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.
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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 1.1
Ali loves Russian authors. He has bought five books by Anton Chekhov and three books by Fyodor Dostoevsky. When going on vacation he randomly selects three books and throws them into his luggage. What is the probability that all three books are written by the same author? Round your answer to the nearest percent.
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Let's represent books by Anton Chekhov with red covers, and represent books by Fyodor Dostoevsky with blue covers.
Ali has bought a total of 5+3=8 books. Since 5 of them are by Chekhov and 3 are by Dostoevsky, we get the following probabilities when Ali is choosing the first book. P(📕)=5/8 and P(📘)=3/8 When the first book is selected, there is one less book on the shelf. Depending on what book was selected, the probability of the second book Ali selects is affected. Let's mark the probabilities along the paths where three books by the same author, Chekhov or Dostoevsky, are selected.
To calculate the probability of either event occurring, we will use the Multiplication Rule of Probability. P(📕,📕,📕)=5/8* 4/7* 3/6=60/336 [1em] P(📘,📘,📘)=3/8* 2/7* 1/6=6/336 If we add these probabilities, we get the combined probability of selecting three books by the same author.
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Ignacio was window shopping online when he found a shirt he really liked on sale. He hurried to order it and complete the purchase but forgot to specify the color. Because the color was not specified, the supplier will select the shirt color randomly. Here is a list of the two colors of shirts in stock for each supplier at the time of the purchase.
| Supplier A | Supplier B |
|---|---|
| 30 Blue | 40 Blue |
| 10 Red | 60 Red |
If Supplier A is currently taking 40% of the orders, what is the probability that Ignacio gets a blue shirt? Make a tree diagram to find this probability and write the answer as a fraction in its simplest form.
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To answer the question it will be helpful to make a tree diagram. The root node of the tree diagram represents the event of choosing a supplier. It can either be Supplier A or Supplier B. We also know that the probability that Supplier A has taken the orders is 40 %= 25. Let's start making the tree diagram using this information.
Recall that the sum of the probabilities of branches originating from one node is always 1. Using this fact, this part of tree diagram can be completed.
The next outcomes correspond to shirt colors.
Recall that Supplier A has 30 blue shirts and 10 red shirts. With this information, the probability that the shirt is of a given color can be calculated. P(Blue shirt|Supplier A)& = 30/30 + 10 = 30/40 = 3/4 [1em] P(Red shirt|Supplier A)& = 10/30 + 10 = 10/40 = 1/4 On the other hand, Supplier B has 40 blue shirts and 60 red shirts. With this information the probability that the shirt is of a given color can be calculated. P(Blue shirt|Supplier B)& = 40/40 + 60 = 40/100 = 2/5 [1em] P(Red shirt|Supplier B)& = 60/40 + 60 = 60/100 = 3/5 Let's add these probabilities to the diagram.
Finally, notice that we need to consider both the probability of getting a blue shirt from Supplier A P(Blue⋂A) or from Supplier B P(Blue⋂B).
Therefore, the probability that Ignacio receives a blue shirt is the sum of the probabilities that he receives the blue shirt from Supplier A or from Supplier B.
Therefore, the probability of receiving a blue shirt is 2750.
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Jordan is getting ready for a party and wants to wear one of her dresses for the occasion. However, she does not want to worry about which one to pick, so she chooses one randomly. Consider the list of colors she has for the following articles of clothing.
| Dresses | Pairs of Shoes |
|---|---|
| 2 Blue | 1 Black |
| 3 Red | 1 white |
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To answer the question, a tree diagram will be made. The root node of the tree diagram represents the event of choosing a dress. It can either be blue or red.
The probability of each outcome should be written on the corresponding branch. Jordan has 2 blue dresses and 3 red dresses. With this information the probability that the dress is of a given color can be calculated. P(blue dress)&=2/2+ 3=2/5 [1em] P(red dress)&=3/2+ 3=3/5 Now, these probabilities will be added to the diagram.
The next outcomes correspond to the colors of shoes.
We know that the probability that Jordan wears white shoes with a blue dress is 45. The probability that Jordan wears white shoes with a red dress is 25. Let's add these probabilities to the diagram.
Recall that 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 probability that Jordan is wearing a blue dress and white shoes, can be calculated by multiplying the probabilities along the corresponding branches of the tree diagram.
Therefore, the probability that Jordan wears a blue dress and white shoes can be calculated as follows. 2/5*4/5 = 8/25
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Jordan doesn't want to worry about what she will wear to go out. She picks a shirt, a pair of trousers, and a pair of shoes randomly. Consider the list of colors she has for each article of clothing.
| Shirts | Trousers | Pair of shoes |
|---|---|---|
| 2 Blue | 4 Blue | 1 Black |
| 3 Red | 2 Red | 2 Violet |
| 2 Pink | 1 Black |
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To answer the questions, a tree diagram will be made. The root node of the tree represents the event of choosing a shirt of a given color. It can either be blue, red, or pink.
The probability of each outcome should be written on the corresponding branch. Jordan has 2 blue shirts, 3 red ones, and 2 orange ones. With this information the probability that the shirt is of a given color can be calculated. P(Blue shirt)&=2/2+ 3+ 2=2/7 [1em] P(Red shirt)&=3/2+ 3+ 2=3/7 [1em] P(Pink shirt)&=2/2+ 3+ 2=2/7 Now, the probabilities will be added to the diagram.
The next outcomes is the color choice for the trousers.
Jordan has 3 blue trousers, 2 red trousers, and 1 black trouser ones. With this information the probability that the trouser is of a given color can be calculated. P(Blue trousers)&=3/3+ 2+1=3/6=1/2 [1em] P(Red trousers)&=2/3+ 2+1=2/6=1/3 [1em] P(Black trousers)&=1/3+ 2+1=1/6 Now, the probabilities will be added to the diagram.
Finally, it is left to consider the possible outcomes for the color choice of the pair of shoes.
Jordan has 1 pair of black shoes and 2 pairs of violet shoes. With this information the probability that the pair of shoes chosen is of a given color can be calculated. P(Black shoes)&=1/1+ 2=1/3 [1em] P(Violet shoes)&=2/1+ 2=2/3 Now, the probabilities will be added to the diagram.
The probability that Jordan is wearing a blue shirt, black trousers, and violet shoes can be calculated by multiplying the probabilities along the corresponding branch of the tree diagram.
Therefore, the probability that Jordan wears a blue shirt, black trousers, and violet shoes is P = 2/7*1/3*2/3 = 4/63
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How a conditional probability can be reversed was taught during the lesson, arriving at the formula shown below. P(A|B) = P(B|A)* P(A)/P(B) This result is known as Bayes' Theorem, and it is very useful when working with conditional probabilities problems. Consider the following situation. Kriz wants to travel to visit their family. They find two options from two different airlines that offer affordable prices. However, according to customers reviews, they both have high chances of being delayed.
| Airline A | Airline B |
|---|---|
| On time 30 % | On time 20 % |
| Delayed 70 % | Delayed 80 % |
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Since we know that Kriz used 4 pieces of paper reading Airline A and 6 reading Airline B to make their decision, we can calculate the probability that they have flown with a given airline. P(airline A) &= 4/4+ 6 = 4/10 = 2/5 [1em] P(airline B) &= 6/4+ 6 = 6/10 = 3/5 Let's summarize all the known probabilities so far.
| Flying with Airline A | Arriving late given that they traveled with Airline A | Arriving on time given that they traveled with Airline A |
|---|---|---|
| P(airline A) = 2/5 | P(late |airline A) = 70 % = 7/10 | P(on time |airline A) = 30 % = 3/10 |
| Flying with Airline B | Arriving late given that they traveled with Airline B | Arriving on time given that they traveled with Airline B |
| P(airline B) = 2/5 | P(late |airline B) = 80 % = 4/5 | P(on time |airline B) = 20 % = 1/5 |
We are interested in finding the probability that Kriz traveled with Airline B, given that they arrived on time. Notice that the probability that Kriz arrived on time provided that they flew with Airline B is known. cc Desired & Given P(airline B|on time) & P(on time|airline B) To reverse the given conditional probability we can use Bayes' Theorem, as mentioned at the beginning of the exercise. P(airline B|& on time) & = P(on time | airline B)* P(airline B)/P(on time) All the probabilities on the right-hand side of the equation are known, except for P(on time). To calculate it, let's represent the situation using a tree diagram.
We need to consider both the probability of arriving on time given that the flight was with Airline A and the probability of arriving on time given that the flight was with Airline B. These can be obtained by multiplying the probabilities along the corresponding branches.
Therefore, the probability that the flight arrives on time P(on time) is the sum of the probability that it arrives on time given that the flight was with Airline A and the probability that it arrives on time given that the flight was with Airline B.
Now, we can go back to the expression for P(airline B | on time) and find the required probability.
Therefore, the probability that Kriz had used Airline B given that they arrived on time is 13.
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Multiplication Rule in the Uniform Probability Model
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▢√
∫▢▢
4
5
6
+
−
≤
<
log
ln
log▢
1
2
3
()
∣
sin
cos
tan
0
.
π
x
y
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