91Ó°ÊÓ

The percentage of infection in operation is $3\%$

The percentage of repair fails in operation is $14\%$

The percentage of both infection and failure in operation is $1\%$

To find the probability that the operation is successful for someone who has an operation that is free from infection.

Let's note that I$=$Infection

R$=$Repair fail

localid="1650258605804" $$\begin{gathered} P\left(I\right)=3\%=0.03 \\ P\left(R\right)=14\%=0.14 \\ P(IandR)=1\backslash \%=0.01 \end{gathered}$$

Using the complement rule

$P(notA)=1-P\left(A\right)$

We get

localid="1650258616160" $$\begin{gathered} P(notRandnotI)=P\left(not\right(RorI\left)\right)=1-P(RorI)=1-P\left(R\right)-P\left(I\right)+P(RandO) \\ P(notI)=1-P\left(I\right)=1-0.03=0.97 \\ =1-0.14-0.03+0.01=0.84 \end{gathered}$$

The conditional probability is given as

$P(Aâˆ\pounds B)=\frac{P\left(A\text{and}B\right)}{P\left(B\right)}$

We want to find the probability of a successful operation given that the operation was free of infection

$P(notRâˆ\pounds notI)=\frac{P(notR\text{and not}I)}{P(notI)}=\frac{0.84}{0.97}≈0.8660.$