Conditional Probability - > A manufacturing process has a 3% defect rate. Inspectors catch 95% of defects but also fail 5% of nondefective parts. If we pick a part at random from all those that pass inspection, what is the probability that part is actually defective? I got my answer wrong. $$ P(defect|pass)=\frac{P(defect \, and \, pass)}{P(pass)} =\frac{P(pass | defect)P(defect)}{P(pass | defect)P(defect) + P(pass | no \, defect)P(no \, defect)} = \frac{3\%\times 5\%}{3\%\times 5\%+97\%\times 95\%} $$ I need to know why **P(pass|defect)** is 5%, instead of 95%. I can't actually decipher the statement.

> Inspectors catch 95% of defects

Catching a defect means identifying that it is no good. 95% is the proportion of the defects **failed** , so the proportion of defects which the inspectors passed must be the rest. $$ \begin{aligned} \mathbf{P}(\text{passed} \vert \text{defect}) &= 1 - \mathbf{P}(\text{failed} \vert \text{defect}),\\\ &= 1 - 0.95, \\\ &= 0.05. \end{aligned} $$ Note that this $0.05$ does **not** come from

> Inspectors [...] also fail 5% of nondefective parts

since this part is saying $$\mathbf{P}(\text{failed} \vert \text{no defect}) = 0.05.$$