In a batch of n products there are k defective items.…

In a batch of $n$ products there are $k$ defective items. Determine the probability that among $m$ randomly chosen products exactly $l$ will be defective.

Step 1: Combinations

The number of ways to choose $r$ objects out of $s$, without regard to order, is

\[ \binom{s}{r} = \frac{s!}{r!\,(s-r)!}. \]

Example: $\binom{5}{2} = \frac{5\cdot4}{2\cdot1} = 10$.

Step 2: All outcomes

An outcome is a group of $m$ products out of $n$. All groups are equally likely:

\[ \text{number of all outcomes} = \binom{n}{m}. \]

Step 3: Favourable outcomes

The batch consists of $k$ defective and $n-k$ good products. A group with exactly $l$ defective products is formed in two independent choices:

• choose the $l$ defective ones from the $k$ defective: $\binom{k}{l}$ ways;
• choose the remaining $m-l$ from the $n-k$ good ones: $\binom{n-k}{m-l}$ ways.

Every way of the first choice combines with every way of the second (multiplication rule):

\[ \text{number of favourable outcomes} = \binom{k}{l}\binom{n-k}{m-l}. \]

Step 4: The probability

\[ \boxed{P = \frac{\dbinom{k}{l}\dbinom{n-k}{m-l}}{\dbinom{n}{m}}.} \]

It makes sense when $l\le k$ and $m-l\le n-k$ (otherwise there are no favourable groups and $P=0$), i.e. $\max(0,\,m-n+k)\le l\le\min(k,\,m)$.

Step 5: Example

A batch of $n=10$ products contains $k=3$ defective. We take $m=4$. Probability of exactly $l=1$ defective:

\[ P = \frac{\binom31\binom73}{\binom{10}{4}} = \frac{3\cdot35}{210} = \frac{105}{210} = 0.5. \]

The whole distribution of the number of defective products in the sample:

The probabilities add up to $1$, because every group of $4$ has some number of defective products (Vandermonde's identity $\sum_l\binom{k}{l}\binom{n-k}{m-l}=\binom{n}{m}$).

Step 6: Where this formula is used

Quality control (defective items in a sample), lotteries (how many of your numbers are drawn), card games (how many aces in a hand) — every time objects are drawn without replacement from a group made of two kinds.