korporasidn

2021-11-12

1.The probability of Christine getting a new job or going on vacation (or both) is 0.85. If the probability of Christine going on vacation is 0.10 and the probability of Christine getting a new job is 0.75, are the events Christine gets a new job and Christine goes on vacation mutually exclusive?
2. The probability that Isabelle will go trick-or-treating is 0.4. The probability that she will go trick-or-treating or watch a scary movie is 0.65. The probability of Isabelle going trick-or-treating and watching a scary movie is 0.20.
A. What is the probability that Isabelle will not watch a scary movie?
B. What is the probability that Isabelle will watch a scary movie but does not go trick-or-treating?

Nicole Keller

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

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