A jewelry company requires for its products to pass three tests before they are sold at stores. For gold rings, 90 % passes the first test, 85 % passes the second test, and 80 % passes the third test. If a product fails any test, the product is thrown away and it will not take the subsequent tests. If a gold ring failed to pass one of the tests, what is the probability that it failed the second test?

Solution.

Let $F$ be the event that a gold ring fails one of the three tests. Let $F_2$ be the event that it fails the second test. Then what we need to compute is the conditional probability
\[P(F_2 \mid F) = \frac{P(F_2 \cap F)}{P(F)}.\]

The numerator is
\[P(F_2 \cap F) = P(F_2) = 0.9 \cdot 0.15.\] (A gold ring passes the first test with probability $0.9$ and fails the second test with probability $1-0.85=0.15$.)

The complement $F^c$ of $F$ is the event that a gold ring passes all the tests. Thus
\[P(F) = 1- P(F^c) = 1 – 0.9 \cdot 0.85 \cdot 0.8.\] It follows that the desired probability is
\begin{align*}
P(F_2 \mid F) &= \frac{0.9 \cdot 0.15}{1 – 0.9 \cdot 0.85 \cdot 0.8}
= \frac{135}{388} \approx 0.348
\end{align*}

Therefore, given that a gold ring failed to pass one of the tests, the probability that it failed the second test is about 34.8 %.

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