Alta- Probability Chapter 3 Exam Questions with Verified Solutions (Rated 100%)
The probability that a student will take loans to pay for their undergraduate education is 0.85, and the
probability that a student will go to graduate school given that the student took loans to pay for their
undergraduate education is 0.13. What is the probability that a student will go to graduate school and
take loans to pay for their undergraduate education?
Round your answer to three decimal places. - Answers $0.111$0.111
Remember the multiplication rule for conditional probability:
P(B AND A)=P(B|A)P(A)
So if we think of A as the event of a student taking loans to pay for their undergraduate education and B
as being the event of a student going to graduate school, then we can plug in the known information to
find
P(B AND A)=(0.13)(0.85)≈0.111
If a police officer pulls over someone for speeding, the police officer can either give a ticket or a warning,
so it is impossible for a police officer to give a ticket and a warning for speeding. If the probability that a
police officer will give a warning for speeding is 0.03, and the probability that a police officer will give a
ticket or a warning for speeding is 0.52, what is the probability that a police officer will give a ticket for
speeding? - Answers $0.49$0.49
Because it is impossible for a police officer to give a ticket and a warning for speeding, we see that they
are mutually exclusive events. Therefore, we know that P(A AND B)=0 (this is the definition of mutually
exclusive events). So for mutually exclusive events, the probability addition rule becomes
P(A OR B)=P(A)+P(B)−P(A AND B)=P(A)+P(B)
Rearranging, we find thatP(A)=P(A OR B)−P(B)So plugging in the given values, we
findP(A)=0.52−0.03=0.49 The probability that a police officer will give a ticket for speeding is 0.49.
If A and B are events with P(A)=0.5, P(A OR B)=0.65, P(A AND B)=0.15, find P(B). - Answers $0.3$0.3
First, it is helpful to write down the addition rule for probabilities:
P(A OR B)=P(A)+P(B)−P(A AND B)
Now, rearranging this, we find thatP(B)=P(A OR B)+P(A AND B)−P(A)Plugging in the known values, we
findP(B)=0.65+0.15−0.5=0.3
There are two known issues with a certain model of new car. The first issue, A, occurs with a probability
of P(A)=0.1. B is another known issue with the car. If it is known that either event occurs with a
,probability of P(A OR B)=0.93, and that both events occur with a probability of P(A AND B)=0.07,
calculate P(B). - Answers $0.9$0.9
Begin with the Addition Rule:
P(A OR B)=P(A)+P(B)−P(A AND B)
Rearranging to solve for P(B), we find that
P(B)=P(A OR B)−P(A)+P(A AND B)=0.93−0.1+0.07=0.9
So P(B)=0.9.
If we were to select one friend at random, what is the probability that they would be in either statistics
or writing, but not both? - Answers To solve this problem, it is pretty simple for us to count the number
of friends that are not in both classes. There are 5 out of the 6 friends who are only in one of the
courses, so
P(NOT both classes)=56
However, there will be many other instances in probability where it isn't as easy to calculate
combinations of events.
If we don't have a Venn diagram, or information about how many friends there are, we have to use
something else to help us find the probability. The following rule will help us to calculate probabilities
for OR :
A group of friends take classes at a university in biology, chemistry, and physics. If you selected a friend
at random, the probabilities that they take certain classes is given below:
The probability that they take biology class is P(B)=13;
The probability that the friend takes chemistry is P(C)=12;
A student takes physics with the probability P(P)=16;
A student takes both biology and physics with a probability of 112.
,What's the probability that a randomly chosen friend takes biology or physics, P(B OR P)? - Answers
Without a diagram, or information about how many friends there are, we have to use the Addition Rule
for Probabilities to P(B OR P):
P(B OR P)=P(B)+P(P)−P(B AND P)=13+16−112=512
So, the probability that a randomly chosen friend takes biology or physics is 512.
How To
Addition Rule for Probabilities
If A and B are events defined on a sample space, then
P(A OR B)=P(A)+P(B)−P(A AND B) - Answers How To
Addition Rule for Probabilities
If A and B are events defined on a sample space, then
P(A OR B)=P(A)+P(B)−P(A AND B)
Employees at a company are randomly assigned certain shifts during the busy holiday season. Let A
represent the shift between the hours of 8 a.m. and 12 p.m., and B represent the shift between the
hours of 12 and 4 p.m. If the shifts are assigned with the following probabilities:
P(A)=0.28;
P(B)=0.83;
P(A OR B)=0.93
What is P(A AND B), the probability that an employee will randomly be assigned both shifts? - Answers
Here, we are given slightly different information than in the rule above, but note that we can rearrange
the rule to solve for P(A AND B):
P(A AND B)=P(A)+P(B)−P(A OR B)=0.28+0.83−0.93=0.18
, So we find that P(A AND B)=0.18. This means that the probability that an employee will randomly be
assigned both shifts is 0.18, or 18%.
Randy wants to either ride share to work or drive his own car to work, but it is impossible for Randy to
ride share and drive his own car in one trip. If the probability that Randy ride shares is 0.22, and the
probability that Randy drives his own car is 0.42, what is the probability that Randy ride shares or drives
his own car to work? - Answers $0.64$0.64
Because it is impossible for Randy to ride share and drive his own car in one trip, we see that they are
mutually exclusive events. Therefore, we know that P(A AND B)=0 (this is the definition of mutually
exclusive events). So for mutually exclusive events, the probability addition rule becomes
P(A OR B)=P(A)+P(B)−P(A AND B)=P(A)+P(B)
So we find thatP(A OR B)=P(A)+P(B)=0.22+0.42=0.64 The probability that Randy ride shares or drives his
own car to work is 0.64.
If A and B are independent events with P(A)=0.90 and P(A AND B)=0.54, find P(B).
Give your answer as a decimal rounded to two decimal places. - Answers $0.60$0.60
Remember that for independent events,
P(A AND B)=P(A)P(B)
Solving for P(B), we find thatP(B)=P(A AND B)P(A)Now, plugging in the values we are given, we find
thatP(B)=0.540.90=0.60
Given that P(A AND B)=0.29 and P(A|B)=0.67, what is P(B)?
Give your answer as a percent. Round to two decimal places. - Answers $43.28\ \%$43.28 %
Remember the multiplication rule for conditional probability:
P(A AND B)=P(A|B)P(B)
Rearranging, we find thatP(B)=P(A AND B)P(A|B)Plugging in the values we were given, we find
thatP(B)=0.290.67≈0.4328
To convert the decimal to a percent, we can multiply by 100:
0.4328×100=43.28%
Adam loves to visit his local coffee shop. When there, his two favorite drinks are cafe latte and espresso.
Let A represent the event that he drinks a cafe latte and B represent the event that he drinks an
espresso. If A and B are independent events with P(A)=0.5 and P(B)=0.8, find P(A AND B).
