A code consists of 3 letters followed by 3 digits. If the letters must be vowels (A, E, I, O, U) and digits cannot repeat, what is the probability that a randomly generated code contains exactly one 'A'?

Official Answer

Certainly, let's delve into the probability question you've posed regarding the creation of a code consisting of 3 letters (all of which are vowels: A, E, I, O, U) followed by 3 digits where repetition is not allowed. The core of this task is to ascertain the probability that a randomly generated code contains exactly one 'A'.

First, let's break down the problem into more manageable components. We have two segments in each code: the letter segment and the digit segment. For the letter segment, given that we need exactly one 'A', there are 3 positions where 'A' can appear. Once 'A' is placed, the remaining two positions can be filled with any of the other 4 vowels (E, I, O, U), with the first position having 4 choices and the second having 3 (due to the constraint of no repetition). This gives us a total of (3 \times 4 \times 3 = 36) ways to arrange the vowels with exactly one 'A'.

For the digit segment, since the code requires 3 digits with no repetition, and considering the digits 0-9 (a total of 10 digits), the first digit can be any of the 10, the second can be any of the remaining 9, and the third can be any of the remaining 8. This gives us (10 \times 9 \times 8 = 720) possible combinations for the digit segment.

Now, to find the total number of possible codes without any constraints on the letter segment (i.e., any vowels can be used in any order), we would have (5^3) ways to arrange the vowels (as each of the 3 positions can be filled by any of the 5 vowels, allowing for repetition) and (10 \times 9 \times 8) ways for the digits, as calculated before. Multiplying these gives us (5^3 \times 720 = 9000) possible codes in total.

Therefore, the probability of generating a code with exactly one 'A' in the vowel segment is the number of ways to arrange the vowels with exactly one 'A' (36) times the number of ways to arrange the digits (720), divided by the total number of possible codes (9000). This simplifies to (\frac{36 \times 720}{9000} = \frac{25920}{9000} = \frac{8}{25}).

In summary, the probability that a randomly generated code contains exactly one 'A' is (\frac{8}{25}), demonstrating a methodical approach to breaking down a complex problem into smaller, more manageable parts. This answer not only highlights my analytical skills as a Data Scientist but also underscores the importance of a systematic approach to problem-solving, which is crucial in data-driven decision-making processes. This framework can be tailored by other job seekers to showcase their problem-solving skills and their ability to apply statistical principles to real-world scenarios.