8. Addition Rule in the Uniform Probability Model
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Chapter 6
8.
Addition Rule in the Uniform Probability Model
Probability is a measure that quantifies the likelihood of an event occurring. The addition rule of probability is one of the core concepts in this domain, especially when considering more than one event. Imagine wanting to know the chance of rain or snow on a given day; this rule helps in computing such combined probabilities. This rule is paramount not only in academic pursuits but also has far-reaching implications in industries such as finance, insurance, and even game design. It assists in making informed decisions, evaluating risks, and predicting future events based on past data. Understanding this rule paves the way for better comprehension of complex probabilistic scenarios.
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Lesson Settings & Tools
| | 9 Theory slides |
| | 8 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Addition Rule in the Uniform Probability Model
Slide of 9
This lesson breaks down the calculation of the probability of real-world compound events using the Addition Rule of Probability.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Challenge
Calculating the Chances of Winning a Card Game
Diego is playing cards using a standard deck.
Diego wins if he draws a card that satisfies any of the following events. &Rule1: The number of the card is even. &Rule2: The number of the card is less than 5. &Rule3: The suit of the card is hearts. If Diego draws a card from a full deck, what is the probability that he wins? Tell Diego his chances of winning by rounding to one decimal place.
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Example
Finding the Probability of a Compound Event When Rolling a Die
Consider the experiment of rolling a regular six-sided die.
It is also possible to consider the event that results from the union of A and B, written as A or B
. Find the probability that the event A or B
occurs. Round the probability to one decimal place.
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Hint
Which are the favorable outcomes for the event A or B?
Solution
To find the probability of these events, the number of favorable outcomes of each event is divided by the total number of outcomes. Since it is known that there are six possible outcomes for rolling a die, the probabilities can be found by identifying the favorable outcomes for the events.
| Event | Favorable Outcomes | Probability |
|---|---|---|
| A | 1, 2 | 2/6 = 1/3 |
| B | 5, 6 | 2/6 = 1/3 |
A diagram can be used to represent the favorable outcomes for the event A or B
— the outcomes that satisfy either A or B.
Since A and B do not share any favorable outcomes, these events are not overlapping events. As can be seen, there are two favorable outcomes for each event. Therefore the total number of favorable outcomes for event A or B
is 4. To find the probability that the event A or B
occurs, the number of favorable outcomes of this event is divided by the number of possible outcomes.
P(Aor B) = 4/6
ReduceFrac
a/b=.a /2./.b /2.
P(Aor B) = 2/3
CalcQuot
Calculate quotient
P(Aor B) = 0.666666...
RoundDec
Round to 1 decimal place(s)
P(Aor B) ≈ 0.7
It is important to understand that the probability of the event A or B
is equal to the sum of the probabilities of the events A and B.
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Example
A Survey, Probabilities, and a Decision
Ramsha took a survey of her school's cinema club members just to get a sense of their taste before she considers joining. She asked 50 members two questions about movie genres: First, do they like classics? And second, do they like romances? The results are displayed in a two-way frequency table.
| Likes Classics | Does Not Like Classics | |
|---|---|---|
| Likes Romances | 8 | 11 |
| Does Not Like Romances | 20 | 11 |
What is the experimental probability that a randomly chosen member likes classics or romances? Round the probability to two decimal places.
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Hint
What can be done about the value for members that like both classics and romances?
Solution
To find the probability that a randomly chosen club member likes classics or romances, it is important to identify the number those who like classics and the number of those who like romances. To do so, a convenient method is to find the marginal frequencies with the aid of a two-way table. The marginal frequencies show the answers of a single question.
| Likes Classics | Does Not Like Classics | Total | |
|---|---|---|---|
| Likes Romances | 8 | 11 | 8+11= 19 |
| Does Not Like Romances | 20 | 11 | 20+11= 31 |
| Total | 8+20= 28 | 11+11= 22 | 50 |
Referencing the table, Ramsha wrote the following data on her notepad.
It is important to notice that if these marginal frequencies are added, the number of club members that like both classics and romances will be counted twice. That create false results. Therefore, it is critical to subtract this number — 8 — from the sum of 28 and 19 to find the actual number of members that like both movie genres. 28 + 19 - 8 = 39 The experimental probability that a randomly chosen members likes either classics or romances is calculated by dividing 39 from the total number of members surveyed.
It should be noted that this probability — expressed as a fraction — can also be expressed in terms of the probabilities of the different events. That can be done if the combined probability's fraction can be written as a sum of fractions. Take the following interpretation as an example.
P(classics or romances) = 39/50
Rewrite
Rewrite 39 as 28 + 19 - 8
P(classics or romances) = 28 + 19 - 8/50
WriteSumFrac
Write as a sum of fractions
P(classics or romances) = 28/50+19/50 - 8/50
These fractions are the probabilities that a randomly chosen cinema club member likes classics, romances, or both, respectively.
P(classics or romances) = 28/50+19/50 - 8/50
SubstituteValues
Substitute values
P(classics or romances) =P(classics) + P(romances) - P(classics and romances)
Well, it looks like for their movie night people might be clamoring for either a classic or a romance! Ramsha decides this is cool with her, and she'll join after all.
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Discussion
Addition Rule of Probability
For two mutually exclusive events A and B, the probability that A or B occur in one trial is the sum of the individual probability of each event.
P(AorB)=P(A)+P(B)
For example, consider rolling a standard six-sided die. Let A be the event that a 3 is rolled and B be the event that a 4 is rolled. The probability of A or B can be found by adding the individual probabilities. P(3 or4) = P(3)+P(4) = 16 + 16 ⇓ P(3 or4)= 26= 13 The formula above can be generalized to events that are not necessarily mutually exclusive. If events are overlapping, the probability of the common outcomes are counted twice in P(A)+P(B), so an adjustment is needed.
P(AorB)=P(A)+P(B)-P(AandB)
For example, consider rolling a standard six-sided die. Let A be the event that an even number is rolled and B be the event that a prime number is rolled.
| Event | Outcome(s) | Probability |
|---|---|---|
| Even | 2, 4, 6 | P(A)=3/6=1/2 |
| Prime | 2, 3, 5 | P(B)=3/6=1/2 |
| Even and prime | 2 | P(AandB)=1/6 |
Using the formula gives the probability that the result of the roll is even or prime. P(AorB)=P(A)+P(B)-P(AandB) ⇓ P(even or prime)= 12+ 12- 16= 56 This probability can be verified by accounting for the five outcomes that are even or prime: 2, 3, 4, 5, and 6.
Proof
Proving the Addition Rule of ProbabilityFor mutually exclusive events, the Addition Rule of Probability is a postulate.
P(AorB)=P(A)+P(B)
Therefore, no proof will be given for mutually exclusive events. Now, consider non-mutually exclusive events A and B.
In the Venn diagram above, it can be seen part of event A does not overlap event B. That part is labeled a. Similarly, the part of event B that does not overlap event A is labeled b. Furthermore, the overlapping part – also known as the intersection — of both events is labeled c. Notation:& P(A⋂ B)= c Meaning:& The probability of eventsA & and B happening is c. Furthermore, in the diagram it can be also seen that a, b, and c are mutually exclusive. Therefore, the union of event A and event B should be considered.
| Notation | Meaning |
|---|---|
| P(A)= a+c | The probability of A happening is a+c. |
| P(B)= b+c | The probability of B happening is b+c. |
| P(A⋃ B)=a+b+c | The probability of A happening or B happening is a+b+c. |
Finally, the fact that P(A⋃ B)=a+b+c will be used to prove the Addition Rule of Probability for non-mutually exclusive events.
P(A ⋃ B)=a+b+c
IdPropAdd
Identity Property of Addition
P(A ⋃ B)=a+b+c+0
Rewrite 0 as c-c
P(A ⋃ B)=a+b+c+(c-c)
CommutativePropAdd
Commutative Property of Addition
P(A ⋃ B)=a+c+b+c-c
AssociativePropAdd
Associative Property of Addition
P(A ⋃ B)=( a+c)+( b+c)- c
SubstituteValues
Substitute values
P(A ⋃ B)= P(A)+ P(B)- P(A⋂ B)
The rule has been proven for non-mutually exclusive events.
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Example
Calculating the Probability of the Union of Two Events
Consider the following probabilities for events A and B.
- P(A) = 0.13
- P(B) = 0.37
- P(Aand B)= 0.09
What is the value of P(Aor B)?
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Hint
Use the formula for the Addition Rule of Probability.
Solution
The Addition Rule of Probability can be used to find the value of P(Aor B).
P(Aor B) = P(A) + P(B) - P(Aand B)
The given values can be substituted in the formula to find the desired probability.
P(Aor B) = P(A) + P(B) - P(Aand B)
SubstituteValues
Substitute values
P(Aor B) = 0.13 + 0.37 - 0.09
AddSubTerms
Add and subtract terms
P(Aor B) = 0.41
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Pop Quiz
Practice Calculating Probabilities of the Union of Two Events
Use the Addition Rule of Probability to answer each question. Write each probability rounded to two decimal places.
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Example
Calculating the Probability of the Union of Three Overlapping Events
As shown in the following diagram, A, B, and C are three overlapping events.
Write an expression for P(A or B or C).
Answer
P(A or BorC) = P(A) + P(B) + P(C)-P(A and B) - P(B and C) - P(A and C) + P(A and BandC)
Hint
Use a Venn diagram to construct the expression.
Solution
To find an expression for this probability, it is convenient to start with the known Addition Rule of Probability for two overlapping events.
P(AorB) = P(A) + P(B) - P(AandB)
This expression can be extended for three events by adding the probabilities of the three events and subtracting the intersections of two of those events.
P(A) + P(B) + P(C) - P(A and B)-P(B and C) -P(A and C)
That expression might appear a bit muddled. To illustrate what happens when these quantities are added and subtracted, a Venn diagram can be used. The numbers on each section illustrate how many times the probability associated with that section has been added to the total.
It can be seen that P(A and BandC) is not added in the total. Therefore, it is needed to add this probability to complete the expression. P(A or BorC) = P(A) + P(B) + P(C) -P(A and B)-P(B and C) -P(A and C) + P(A and BandC) This expression can be illustrated using a Venn diagram.
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Closure
Probability of Winning a Card Game with Multiple Winning Conditions
The challenge presented at the beginning of this lesson asked for the probability that Diego draws a card that satisfies either of these events. &Rule1: The number of the card is even. &Rule2: The number of the card is less than 5. &Rule3: The suit of the card is hearts. It can be seen that these three events are overlapping events because there are cards that satisfy more than one event at a time. Previously, an expression for P(A or B or C) was found. This expression can be used to find the probability that Diego wins. P(A or B or C) = P(A) + P(B) + P(C) -P(A and B)-P(B and C) -P(A and C) + P(A and BandC) A letter can be associated to each of the events described. &EventA: The number of the card is even. &EventB: The number of the card is less than 5. &EventC: The suit of the card is hearts. A standard deck of cards has 52 cards. Knowing this, it is possible to find the probability of each event — including compound events — by considering the favorable outcomes. Note that aces will be considered as a number 1, while face cards will not be assigned a number value in this game.
| Event | Favorable Outcomes | Probability |
|---|---|---|
| A | 2♣, 4♣, 6♣, 8♣, 10♣, 2♠, 4♠, 6♠, 8♠, 10♠, 2◊, 4◊, 6◊, 8◊, 10◊, 2♡, 4♡, 6♡, 8♡, 10♡ | 20/52 |
| B | A♣, 2♣, 3♣, 4♣, A♠, 2♠, 3♠, 4♠, A◊, 2◊, 3◊, 4◊, A♡, 2♡, 3♡, 4♡ | 16/52 |
| C | A♡, 2♡, 3♡, 4♡, 5♡, 6♡, 7♡, 8♡, 9♡, 10♡, J♡, Q♡, K♡ | 13/52 |
| A andB | 2♣, 4♣, 2♠, 4♠, 2◊, 4◊, 2♡, 4♡ | 8/52 |
| A andC | 2♡, 4♡, 6♡, 8♡, 10♡ | 5/52 |
| B andC | A♡, 2♡, 3♡, 4♡ | 4/52 |
| A andB andC | 2♡, 4♡ | 2/52 |
Now the probabilities from the table can be substituted into the expression for P(A orB orC).
P(A orB orC) = P(A) + P(B) + P(C) -P(A and B)-P(B and C) -P(A and C) +P(A and BandC)
▼
Evaluate
SubstituteValues
Substitute values
P(A orB orC) = 20/52 + 16/52 + 13/52 - 8/52 - 5/52 - 4/52 + 2/52
AddFrac
Add fractions
P(A orB orC) = 34/52
ReduceFrac
a/b=.a /2./.b /2.
P(A orB orC) = 17/26
CalcQuot
Calculate quotient
P(A orB orC) = 0.653846 ...
RoundDec
Round to 1 decimal place(s)
P(A orB orC) ≈ 0.7
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Addition Rule in the Uniform Probability Model
Exercise 3.1
File:Wheeloffortune.png
Find the probability of a contestant winning the game if either event A or B pays out a prize.
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We want to determine the probability of the compound event P(Aor B). To do this we can use the Addition Rule of Probability. P(Aor B)=P(A)+P(B) -P(A and B) Now let's calculate the required probabilities appearing on the right-hand side of the equation. Recall that a prime number is any number that is greater than 1 that is only divisible by 1 or itself. Below all of the prime numbers in the sample space are indicated in red. S = { 1,2,3,4,5,6,7,8,9,10,11 12,13,14,15,16,17,18,19,20 } Now we will indicate all the multiples of 7 in blue. S = { 1,2,3,4,5,6,7,8,9,10,11 12,13,14,15,16,17,18,19,20 } We have 8 prime numbers and 2 multiples of 7. Additionally, 7 is both a prime number and a multiple of 7. With this information we can find the needed probabilities. P(A) &= 8/20 P(B) &= 2/20 P(A and B) &= 1/20 Now we can use the Addition Rule of Probability to calculate the probability of either A or B occurring.
Let's start by marking all of the numbers in the sample space that contain a 2.
S = { 1,2,3,4,5,6,7,8,9,10,11
12,13,14,15,16,17,18,19,20 }
Next, we will mark all of the numbers containing a 0.
S = { 1,2,3,4,5,6,7,8,9,10,11
12,13,14,15,16,17,18,19,20 }
We have 3 numbers containing a 2, and 2 numbers containing a 0. Additionally, 20 contains both a 2 and a 0. With this information we can find the needed probabilities.
P(A) &= 3/20
P(B) &= 2/20
P(A and B) &= 1/20
Now let's calculate the probability of either A or B occurring using the Addition Rule of Probability.
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Addition Rule in the Uniform Probability Model
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