From a batch of products, among which there are n…
From a batch of products, among which there are $n$ non-defective and $m$ defective, $s$ specimens are taken for inspection. It was observed that the first $k$ of the $s$ specimens taken are non-defective. Determine the probability that the next specimen will also be non-defective.
Explanation of Method
The specimens are taken one after another without replacement. Once the first $k$ are known to be good, we simply look at what is left in the batch: $n-k$ good and $m$ defective products. The next specimen is equally likely to be any of the remaining products, so the classical definition gives the answer. We confirm it with the definition of conditional probability.
Step 1: What remains after $k$ good specimens
At the start the batch has $n+m$ products: $n$ good, $m$ defective. We have taken $k$ good ones (and $k\lt s$, so there is a next specimen). Remaining:
| Good | Defective | Total | |
|---|---|---|---|
| At the start | $n$ | $m$ | $n+m$ |
| After $k$ good specimens | $n-k$ | $m$ | $n+m-k$ |
Step 2: The next specimen
The $(k+1)$-th specimen is drawn at random from the remaining $n+m-k$ products, each equally likely. Favourable: the $n-k$ good ones.
\[ \boxed{P = \frac{n-k}{n+m-k}.} \]
The total number $s$ of inspected specimens does not affect the answer (it only guarantees that a next specimen is taken).
Step 3: Check with conditional probability
Let $G_i$ = "the $i$-th specimen is good". Drawing one by one,
\[ P(G_1\cdots G_k) = \frac{n}{n+m}\cdot\frac{n-1}{n+m-1}\cdots\frac{n-k+1}{n+m-k+1}, \]
and $P(G_1\cdots G_kG_{k+1})$ has one more factor $\dfrac{n-k}{n+m-k}$. Therefore
\[ P(G_{k+1}\mid G_1\cdots G_k) = \frac{P(G_1\cdots G_{k+1})}{P(G_1\cdots G_k)} = \frac{n-k}{n+m-k}. \ \checkmark \]
Step 4: Example
A batch has $n=8$ good and $m=2$ defective products. The first $k=3$ inspected are good. Then $5$ good and $2$ defective remain:
\[ P = \frac{8-3}{8+2-3} = \frac{5}{7} \approx 0.714. \]
Before any inspection the probability that a specimen is good was $8/10=0.8$. After removing good ones, the defective products make up a larger share of what remains, so the chance of another good one decreases.
Result
| Answer | |
|---|---|
| Remaining after $k$ good | $n-k$ good, $m$ defective |
| Probability the next is good | $P=\dfrac{n-k}{n+m-k}$ (independent of $s$) |