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- Homework Statement
- With the high school distance learning, I'm helping my daughter with her 9th grade math problems (probability):

A lock has a 4 "digit" code without repetition. The possible digits are the numbers 0-9, and the letters A,B,C,D,E and F. The question is what is the probability of a code which has only one numeric digit (0-9).

- Relevant Equations
- Joint Probabilities. Conditional Probability.

The lessons show using Permutations and Combinations for other problems in this section, but we were having difficulty what to choose to put in those formulas. I used this approach:

We have 10 numeric and 6 alphabetic = 16 total possible.

Let's look at the first digit numeric, and the next 3 alpha:

$$ P(First Digit Numeric) = \frac{10}{16} $$ and

## P(Next Digits Alpha) = \frac{6}{15} ## and ## \frac{5}{14} ## and ## \frac{4}{13} ##

Since all four of these have to occur, we multiply them to get ## \frac{1200}{43680} ## which reduces to ## \frac{5}{182} ##

I go through the probabilities for (2nd digit numeric) and then 3rd and 4th, with the same probability for each situation. So we can multiply by 4 to get: ## \frac{20}{182} = \frac{10}{91} ##

This was one of the choices in the multiple choice, but the autograder marked it wrong. I looked it over several times, but come up with the same thing. Then I even programmed a little simulation, and after over 1 million random samples, get nearly the same result. I've known these online homeworks to have mistakes, sometimes, but I wanted to run by here before emailing the teacher.

We have 10 numeric and 6 alphabetic = 16 total possible.

Let's look at the first digit numeric, and the next 3 alpha:

$$ P(First Digit Numeric) = \frac{10}{16} $$ and

## P(Next Digits Alpha) = \frac{6}{15} ## and ## \frac{5}{14} ## and ## \frac{4}{13} ##

Since all four of these have to occur, we multiply them to get ## \frac{1200}{43680} ## which reduces to ## \frac{5}{182} ##

I go through the probabilities for (2nd digit numeric) and then 3rd and 4th, with the same probability for each situation. So we can multiply by 4 to get: ## \frac{20}{182} = \frac{10}{91} ##

This was one of the choices in the multiple choice, but the autograder marked it wrong. I looked it over several times, but come up with the same thing. Then I even programmed a little simulation, and after over 1 million random samples, get nearly the same result. I've known these online homeworks to have mistakes, sometimes, but I wanted to run by here before emailing the teacher.