1.The probability of Christine getting a new job or going on vacation (or both) is...

Step 1
1.Given that
The probability of Christine getting a new job ${\displaystyle =0.75}$
the probability of Christine going on vacation ${\displaystyle =0.10}$
The probability that getting a new job or going on vacation ${\displaystyle =0.85}$
Are the events mutually exclusive
2.
The probability that Isabelle will go trick-or-treating ${\displaystyle =0.4}$
The probability that she will trick or treating and watching a scary movie ${\displaystyle =0.20}$
The probability that she will trick and treating and watching a scary movie ${\displaystyle =0.65}$
A. Find The probability that she will not watch a scary movie
B. Find the probability that she will watch a scary movie but not go trick or treating
Step 2
1.
Let A be the event that Christine getting a new job
Let B be the event that Christine going on a vacation
From the given probabilities
${\displaystyle P\left(A\right)=0.75}$
${\displaystyle P\left(B\right)=0.10}$
${\displaystyle P(A\cup B)=0.85}$
Events A and B are mutually exclusive
Because the occurrence of event A will not affect the Occurrence of B
From the formula, if ${\displaystyle P(A\cup B)=P\left(A\right)+P\left(B\right)}$ A and B can be mutually exclusive events
here ${\displaystyle 0.75+0.10=0.85}$
Step 3
2.
Let A be an event that Isabelle will go on trick or treat
Let B be an event that Isabelle will go on watching a scary movie
From the given probabilities
${\displaystyle P\left(A\right)=0.4}$
${\displaystyle P(A\cup B)=0.65}$
${\displaystyle P(A\cap B)=0.20}$
a) The probability that she will not watch a scary movie
${\displaystyle P\left(\bar{B}\right)=1-P\left(B\right)}$
P(B) can be calculated from the relation
${\displaystyle P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)}$
${\displaystyle P\left(B\right)=P(A\cup B)-P\left(A\right)+P(A\cap B)}$
${\displaystyle =0.65-0.4+0.20}$
${\displaystyle =0.45}$
${\displaystyle P\left(\bar{B}\right)=1-0.45}$
${\displaystyle =0.55}$
b) The probability that she will watch a scary movie but not go trick or treat
${\displaystyle P(\bar{A}\cap B)=P\left(B\right)-P(A\cap B)}$
${\displaystyle =0.45-0.20}$
${\displaystyle =0.25}$