Reconsider Example 2.20. The conditional probability that a high level of contamination was present when a failure occurred is to be determined. The information from Example 2.20 is summarized here.

The probability of \(P(H \mid F)\) is determined from \[
P(H \mid F)=\frac{P(F \mid H) P(H)}{P(F)}=\frac{0.10(0.20)}{0.024}=0.83 \]

The value of \(P(F)\) in the denominator of our solution was found from \(P(F)=P(F \mid H) P(H)+P\left(F \mid H^{\prime}\right) P\left(H^{\prime}\right)\).

Example 2.20

Consider the contamination discussion at the start of this section. The information is summarized here.

Probability of Failure Level of Probability of Contamination Level 0.1 High 0.2 0.005 Not high 0.8