Respuesta :
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
[tex]P(B^{\prime})=0.05[/tex]The probability that an audit report is complete given it was submitted late is 0.80
[tex]P(B|A^{\prime})=0.80[/tex]and the probability that an audit report is on-time is 0.90
[tex]P(A)=0.90[/tex]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
[tex]\begin{gathered} P(A)=0.90\Rightarrow P(A^{\prime})=0.10 \\ P(B^{\prime})=0.05\Rightarrow P(B)=0.95 \end{gathered}[/tex]We want to calculate the probability that the audit report is complete and not late
[tex]P(A\cap B)[/tex]Using the conditional probability formula, this probability is given by
[tex]P(A\cap B)=P(A|B)P(B)[/tex]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).
[tex]P(A|B)+P(A^{\prime}|B)=1[/tex]Calculating P(A'|B) using the Bayes Theorem, we have
[tex]P(A^{\prime}|B)=\frac{P(B|A^{\prime})P(A)}{P(B)}[/tex]Then, P(A|B) is
[tex]P(A|B)=1-\frac{P(B|A^{\prime})P(A)}{P(B)}[/tex]Then, the probability that the audit report is complete and not late is
[tex]\begin{gathered} P(A\cap B)=(1-\frac{P(B|A^{\prime})P(A)}{P(B)})\times P(B) \\ =P(B)-P(B|A^{\prime})P(A) \\ =0.95-0.80\cdot0.90 \\ =0.23 \end{gathered}[/tex]