Chapter 13.5, Problem 12CFU

a.

To determine

To calculate: The probability that disease is prevented in a volunteer chosen at random.

Answer to Problem 12CFU

The probability that disease is prevented in a volunteer chosen at random is $\frac{13}{20}$ .

Explanation of Solution

Given information:

100 volunteers were given conventional treatment for a particular disease. 100 volunteers were given a new vaccine to test the effectiveness of new vaccine.

$\begin{array}{ccc}\text{Treatment}& \begin{array}{l}\text{Disease }\\ \text{Prevented}\end{array}& \begin{array}{l}\text{Disease not }\\ \text{Prevented}\end{array}\\ \text{New vaccine}& 68& 32\\ \begin{array}{l}\text{Conventional }\\ \text{Treatment}\end{array}& 62& 38\end{array}$

Formula used:

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

Calculation:

Consider the provided information that 100 volunteers were given conventional treatment for a particular disease. 100 volunteers were given a new vaccine to test the effectiveness of new vaccine.

$\begin{array}{ccc}\text{Treatment}& \begin{array}{l}\text{Disease }\\ \text{Prevented}\end{array}& \begin{array}{l}\text{Disease not }\\ \text{Prevented}\end{array}\\ \text{New vaccine}& 68& 32\\ \begin{array}{l}\text{Conventional }\\ \text{Treatment}\end{array}& 62& 38\end{array}$

Total number of volunteers are 200.

Let A be an event that disease is prevented in a volunteer.

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 disease is prevented in a volunteer,

  $\begin{array}{c}P\left(A\right)=\frac{68+62}{200}\\ =\frac{130}{200}\\ =\frac{13}{20}\end{array}$

Thus, the probability that disease is prevented in a volunteer chosen at random is $\frac{13}{20}$ .

b.

To determine

To calculate: The probability that disease is prevented in a volunteer provided that volunteer was given new vaccine.

Answer to Problem 12CFU

The probability that disease is prevented in a volunteer provided that volunteer was given new vaccine is $\frac{17}{25}$ .

Explanation of Solution

Given information:

100 volunteers were given conventional treatment for a particular disease. 100 volunteers were given a new vaccine to test the effectiveness of new vaccine.

$\begin{array}{ccc}\text{Treatment}& \begin{array}{l}\text{Disease }\\ \text{Prevented}\end{array}& \begin{array}{l}\text{Disease not }\\ \text{Prevented}\end{array}\\ \text{New vaccine}& 68& 32\\ \begin{array}{l}\text{Conventional }\\ \text{Treatment}\end{array}& 62& 38\end{array}$

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 100 volunteers were given conventional treatment for a particular disease. 100 volunteers were given a new vaccine to test the effectiveness of new vaccine.

  $\begin{array}{ccc}\text{Treatment}& \begin{array}{l}\text{Disease }\\ \text{Prevented}\end{array}& \begin{array}{l}\text{Disease not }\\ \text{Prevented}\end{array}\\ \text{New vaccine}& 68& 32\\ \begin{array}{l}\text{Conventional }\\ \text{Treatment}\end{array}& 62& 38\end{array}$

Total number of volunteers are 200.

Let A be an event that disease is prevented and B be an event that volunteer was given new vaccine.

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 volunteer was given new vaccine,

  $P\left(B\right)=\frac{100}{200}$

Probability that disease is prevented and volunteer was given new vaccine,

  $P\left(A\text{ and B}\right)=\frac{68}{200}$

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.

  $\begin{array}{c}P\left(A|B\right)=\frac{\frac{68}{200}}{\frac{100}{200}}\\ =\frac{68}{100}\\ =\frac{17}{25}\end{array}$

Thus, the probability that disease is prevented in a volunteer chosen at random is $\frac{17}{25}$ .

c.

To determine

To calculate: The probability that disease is prevented in a volunteer provided that volunteer was not given new vaccine.

Answer to Problem 12CFU

The probability that disease is prevented in a volunteer provided that volunteer was not given new vaccine is $\frac{31}{50}$ .

Explanation of Solution

Given information:

100 volunteers were given conventional treatment for a particular disease. 100 volunteers were given a new vaccine to test the effectiveness of new vaccine.

$\begin{array}{ccc}\text{Treatment}& \begin{array}{l}\text{Disease }\\ \text{Prevented}\end{array}& \begin{array}{l}\text{Disease not }\\ \text{Prevented}\end{array}\\ \text{New vaccine}& 68& 32\\ \begin{array}{l}\text{Conventional }\\ \text{Treatment}\end{array}& 62& 38\end{array}$

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 100 volunteers were given conventional treatment for a particular disease. 100 volunteers were given a new vaccine to test the effectiveness of new vaccine.

$\begin{array}{ccc}\text{Treatment}& \begin{array}{l}\text{Disease }\\ \text{Prevented}\end{array}& \begin{array}{l}\text{Disease not }\\ \text{Prevented}\end{array}\\ \text{New vaccine}& 68& 32\\ \begin{array}{l}\text{Conventional }\\ \text{Treatment}\end{array}& 62& 38\end{array}$

Total number of volunteers are 200.

Let A be an event that disease is prevented and B be an event that volunteer was given new vaccine.

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 volunteer was not given new vaccine means probability that volunteer was given conventional treatment.

  $P\left(B\right)=\frac{100}{200}$

Probability that disease is prevented and volunteer was not given new vaccine, this means Probability that disease is prevented and volunteer was not given conventional treatment is,

  $P\left(A\text{ and B}\right)=\frac{62}{200}$

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 disease is prevented in a volunteer provided that volunteer was not given new vaccine.

  $\begin{array}{c}P\left(A|B\right)=\frac{\frac{62}{200}}{\frac{100}{200}}\\ =\frac{62}{100}\\ =\frac{31}{50}\end{array}$

Thus, the probability that disease is prevented in a volunteer provided that volunteer was not given new vaccine is $\frac{31}{50}$ .