9. Multiplication Rule in the Uniform Probability Model
+ Customize lesson
0
0
Sign In
An error occurred, try again later!
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.
Show less Show more expand_more| Student Learning Objectives: |
|---|
|
| | 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
Slide of 10
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.
Share
Report error
Translate
Explore
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?
Share
Report error
Translate
Challenge
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?
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["0.000003"]}}
b
What is the probability that a person is actually infected when they have a positive test result?
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["0.5"]}}
c
Finally, what is the probability that a person is not infected when they have a positive test result?
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["0.5"]}}
Share
Report error
Translate
Discussion
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)
Share
Report error
Translate
Example
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.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":"P(a∈ Canda>0)=","formTextAfter":null,"answer":{"text":["\\dfrac{3}{10}","0.3"]}}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":"P(a∈ Danda<0)=","formTextAfter":null,"answer":{"text":["\\dfrac{1}{10}","0.1"]}}
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.
Share
Report error
Translate
Pop Quiz
Finding Probabilities Along Branches of a Tree Diagram
For the following questions, round the answers to two decimal places.
Share
Report error
Translate
Example
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.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":"P(Dylan wins)\u2248","formTextAfter":"%","answer":{"text":["37"]}}
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.
The sum of the probabilities of these outcomes is the probability that 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%
Share
Report error
Translate
Example
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?
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["3\/5","\\dfrac{3}{5}","0.6"]}}
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
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
The probability that the marble is from box A if it is orange equals 35.
Share
Report error
Translate
Pop Quiz
Practice Reversing Conditional Probability
For the following questions, round the answers to two decimal places.
Share
Report error
Translate
Closure
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.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":"P(infected|negative)=","formTextAfter":null,"answer":{"text":["0.000003"]}}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":"P(infected|positive)=","formTextAfter":null,"answer":{"text":["0.5"]}}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":"P(not infected|positive)=","formTextAfter":null,"answer":{"text":["0.5"]}}
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
.
Now, the probability that a person is infected if the test is negative can be calculated.
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
Finally, the desired conditional probability will be found.
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.
All of the information needed to calculate the conditional probability has been found.
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.
Share
Report error
Translate
Multiplication Rule in the Uniform Probability Model
Exercise 2.1
A knight has managed to escape and is running through a dark dungeon with guards right behind him.
He cannot afford to turn around, as he will get caught. What is the probability that the knight makes it to freedom? Give an exact answer.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":""},"formTextBefore":"P(freedom)=","formTextAfter":null,"answer":{"text":["\\dfrac{1}{6}"]}}
security
report
code
security
report
code
We can illustrate the knight's escape as a tree diagram. When he reaches the first junction, there are two paths which will result in him getting caught (C) and one path where the knight still has a chance at freedom (F). The probability that the knight takes any of these paths is 13.
If the knight reaches the second junction, there is a fifty-fifty chance that the knight chooses the path to freedom. Notice that if the knight turned either left or right at the first junction, the probability of getting caught is 1. Now we can draw the second level of the tree diagram.
By using the Multiplication Rule of Probability, we can calculate the probability of the knight reaching freedom.
The probability of the knight reaching freedom is 16.
security
report
code
E
Hint & Answer
S
Solution
In poker, a royal straight flush is a hand where you have been dealt an ace, all the face cards, and a 10 of the same suit. We see an example of a royal straight flush below.
What is the probability of being dealt a royal straight flush? Answer with a fraction.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":""},"formTextBefore":"P(RSF)=","formTextAfter":null,"answer":{"text":["\\dfrac{1}{649740}"]}}
security
report
code
security
report
code
To calculate the probability of obtaining a royal straight flush, we will consider the first card we are dealt and then the remaining cards that we need.
First Card
A standard deck contains 52 cards. Of these cards, we need to be dealt one of the five cards in a royal straight flush. Notice that we can obtain a royal straight flush with all four suits. Therefore, the first card can be any of the five cards from any of the suits. We have a total of 4* 5=20 cards that are favorable outcomes. P(first card)=20/52
Remaining Cards
When the first card is dealt, the remaining cards must have the same suit as the first card. This means the probability of our second card being any of the remaining cards we need is 451. Let's write all the probabilities for our remaining cards in this way. P(second card)&=4/51 [1em] P(third card)&=3/50 [1em] P(fourth card)&=2/49 [1em] P(fifth card)&=1/48
Probability of a Royal Straight Flush
Now we can calculate the probability of getting a royal straight flush by using the Multiplication Rule of Probability. 20/52* 4/51 * 3/50 * 2/49 * 1/48 =1/649 740
security
report
code
E
Hint & Answer
S
Solution
Multiplication Rule in the Uniform Probability Model
Recommended exercises
To get personalized recommended exercises, please login first.
Loading content
Edit Lesson
≥
>
■2
■▢
e▢
7
8
9
×
÷■1
=
=
√▢
▢√
∫▢▢
4
5
6
+
−
≤
<
log
ln
log▢
1
2
3
()
∣
sin
cos
tan
0
.
π
x
y
Loading content