Chapter 13, Problem 13.47E

To determine

To find: The conditional probability that he or she also smokes cigarettes given that a student smoke electronic cigarette and the conditional probability that he or she also smokes electronic cigarettes given that a student smoke cigarette.

Answer to Problem 13.47E

The conditional probability that he or she also smokes cigarettes given that a student smoke electronic cigarette is 0.1875.

The conditional probability that he or she also smokes electronic cigarettes given that a student smoke cigarette is 0.3333.

Explanation of Solution

Given info:

The youth of U.S. has declined cigarette smoking in recent years while the use of some other tobacco products has increased. The high school students used several tobacco products in last 30 days, more than who had used more than half and multiple tobacco products. Let A, B and C denotes the events corresponding to the different types of tobacco products in last 30 days and these are given below:

Events A denotes cigarette, B denotes electronic cigarette and C denotes other tobacco products including cigars, pipes, smokeless tobacco and hookahs.

The probabilities that a randomly selected high school student used these different tobacco products are shown below:

$P\left(A\right)=0.09$, $P\left(B\right)=0.16$, $P\left(C\right)=0.12$, $P\left(A\text{and}B\right)=0.03$, $P\left(A\text{and}C\right)=0.04$, $P\left(B\text{and}C\right)=0.04$ and $P\left(A\text{and}B\text{and}C\right)=0.01$.

Calculation:

From the given information,

$P\left(A\right)=0.09$, $P\left(B\right)=0.16$, $P\left(C\right)=0.12$, $P\left(A\text{and}B\right)=0.03$, $P\left(A\text{and}C\right)=0.04$, $P\left(B\text{and}C\right)=0.04$ and $P\left(A\text{and}B\text{and}C\right)=0.01$.

The remaining probabilities are given below:

The probability of $\left(A\text{and}B\text{and}\text{not}C\right)$ is given as,

$\begin{array}{c}P\left(A\text{and}B\text{and}\text{not}C\right)=P\left(A\right)-\left[\begin{array}{l}P\left(A\text{and}B\right)\\ +P\left(A\text{and}C\right)\end{array}\right]\\ =0.09-\left[0.03+0.04\right]\\ =0.09-0.07\\ =0.02\end{array}$

The probability of $\left(A\text{and}\text{not}B\text{and}\text{not}C\right)$ is given as,

$\begin{array}{c}P\left(\begin{array}{l}A\text{and}\text{not}\\ B\text{and}\text{not}C\end{array}\right)=P\left(A\right)-\left[\begin{array}{l}P\left(A\text{and}C\right)+P\left(A\text{and}B\text{and}C\right)\\ +P\left(A\text{and}B\text{and}\text{not}C\right)\end{array}\right]\\ =0.09-\left[0.03+0.01+0.02\right]\\ =0.09-0.06\\ =0.03\end{array}$

The probability of $\left(A\text{and}C\text{and}\text{not}B\right)$ is given as,

$\begin{array}{c}P\left(\begin{array}{l}A\text{and}C\\ \text{and}\text{not}B\end{array}\right)=P\left(A\text{and}C\right)-P\left(\begin{array}{l}A\text{and}B\\ \text{and}C\end{array}\right)\\ =0.04-0.01\\ =0.03\end{array}$

The probability of $\left(B\text{and}A\text{and}\text{not}C\right)$ is given as,

$\begin{array}{c}P\left(\begin{array}{l}B\text{and}C\\ \text{and}\text{not}A\end{array}\right)=P\left(B\text{and}C\right)-P\left(\begin{array}{l}A\text{and}\\ B\text{and}C\end{array}\right)\\ =0.04-0.01\\ =0.03\end{array}$

The probability of $\left(B\text{and}\text{not}A\text{and}\text{not}C\right)$ is given as,

$\begin{array}{c}P\left(B\text{and}\text{not}A\text{and}\text{not}C\right)=P\left(B\right)-\left[\begin{array}{l}P\left(A\text{and}B\text{and}C\right)\\ +P\left(A\text{and}B\text{and}\text{not}C\right)\\ +P\left(B\text{and}C\text{and}\text{not}A\right)\end{array}\right]\\ =0.16-\left[0.01+0.02+0.03\right]\\ =0.16-0.06\\ =0.10\end{array}$

The probability of $\left(C\text{and}\text{not}A\text{and}\text{not}B\right)$ is given as,

$\begin{array}{c}P\left(\begin{array}{l}C\text{and}\text{not}A\\ \text{and}\text{not}B\end{array}\right)=P\left(C\right)-\left[\begin{array}{l}P\left(A\text{and}B\text{and}C\right)\\ +P\left(A\text{and}C\text{and}\text{not}B\right)\\ +P\left(B\text{and}C\text{and}\text{not}A\right)\end{array}\right]\\ =0.12-\left[0.01+0.03+0.03\right]\\ =0.12-0.07\\ =0.05\end{array}$

Venn diagram:

The Venn diagram of the events A, B and C and mark the probabilities of all combinations of school students used different tobacco product is given below:

Probability:

The conditional probability that he or she also smokes cigarettes given that a student smoke electronic cigarette and the conditional probability that he or she also smokes electronic cigarettes given that a student smoke cigarette are obtained as shown below:

The conditional probability that he or she also smokes cigarettes given that a student smoke electronic cigarette is given as,

$\begin{array}{c}P\left(A|B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}\\ =\frac{\left[\begin{array}{l}P\left(A\text{and}B\text{and}C\right)+\\ P\left(A\text{and}B\text{and}\text{not}C\right)\end{array}\right]}{\left[\begin{array}{l}P\left(A\text{and}B\text{and}C\right)+\\ P\left(A\text{and}B\text{and}\text{not}C\right)+\\ P\left(B\text{and}C\text{and}\text{not}A\right)+\\ P\left(B\text{and}\text{not}A\text{and}\text{not}C\right)\end{array}\right]}\\ =\frac{\left[0.02+0.01\right]}{\left[0.01+0.02+0.03+0.10\right]}\\ =\frac{0.03}{0.16}\\ =0.1875\end{array}$

Thus, the conditional probability that a student smokes cigarettes given that he or she smokes electronic cigarette is 0.1875.

The conditional probability that a student smokes electronic cigarettes given that he or she smokes cigarette is given as,

$\begin{array}{c}P\left(B|A\right)=\frac{P\left(A\cap B\right)}{P\left(A\right)}\\ =\frac{\left[\begin{array}{l}P\left(A\text{and}B\text{and}C\right)+\\ P\left(A\text{and}B\text{and}\text{not}C\right)\end{array}\right]}{\left[\begin{array}{l}P\left(A\text{and}B\text{and}C\right)+\\ P\left(A\text{and}B\text{and}\text{not}C\right)+\\ P\left(A\text{and}C\text{and}\text{not}B\right)+\\ P\left(A\text{and}\text{not}B\text{and}\text{not}C\right)\end{array}\right]}\\ =\frac{\left[0.01+0.02\right]}{\left[0.01+0.02+0.03+0.03\right]}\\ =\frac{0.03}{0.09}\\ =0.3333\end{array}$

Thus, the conditional probability that a student smokes electronic cigarette given that he or she smokes cigarette is 0.3333.

Interpretation:

Among the different types of tobacco products used by the high school students, the percentage of students who smoke cigarette given that electronic cigarette is 18.75% and the percentage of students who smoke electronic cigarette given that smoke cigarette is 33.33%.