Chapter 7, Problem 9PT2

To determine

To find: The probability that operation is successful for the individual that has infection free operation.

Answer to Problem 9PT2

The probability is 0.866.

Explanation of Solution

Given:

Probability of infection is 0.03.

Probability of repair falls is 0.14.

Probability of infection and refection falls is 0.01.

Calculation:

Consider, X and Y be the events that depict the infection and repair fall.

Thus,

  $\begin{array}{l}P\left(X\right)=0.03\\ P\left(Y\right)=0.14\\ P\left(X\text{}\text{ and }Y\right)=0.01\end{array}$

The probability of not infection can be calculated as:

  $\begin{array}{c}P\left(X^c\right)=1-P\left(C\right)\\ =1-0.03\\ =0.97\end{array}$

Now,

  $\begin{array}{c}P\left(X^C\text{ and }{Y}^C\right)=1-\left[P\left(X\right)+P\left(Y\right)-P\left(X\text{ and }Y\right)\right]\\ =1-\left[0.14+0.03-0.01\right]\\ =0.84\end{array}$

The probability that operation is successful for the individual that has infection free operation can be calculated as:

  $\begin{array}{c}P\left(Y^C|X^C\right)=\frac{P\left(Y^C\text{ and }{X}^C\right)}{P\left(X^C\right)}\\ =\frac{0.84}{0.97}\\ =0.866\end{array}$

Thus, the required probability is 0.866.

Hence, the correct option is (d).