Sampling probability permutation

Question

Question 1) I have a random 4-digit number given to me (which may begin with one or more zeros). What is the probability tha tmy number has no digits occuring more than once?

My answer:

$$\dfrac{10\cdot9\cdot8\cdot7}{10^4}$$ is the probability

question 2) A box of 20 spare parts contains 15 good parts and 5 defective. If four parts are selected at random, what is the probability that exactly $k$ will be good, as a function of $k$ ($0\leq k \leq 4$)

We are choosing $k$ parts out of a possible 20, this gives us $20 \choose k$ possible combinations. The probability that $k$ will be "good" is given by $(15/20)^k$ and the remaining will be bad, so $(1-15/20)^{20-k}$. Multiplying all these three together gives the probability I think

Hoping someone can double check these for me

Answer 1

Your answer to 1) is fine, but your answer to 2) is incorrect. This is because the events are not independent.

It's easiest here to think of this entirely in terms of subsets. We want the ratio of the number of subsets with exactly $k$ defective to the total number of subsets. The total is $\binom{20}4$. To count the desired subsets, note that a set with exactly $k$ defective consists of one set of $k$ defective elements and one set of $4-k$ good elements. Compute these two numbers. The result is their product, representing all possibilities.

Answer 2

Question 1 looks good to me.

For Question 2, it looks as though you're assuming that the probabilities of being good will remain the same for each part. However, this is not the case - this probability changes depending on the parts you've previously drawn. Instead, try thinking about what event you want to consider in this case and try to count using that (maybe try small values of $k$ explicitly to test).