Probability of Repeated Digits in a 4-Digit Code
When generating a 4-digit code by randomly selecting digits from a set {1239} with replacement, we often encounter the question: what is the probability that the code number contains at least one repeated digit? This problem can be approached in several ways, and we will explore these methods step-by-step.
Approach 1: Direct Computation
First, let's determine the number of possible 4-digit codes with repetition allowed. Since we have 4 positions and each position can be filled by any of the 4 digits (1, 2, 3, 9), the total number of such codes is:
Total number of codes: 4^4 6561
Next, we calculate the number of codes where no digit is repeated. Since we select from a set of 4 digits, the first digit can be chosen in 4 ways, the second in 3 ways, the third in 2 ways, and the fourth in 1 way. Therefore, the number of such codes is:
Number of codes with no digit repeated: 4! 4 × 3 × 2 × 1 24
However, these 24 codes are for a set of 4 digits. If the set is {1, 2, 3, 9}, we need to calculate the number of codes without repetition for the set of 9 digits (1 to 9). Thus:
Number of codes with no digit repeated using digits 1 to 9: 9 × 8 × 7 × 6 3024
The probability that a randomly generated code contains at least one repeated digit is the complement of the probability that no digit is repeated. Hence:
Probability of at least one repeated digit: (6561 - 3024) / 6561 3537 / 6561 ≈ 0.539 or 53.9%
Approach 2: Complement Probability
Another way to solve this is to find the opposite probability, which is the probability that no digits are repeated and then subtract from 1. For a set {1, 2, 3, 9}, the complementary calculation is:
Complement probability (no digit repeated): 9/9 × 8/9 × 7/9 × 6/9 336/729 112/243
Therefore, the probability that at least one digit is repeated is:
Probability of at least one repeated digit: 1 - 112/243 131/243 ≈ 0.539 or 53.9%
Approach 3: Simplification of the Problem
If we interpret the problem as generating a 4-digit code from the set {1, 2, 3, 9} (a more straightforward approach), we can simplify the calculation:
Total number of codes with repetition allowed (from 1 to 9): 9^4 6561
Number of codes without repetition (from 1 to 9): 9 × 8 × 7 × 6 3024
Thus:
Probability of at least one repeated digit: (6561 - 3024) / 6561 3537 / 6561 ≈ 0.539 or 53.9%
Alternative Solution
An alternative way to view this problem is by considering the probability that all digits are different. If all digits are different, the probability is:
Probability of all digits different (from 1 to 9): 8/9 × 7/9 × 6/9 × 5/9 280/729 ≈ 0.384
The probability that at least one digit is repeated is the complement of this:
Probability of at least one repeated digit: 1 - 280/729 449/729 ≈ 0.616
However, this solution seems to have an error in the calculation, as the correct complement based on the first approach would be 131/243 ≈ 0.539.
Conclusion
The probability that a randomly generated 4-digit code, with digits selected from {1, 2, 3, 9} with replacement, contains at least one repeated digit is approximately 53.9%. This can be derived from multiple approaches, all yielding the same result.