We have a total of $N$ products, among which $D$ are defective. We randomly select $n$ products. We want to find the probability that exactly $k$ of the selected products are defective, where $k \le D$.

1. Problem Description

We have a total of $N$ products, among which $D$ are defective. We randomly select $n$ products. We want to find the probability that exactly $k$ of the selected products are defective, where $k \le D$.

2. Solution Steps

This is a problem involving combinations. The total number of ways to choose $n$ products from $N$ is given by the combination formula:

$C(N, n) = \frac{N!}{n!(N-n)!}$

We want to choose exactly $k$ defective products out of the $D$ defective products, which can be done in $C(D, k)$ ways:

$C(D, k) = \frac{D!}{k!(D-k)!}$

We also need to choose the remaining $n - k$ non-defective products from the $N - D$ non-defective products. This can be done in $C(N-D, n-k)$ ways:

$C(N-D, n-k) = \frac{(N-D)!}{(n-k)!(N-D-n+k)!}$

The number of ways to choose $k$ defective products and $n - k$ non-defective products is the product of these two combinations:

$C(D, k) \cdot C(N-D, n-k) = \frac{D!}{k!(D-k)!} \cdot \frac{(N-D)!}{(n-k)!(N-D-n+k)!}$

The probability of selecting exactly $k$ defective products is the ratio of the number of ways to choose $k$ defective and $n-k$ non-defective products to the total number of ways to choose $n$ products:

$P(\text{exactly } k \text{ defective}) = \frac{C(D, k) \cdot C(N-D, n-k)}{C(N, n)} = \frac{\frac{D!}{k!(D-k)!} \cdot \frac{(N-D)!}{(n-k)!(N-D-n+k)!}}{\frac{N!}{n!(N-n)!}}$

3. Final Answer

The probability that exactly $k$ of the selected $n$ products are defective is:

$\frac{C(D, k) \cdot C(N-D, n-k)}{C(N, n)} = \frac{\frac{D!}{k!(D-k)!} \cdot \frac{(N-D)!}{(n-k)!(N-D-n+k)!}}{\frac{N!}{n!(N-n)!}}$