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Harif Velarde · Answer 1 · 2023-12-04T09:27:58+0000

On this problem, we have two sets of criteria. Let's call the event A as the event of the audit report to be on-time, consequently being late is the complement of A(noted as A'), and the event B as the event of the audit report to be

complete(incomplete reports noted as B').

From the text, we know that the probability that an audit report is incomplete is 0.05

$P(B^(\prime))=0.05$

The probability that an audit report is complete given it was submitted late is 0.80

$P(B|A^(\prime))=0.80$

and the probability that an audit report is on-time is 0.90

The probability of an event and its complement must add up to 1, because only one of them can happen. This gives to us the following probabilities

$\begin{gathered} P(A)=0.90\Rightarrow P(A^(\prime))=0.10 \\ P(B^(\prime))=0.05\Rightarrow P(B)=0.95 \end{gathered}$

We want to calculate the probability that the audit report is complete and not late

$P(A\cap B)$

Using the conditional probability formula, this probability is given by

$P(A\cap B)=P(A|B)P(B)$

We already know the probability for B to happen, now we need to determinate the probability for A to happen given B. We already have the probability for B to happen given the A'. Using the Bayes Theorem we can calculate the probability of A' happening given B, and this will be the complement of P(A|B).

$P(A|B)+P(A^(\prime)|B)=1$

Calculating P(A'|B) using the Bayes Theorem, we have

$P(A^(\prime)|B)=(P(B|A^(\prime))P(A))/(P(B))$

Then, P(A|B) is

$P(A|B)=1-(P(B|A^(\prime))P(A))/(P(B))$

Then, the probability that the audit report is complete and not late is

$\begin{gathered} P(A\cap B)=(1-(P(B|A^(\prime))P(A))/(P(B)))* P(B) \\ =P(B)-P(B|A^(\prime))P(A) \\ =0.95-0.80\cdot0.90 \\ =0.23 \end{gathered}$

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