Conditional Probability -

Question

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.

Answer 1

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 $$\mathbf{P}(\text{failed} \vert \text{no defect}) = 0.05.$$

Answer 2

"Inspectors catch $95\%$ of defects" means that if a part has a defect then

• the inspectors say "fail" with probability $0.95$, since they spot the defect
• the inspectors say "pass" with probability $0.05$, since they wrongly think there is no defect

so $\mathbb P (\text{pass} \mid \text{defect}) = 0.05$

Answer 3

The $95\%$ is referring to the chance that a defect will be recognized by an inspector. So P(pass|defect)=$1-95\%=1-0.95=0.05$ or $5\%$. Note that this comes from:

...95% of defects

and not:

...5% of nondefective parts

as the 5% is talking about the detection of nondefective parts, or working parts.

Answer 4

Referring to the probability tree diagram:

and using the Bayes's theorem we get: $$P(D|P)=\frac{P(D\cap P)}{P(P)}=\frac{P(D\cap P)}{P(D\cap P)+P(N\cap P)}=\frac{0.03\cdot 0.05}{0.03\cdot 0.05+0.97\cdot 0.95}.$$