Chapter 13.5, Problem 38E

a.

To determine

To graph: The Venn diagram that represents the number of patients of lung cancer and patients of chronic smokers.

Explanation of Solution

Given information:

The number of patients diagnosed with lung cancer are 40, number of patient who are chronic smokers are 30. 25 patients have both lung cancer and smoke.

Graph:

Consider the information that the number of patients diagnosed with lung cancer are 40, number of patient who are chronic smokers are 30. 25 patients have both lung cancer and smoke.

Let A represents the patient only with lung cancer.

So,

  $\begin{array}{c}P\left(A\text{ only}\right)=P\left(A\right)-P\left(A\cap B\right)\\ =40-25\\ =15\end{array}$

Let C represents only the chronic smoker patients.

So,

  $\begin{array}{c}P\left(\text{B only}\right)=P\left(B\right)-P\left(A\cap B\right)\\ =30-25\\ =5\end{array}$

The Venn diagram that represent the situation is

  

The portion A represents the patient only with lung cancer.

The portion C represents only the chronic smoker patients.

The portion B represents the both that is patients have both lung cancer and smoke.

Interpretation:

The sum total of portion A and B is equal to patient with lung cancer. The sum total of portion B and C is equal to chronic smoker patients.

b.

To determine

To calculate: The probability that patient has lung cancer provided that patient smokes.

Answer to Problem 38E

The probability that patient has lung cancer provided that patient smokes is $\frac{5}{6}$ .

Explanation of Solution

Given information:

The number of patients diagnosed with lung cancer are 40, number of patient who are chronic smokers are 30. 25 patients have both lung cancer and smoke.

Formula used:

Probability of an event E is $P\left(\text{E}\right)=\frac{\text{number of outcome}}{\text{total number of outcome}}$ .

Given an event Q, the conditional probability of event A is $P\left(A|Q\right)=\frac{P\left(A\text{ and Q}\right)}{P\left(Q\right)}$ where $P\left(Q\right)\ne 0$ .

Calculation:

Consider the provided information that the number of patients diagnosed with lung cancer are 40, number of patient who are chronic smokers are 30. 25 patients have both lung cancer and smoke.

Total number of patients are 200.

Let A be an event that patient has lung cancer and B be an event that patient smokes.

Recall that probability of an event E is $P\left(\text{E}\right)=\frac{\text{number of outcome}}{\text{total number of outcome}}$ .

Probability that patient smokes,

  $P\left(B\right)=\frac{30}{200}$

Probability that patient has lung cancer and patient smokes,

  $P\left(A\text{ and B}\right)=\frac{25}{200}$

Recall that given an event Q, the conditional probability of event A is $P\left(A|Q\right)=\frac{P\left(A\text{ and Q}\right)}{P\left(Q\right)}$ where $P\left(Q\right)\ne 0$ .

The probability that patient has lung cancer provided that patient smokes.

  $\begin{array}{c}P\left(A|B\right)=\frac{\frac{25}{200}}{\frac{30}{200}}\\ =\frac{25}{30}\\ =\frac{5}{6}\end{array}$

Thus, the probability that patient has lung cancer provided that patient smokes is $\frac{5}{6}$ .