Chapter 13, Problem 13.44E

To determine

The probability that any of the three jobs is not offered to Julie.

Answer to Problem 13.44E

The probability that any of the three jobs is not offered to Julie is 0.1.

Explanation of Solution

Given info:

Julie is offered a job by three events A, B and C. A be the event for Connecticut office of the Chief Medical Examiner, B be the event for the New Jersey Division of Criminal Justice and C be the event for Federal Disaster Mortuary Operations Response Team.

Julie writes some probabilities $P\left(A\right)$ is 0.5, $P\left(B\right)$ is 0.4, $P\left(C\right)$ is 0.2, $P\left(A\text{and}B\right)$ is 0.1, $P\left(A\text{and}C\right)$ is 0.05, $P\left(B\text{and}C\right)$ is 0.05 and $P\left(A\text{and}B\text{and}C\right)$ is 0.

Calculation:

To draw the Venn diagram, find the probabilities.

Calculate the probability for event $\left(A\text{and}\left(\text{not}B\right)\text{and}\left(\text{not}C\right)\right)$.

$P\left(A\text{and}\left(\text{not}B\right)\text{and}\left(\text{not}C\right)\right)=P\left(A\right)-P\left(A\text{and}B\right)-P\left(A\text{and}C\right)$

Substitute 0.5 for $P\left(A\right)$, 0.1 for $P\left(A\text{and}B\right)$ and 0.05 for $P\left(A\text{and}C\right)$ in above formula.

$\begin{array}{c}P\left(A\text{and}\left(\text{not}B\right)\text{and}\left(\text{not}C\right)\right)=0.5-0.1-0.05\\ =0.35\end{array}$

Calculate the probability for event $\left(B\text{and}\left(\text{not}A\right)\text{and}\left(\text{not}C\right)\right)$.

$P\left(B\text{and}\left(\text{not}A\right)\text{and}\left(\text{not}C\right)\right)=P\left(B\right)-P\left(A\text{and}B\right)-P\left(B\text{and}C\right)$

Substitute 0.4 for $P\left(B\right)$, 0.1 for $P\left(A\text{and}B\right)$ and 0.05 for $P\left(B\text{and}C\right)$ in above formula.

$\begin{array}{c}P\left(B\text{and}\left(\text{not}A\right)\text{and}\left(\text{not}C\right)\right)=0.4-0.1-0.05\\ =0.25\end{array}$

Calculate the probability for event $\left(C\text{and}\left(\text{not}A\right)\text{and}\left(\text{not}\text{B}\right)\right)$.

$P\left(C\text{and}\left(\text{not}A\right)\text{and}\left(\text{not}B\right)\right)=P\left(C\right)-P\left(A\text{and}C\right)-P\left(B\text{and}C\right)$

Substitute 0.2 for $P\left(C\right)$, 0.05 for $P\left(A\text{and}C\right)$ and 0.05 for $P\left(B\text{and}C\right)$ in above formula.

$\begin{array}{c}P\left(C\text{and}\left(\text{not}A\right)\text{and}\left(\text{not}B\right)\right)=0.2-0.05-0.05\\ =0.1\end{array}$

Calculate the probability for event $\left(\text{neither}A\text{nor}B\text{nor}C\right)$.

$P\left(\text{neither}A\text{nor}B\text{nor}C\right)=1-\left[\begin{array}{l}P\left(A\text{and}\left(\text{not}B\right)\text{and}\left(\text{not}C\right)\right)\\ +P\left(B\text{and}\left(\text{not}A\right)\text{and}\left(\text{not}C\right)\right)\\ +P\left(C\text{and}\left(\text{not}A\right)\text{and}\left(\text{not}B\right)\right)+P\left(A\text{and}B\right)\\ +P\left(A\text{and}C\right)+P\left(B\text{and}C\right)+P\left(A\text{and}B\text{and}C\right)\end{array}\right]$

Substitute 0.35 for

$P\left(A\text{and}\left(\text{not}B\right)\text{and}\left(\text{not}C\right)\right)$,

0.25 for $P\left(B\text{and}\left(\text{not}A\right)\text{and}\left(\text{not}C\right)\right)$,

0.1 for $P\left(C\text{and}\left(\text{not}A\right)\text{and}\left(\text{not}B\right)\right)$, 0.1 for $P\left(A\text{and}B\right)$, 0.05 for $P\left(A\text{and}C\right)$, 0.05 for $P\left(B\text{and}C\right)$ and 0 for $P\left(A\text{and}B\text{and}C\right)$ in above formula.

$\begin{array}{c}P\left(\text{neither}A\text{nor}B\text{nor}C\right)=1-\left[0.35+0.25+0.1+0.1+0.05+0.05+0\right]\\ =1-\left[0.9\right]\\ =0.1\end{array}$

Instructions:

• Draw a Venn diagram form all the above information.

From the Venn diagram it can be noticed that the probability probability that any of the three jobs is not offered to Julie is 0.1.

Thus, the probability that any of the three jobs is not offered to Julie is 0.1.