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.

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