Determine the probability that a batch of one hundred…

Determine the probability that a batch of one hundred products, five of which are defective, will be accepted when a randomly selected half of the batch is tested, if the acceptance condition allows at most one defective item among the fifty tested.

Step 1: All outcomes

\[ \binom{100}{50} \text{ equally likely groups of } 50 \text{ tested items}. \]

Step 2: No defective item

All $50$ from the $95$ good items:

\[ P_0=\frac{\binom{95}{50}}{\binom{100}{50}}\approx0.0281. \]

Draw by draw this is $\frac{50}{100}\cdot\frac{49}{99}\cdot\frac{48}{98}\cdot\frac{47}{97}\cdot\frac{46}{96}$ (the chance that none of the $5$ defective items falls into the tested half).

Step 3: Exactly one defective item

One of the $5$ defective and $49$ of the $95$ good ones:

\[ P_1=\frac{\binom51\binom{95}{49}}{\binom{100}{50}}\approx0.1529. \]

Step 4: The answer

\[ \boxed{P(\text{accepted})=\frac{\binom{95}{50}+5\binom{95}{49}}{\binom{100}{50}}\approx0.1811.} \]

Step 5: Interpretation

A batch with $5\%$ defective products is accepted in only about $18\%$ of cases: testing half of the batch is a strict check. On average the tested half contains $2.5$ defective items, so finding at most one is unlikely.