correct in a single attempt.

Given Data / Assumptions:

• Total number of digits in the phone number = 7.
• The first four digits are known and dialed correctly.
• The remaining three digits are unknown and must be guessed.
• Each of the remaining three positions can be any digit from 0 to 9.
• All 3 digit combinations are equally likely to be chosen.

Concept / Approach:
We treat the last three digits as a three digit code that can take values from 000 to 999. There are 10 possibilities for each digit, so there are 10^3 possible combinations. Only one of these combinations corresponds to the actual correct last three digits. The probability of dialing the correct combination in one random attempt is therefore 1 divided by the total number of possible combinations.

Step-by-Step Solution:
Number of possible values for each unknown digit = 10 (0 to 9).Total number of different three digit strings = 10 * 10 * 10 = 10^3 = 1000.Out of these 1000 possibilities, exactly one is correct.Probability of choosing the correct three digit string = 1 / 1000.Therefore, the probability of dialing the full seven digit number correctly on that attempt is 1 / 1000.

Verification / Alternative check:
We can also think of the last three digits as a number from 0 to 999 inclusive. There are 1000 distinct numbers in this range. A random choice is equally likely to be any one of these 1000 possibilities, so the chance of selecting the unique correct number is 1 out of 1000, again confirming the answer.

Why Other Options Are Wrong:
The probability 1/999 and 1/1001 come from assuming slightly different counts of possible combinations, for example by mistakenly excluding 000 or including an extra invalid option. The value 4/1000 would suggest that there are four correct combinations, which is not true. There is precisely one correct triple of digits, so only 1/1000 is valid.

Common Pitfalls:
Students sometimes believe that phone numbers cannot start with zero in the last three digits and incorrectly count combinations from 100 to 999, producing only 900 possibilities. This is not justified unless explicitly stated. Others might overlook that each digit has 10 independent choices and forget to use exponentiation. Carefully counting all allowed combinations is the key.

Final Answer:
The probability of dialing the correct full number on that attempt is 1/1000.