9. Multiplication Rule in the Uniform Probability Model
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Chapter 6
9.
Multiplication Rule in the Uniform Probability Model
This lesson delves into the Multiplication Rule of Probability, a fundamental concept in statistics. This rule is pivotal in determining the likelihood of two dependent events occurring simultaneously. Through various scenarios, like determining the probability of a specific test result or the chances of drawing a particular ticket from a hat, the rule's practical applications are showcased. By using tree diagrams and conditional probabilities, the lesson simplifies complex problems, making them more comprehensible. Whether it's understanding the chances of a disease's presence based on a test or the likelihood of drawing a specific colored marble from a box, the multiplication rule offers clarity.
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Lesson Settings & Tools
| | 10 Theory slides |
| | 8 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Multiplication Rule in the Uniform Probability Model
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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.
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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. |
|---|---|
| Event B | 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 P(A|B) 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. P(A|B) is small ? ⇒ P(B|A) is small
This reasoning is based on the assumption that P(A|B) is equal to P(B|A). Is this assumption, and therefore the reasoning for it, correct or incorrect? Why?Challenge
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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 10 000 cases. P(infected)&=0.0001 P(positive|infected)&=0.97 P(positive|not infected)&=0.0001 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?
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b What is the probability that a person is actually infected when they have a positive test result?
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c Finally, what is the probability that a person is not infected when they have a positive test result?
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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.
P(AandB)=P(A)* P(B|A) or P(AandB)=P(B)* P(A|B)
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. P(A|B)=P(A⋂ B)/P(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. P(A⋂ B)=P(A|B)* P(B) Following similar reasoning, an equivalent form of the rule can be obtained. P(A⋂ B)=P(B|A)* P(A)
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Using a Tree Diagram to Determine Probability
Consider two sets of integer numbers.
C&={- 5,- 2,- 1,2,6,7} D&={- 4,1,3,5}
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. 
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Hint
Outcomes that should be included in the tree diagram are a∈ C, a∈ D, 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 C⋃ D. 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. P(a∈ C)&=6/6+ 4=6/10 [1em] P(a∈ D)&=4/6+ 4=4/10 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 3 6 and the probability it is negative is also 3 6.
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 3 4 and the probability it is negative is 1 4.
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.
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Finding Probabilities Along Branches of a Tree Diagram
For the following questions, approximate the answers to two decimal places.
Example
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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.
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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, 12 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 1 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.
P(win)=P(box A)* P(win|box A)+P(box B)* P(win|box B)+P(box C)* P(win|box C)
SubstituteValues
Substitute values
P(win)=12/30*1/5+10/30*2/5+8/30*3/5
▼
Evaluate right-hand side
ReduceFrac
a/b=.a /6./.b /6.
P(win)=2/5*1/5+10/30*2/5+8/30*3/5
ReduceFrac
a/b=.a /10./.b /10.
P(win)=2/5*1/5+1/3*2/5+8/30*3/5
ReduceFrac
a/b=.a /2./.b /2.
P(win)=2/5*1/5+1/3*2/5+4/15*3/5
MultFrac
Multiply fractions
P(win)=2/25+2/15+12/75
ExpandFrac
a/b=a * 3/b * 3
P(win)=6/75+2/15+12/75
ExpandFrac
a/b=a * 5/b * 5
P(win)=6/75+10/75+12/75
AddFrac
Add fractions
P(win)=28/75
FracToDiv
a/b=a÷ b
P(win)=0.373333...
WritePercent
Convert to percent
P(win)=37.3%
RoundInt
Round to nearest integer
P(win)≈ 37%
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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.
P(orange)&=5/12 [0.8em]
P(box A)&=7/12 [0.8em]
P(orange|box A)&=3/7
Knowing that the drawn marble is orange, what is the probability that it was taken from box A?
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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.
cc
Desired & Given P( box A| orange) & P( orange| box A)
By the definition of a conditional probability, the desired conditional probability can be expressed as follows.
P(box A|orange) = P(box A and orange)/P(orange)
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
The probability that the marble is from box A if it is orange equals 35.
marble from box Aand
orange marble,it can be rewritten using the Multiplication Rule of Probability. P(box A and orange) = P(orange|box A)* P(box A) The above expression can be substituted into the formula for the probability that the marble is from box A knowing it is orange. P(box A|orange) = P(box A and orange)/P(orange) ⇕ P(box A|orange)= P(orange|box A) * P(box A)/P(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.
P(box A|orange)=P(orange|box A)* P(box A)/P(orange)
SubstituteValues
Substitute values
P(box A|orange)=37* 712/512
▼
Evaluate right-hand side
MultFrac
Multiply fractions
P(box A|orange)=3* 77* 12/512
CancelCommonFac
Cancel out common factors
P(box A|orange)=3* 77* 12/512
SimpQuot
Simplify quotient
P(box A|orange)=.3 /12./.5 /12.
DivFracByFracSameDenom
.a /12./.b /12.=a/b
P(box A|orange)=3/5
CalcQuot
Calculate quotient
P(box A|orange)=0.6
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Practice Reversing Conditional Probability
For the following questions, approximate the answers to two decimal places.
Closure
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Investigating Conditional Probability
The following probabilities are known.
P(infected)&=0.0001 P(positive|infected)&=0.97 P(positive|not infected)&=0.0001
Using this information, calculate three other probabilities. Write the answers correct to one significant digit. 
Now, the probability that a person is infected if the test is negative can be calculated.
Finally, the desired conditional probability will be found.
All of the information needed to calculate the conditional probability has been found.
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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. P(infected|negative) = P(infected ⋂ negative)/P(negative) The probability of the intersection of two events can be rewritten using the Multiplication Rule of Probability. P(infected|negative) = P(negative|infected)* P(infected)/P(negative) 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
.
P(infected|negative)=P(negative|infected)* P(infected)/P(negative)
SubstituteValues
Substitute values
P(infected|negative)=0.03* 0.0001/0.99980301
▼
Evaluate right-hand side
Multiply
Multiply
P(infected|negative)=0.000003/0.99980301
UseCalc
Use a calculator
P(infected|negative)=0.0000030005...
RoundSigDig
Round to 1 significant digit(s)
P(infected|negative)≈ 0.000003
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. P(infected|positive) = P(infected ⋂ positive)/P(positive) Again, the probability in the numerator can be rewritten using the Multiplication Rule of Probability. P(infected|positive) = P(positive|infected)* P(infected)/P(positive) It is known that P(infected)= 0.0001 and P(positive|infected)= 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)=1-P(negative)
Substitute
P(negative)= 0.99980301
P(positive)=1- 0.99980301
SubTerm
Subtract term
P(positive)=0.00019699
P(infected|positive)=P(positive|infected)* P(infected)/P(positive)
SubstituteValues
Substitute values
P(infected|positive)=0.97* 0.0001/0.00019699
▼
Evaluate right-hand side
Multiply
Multiply
P(infected|positive)=0.000097/0.00019699
UseCalc
Use a calculator
P(infected|positive)=0.49241...
RoundSigDig
Round to 1 significant digit(s)
P(infected|positive)≈ 0.5
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. P(not infected|positive) = P(positive|not infected)* P(not infected)/P(positive) It is known that P(positive)=0.00019699. The remaining two probabilities in the above expression are represented on the tree diagram.
P(not infected|positive)=P(positive|not infected)* P(not infected)/P(positive)
SubstituteValues
Substitute values
P(not infected|positive)=0.0001* 0.9999/0.00019699
▼
Evaluate right-hand side
Multiply
Multiply
P(not infected|positive)=0.00009999/0.00019699
UseCalc
Use a calculator
P(not infected|positive)=0.50758...
RoundSigDig
Round to 1 significant digit(s)
P(not infected|positive)≈ 0.5
Extra
Comparing Reversed Conditional ProbabilitiesThe two given conditional probabilities have been reversed. Based on that, the example values of P(A|B) and P(B|A) 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(A|B) and P(B|A) are not equal.
Multiplication Rule in the Uniform Probability Model
Exercise 3.1
A bag contains red and blue balls. The following tree diagram shows the probability of drawing two blue balls from the bag without replacement.
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We will start by finding the value of x. Then, we will write a system of equations that can be solved for the total numbers of balls in the bag.
Finding x
According to the Multiplication Rule of Probability we can write the following equation. 17/20 * x =187/260 Here, x represents the probability of drawing a second blue ball if the first ball drawn is blue, as shown on the tree diagram. Let's solve our equation for x.
The probability of drawing a blue ball if the first ball drawn is blue is 1113. Let's update the tree diagram.
Finding the Total Number of Balls
Let's label the total number of balls T and the number of blue balls B. To determine the number of balls, we must write a system of two equations for T and B. From the tree diagram, we know the probability of drawing a blue ball to be 1720. Therefore, if we divide B by T we can write the following equation. B/T=17/20 We know the probability of drawing a second blue ball if the first ball drawn is blue. Notice that at this point we have one less blue ball in the bag and the total number of balls in the bags is also one less. We obtain the following equation. B-1/T-1=11/13 Let's rewrite this equation.
Now we have two equations which describe B and T. These equations form a system of equations. BT= 1720 & (I) 13B-11T=2 & (II) Let's solve this system for T using the Substitution Method.
The total number of balls is 40. Notice that we are not required to solve for the number of blue balls.
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Multiplication Rule in the Uniform Probability Model
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