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{{ printedBook.courseTrack.name }} {{ printedBook.name }} Multiplication Rule in the Uniform Probability Model

In this lesson the Multiplication Rule of Probability will be introduced. Also, this lesson will explore how this rule can be used to find the probability of the intersection of two events and how it can be used to reverse conditional probability.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Prosecutor's Fallacy

Consider the following two events.

 Event A The DNA from a crime scene is concluded to be the DNA of the defendant. The defendant is not guilty.
If the defendant is not guilty — event B occurs — then the probability that their DNA was found on the crime scene is very small. Therefore, the conditional probability is small. Often, based on this information, the prosecutor claims that the probability that the defendant is not guilty given that A occurs is also small.
This reasoning is based on the assumption that is equal to Is this assumption, and therefore the reasoning for it, correct or incorrect? Why?

Investigating Conditional Probability

The probability of being infected with a certain disease is 0.0001. There is a test used to detect this disease. If someone is infected, the test comes back positive 97 times out of 100 cases. If someone is not infected, the test comes back positive 1 time out of 10000 cases.
Express the answers as decimal numbers approximated to one significant figure.
a What is the probability that a person is infected when they have a negative test result?
b What is the probability that a person is actually infected when they have a positive test result?
c Finally, what is the probability that a person is not infected when they have a positive test result?

Multiplication Rule of Probability

Conversely, if A and B are dependent events, a rearrangement of the Conditional Probability Formula can be used to find the probability of the intersection of the events.

Proof

Consider two dependent events A and B. The conditional probability of A given B is the ratio of the probability of the intersection of A and B to the probability of B.
The Multiplication Rule of Probability is obtained by multiplying both sides of the above formula by P(B) and using the Symmetric Property of Equality.
Following similar reasoning, an equivalent form of the rule can be obtained.

Using a Tree Diagram to Determine Probability

Consider two sets of integer numbers.
A number a is chosen from the union of C and D. What is the probability that the number is a positive element of C? What is the probability that the number is a negative element of D? Use a tree diagram to determine the probabilities.

Hint

Outcomes that should be included in the tree diagram are a>0, and a<0.

Solution

A tree diagram will be made to answer the questions. The root node of the tree represents the event of choosing a number a from The chosen number can be either from C or from D. The probability of each outcome should be written on the corresponding branch. Set C has 6 elements, D has 4 elements, and the sets do not have any elements in common. With this information the probability that a is an element of C and that a is an element of D can be calculated.
Now the probabilities will be added to the diagram. The other outcomes that need to be considered to answer the questions are that a can be positive or negative. Set C has 6 elements. 3 of them are positive and 3 of them are negative. Therefore, knowing that a is an element of C, the probability that it is positive is and the probability it is negative is also Likewise, D has 4 elements: 3 of them positive and 1 of them negative. Knowing that a is from D, the probability it is positive is and the probability it is negative is The probability that a is an element of C and positive can be calculated by multiplying the probabilities along the corresponding branch of the tree diagram. Similarly, the probability that a is an element of D and negative can be found. Finding Probabilities Along Branches of a Tree Diagram

For the following questions, approximate the answers to two decimal places. Using a Tree Diagram to Determine Probability

Dylan participates in a school lottery that consists of two stages. He starts by drawing a ticket from a hat. This ticket tells Dylan from which of three boxes he will be drawing a second ticket. There are 30 tickets in the hat: 12 for box A, 10 for box B, and 8 for box C. Each box has 5 tickets. Box A has 1 winning ticket, box B has 2 winning tickets, and box C has 3. What is the probability that Dylan wins? Express the answer as a percent rounded to the nearest whole number.

Hint

Use a tree diagram to represent the given information.

Solution

First make a tree diagram. The root node of the tree represents the event of drawing a ticket from the hat. There are 30 tickets in the hat. From those, correspond to box A, 10 correspond to box B, and 8 correspond to box C. With this information, the probabilities for the first three branches of the tree can be written. A ticket drawn from any of the boxes can be a winning or a loosing ticket. Each box contains 5 tickets. Box A has winning ticket, box B has 2, and box C 3. With this information, the probabilities of drawing a winning ticket from each box can be written. The sum of the probabilities of the branches that come out of the same node is equal to 1. Knowing this, the probabilities of not drawing a winning ticket from each box can be calculated. There are three possible outcomes in which Dylan draws a winning ticket. The sum of the probabilities of these outcomes is the probability that Dylan draws a winning ticket.
P(win)=P(box A)P(winbox A)+P(box B)P(winbox B)+P(box C)P(winbox C)
Evaluate right-hand side

Reversing Conditional Probability

Consider two boxes A and B that both contain orange and purple marbles. A box is randomly chosen and then one marble is drawn from this box. The following probabilities are known.
Knowing that the drawn marble is orange, what is the probability that it was taken from box A?

Hint

How can the probability that the marble is from box A and it is orange be calculated using the given information?

Solution

The probability that a marble is from box A knowing that it is orange is a conditional probability. It is the reverse of the given conditional probability.
By the definition of a conditional probability, the desired conditional probability can be expressed as follows.
The probability in the denominator is known. However, the probability in the numerator is not. Since the expression in the numerator is the probability of the intersection of the events marble from box A and orange marble, it can be rewritten using the Multiplication Rule of Probability.
The above expression can be substituted into the formula for the probability that the marble is from box A knowing it is orange.
The conditional probability has been expressed in terms of the known probabilities. Now, the values of the known probabilities can be substituted into the formula.
Evaluate right-hand side
P(box Aorange)=0.6
The probability that the marble is from box A if it is orange equals

Practice Reversing Conditional Probability

For the following questions, approximate the answers to two decimal places. Investigating Conditional Probability

The following probabilities are known.
Using this information, calculate three other probabilities. Write the answers correct to one significant digit.

Solution

First, the given information will be represented in a tree diagram. The sum of the probabilities of branches coming out of the same node is always 1. Using this fact, the tree diagram can be completed. The tree diagram gathers all of the known information. Now, the desired unknown probabilities will be calculated one by one.

Probability That a Person Is Infected if the Test Is Negative

By the definition of conditional probability, the conditional probability that a person is infected given their test result is negative can be written as a ratio.
The probability of the intersection of two events can be rewritten using the Multiplication Rule of Probability.
Both probabilities in the numerator are represented in the tree diagram. The probability that the test comes back negative can be found using the tree diagram as well. To do so, the probabilities of all outcomes that include receiving a negative test result should be added. There are two such outcomes. These are infected with a negative test result and not infected with a negative test result. Now, the probability that a person is infected if the test is negative can be calculated.
Evaluate right-hand side

Probability That a Person Is Infected if the Test Is Positive

The probability that a person is infected if the test is positive can be written as a ratio.
Again, the probability in the numerator can be rewritten using the Multiplication Rule of Probability.
It is known that P(infected)=0.0001 and P(positiveinfected)=0.97. Now, the probability that the test result is positive has to be found. Since the result of the test can be either positive or negative, the probability that it is positive is the difference of 1 and the probability that the result is negative.
P(positive)=1P(negative)
P(positive)=10.99980301
Finally, the desired conditional probability will be found.
Evaluate right-hand side

Probability That a Person Is Not Infected if the Test Is Positive

The probability that a person is not infected given the test result is positive can be found by reversing the conditional probability.
It is known that The remaining two probabilities in the above expression are represented on the tree diagram. All of the information needed to calculate the conditional probability has been found.
Evaluate right-hand side

Extra

Comparing Reversed Conditional Probabilities

The two given conditional probabilities have been reversed. Based on that, the example values of P(AB) and P(BA) can be compared.

P(positive|infected) P(infected|positive)
0.97 0.5
P(positive|not infected) P(not infected|positive)
0.0001 0.5

It is seen that P(AB) and P(BA) are not equal.