Probability of a 7-Digit Phone Number Containing Exactly One '3'

Imagine a randomly selected 7-digit phone number. What is the probability that it contains exactly one '3'? This question delves into the realm of combinatorics and probability theory, providing insights into the intricacies of telephone number configurations.

Calculating the Probability

Firstly, we need to establish the total number of possible 7-digit phone numbers. Since each digit can be any number from 0 to 9, the total number of 7-digit phone numbers is 10^7 10,000,000.

Now, to find the number of 7-digit phone numbers that contain exactly one '3', we consider the positions where the '3' can appear. There are 7 positions, and for each of these positions, the other 6 positions can be any of the remaining 9 digits (0, 1, 2, 4, 5, 6, 7, 8, or 9).

The number of valid configurations is calculated as:

7 times; 96 531,441

Therefore, the probability of a 7-digit phone number having exactly one '3' is:

7 times; 96 / 107 531,441 / 10,000,000 0.0531441

Expressed as a percentage, this is approximately 5.31441%, or simply over 5.3%.

Real-World Application

In the North American telephone numbering plan, 7-digit phone numbers can have restrictions on their first and second digits. Prior to the 1990s, the second digit could not be '0' or '1'. This significantly alters the possible configurations of 7-digit phone numbers, which are approximately 8 million in total.

Let's consider two cases:

Case 1: First Digit is '3'

In this case, the remaining six digits can be any of the 9 other possible digits. The number of configurations is:

96 531,441

Case 2: First Digit is Not '3'

In this case, the first digit can be any of the 8 other digits (1, 2, 4, 5, 6, 7, 8, 9), and the '3' must appear in one of the remaining 6 positions. The number of configurations is:

8 times; 6 times; 95 2,480,058

The total number of possible 7-digit phone numbers with exactly one '3' is:

531,441 2,480,058 3,011,499

The probability of selecting such a number is:

3,011,499 / 8,000,000 0.376437375

This translates to approximately 37.64%.

Combinatorial Interpretation

Another approach involves considering the string of digits without worrying about their order. We have one digit that must be '3' and six digits that can be any of the 9 other digits. The total number of such strings is:

7 times; (1 9 92 93 94 95) 1,999,999

Since the '3' can be in any of the 7 positions, the total number of valid configurations is:

7 times; 1,999,999 3,720,087

With approximately 10 million possible 7-digit phone numbers, the probability of picking a number with exactly one '3' is:

3,720,087 / 10,000,000 0.3720087, or just over 37.2%.

Conclusion

The probability of selecting a 7-digit phone number with exactly one '3' is significant, ranging from 5.31441% to 37.64%, depending on the conditions and restrictions applied. This problem not only highlights the application of combinatorial and probabilistic principles but also underscores the impact of real-world constraints on such configurations.