Here are a few recommended readings before getting started with this lesson.
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 than5.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.
$A$ or $B$. Find the probability that the event
$A$ or $B$occurs. Round the probability to one decimal place.
Which are the favorable outcomes for the event $A$ or $B?$
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 |
---|---|---|
$A$ | $1,$ $2$ | $62 =31 $ |
$B$ | $5,$ $6$ | $62 =31 $ |
A diagram can be used to represent the favorable outcomes for the event $A$ or $B$
— the outcomes that satisfy either $A$ or $B.$
$A$ 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.
$ba =b/2a/2 $
Calculate quotient
Round to ${\textstyle 1 \, \ifnumequal{1}{1}{\text{decimal}}{\text{decimals}}}$
$A$ or $B$is equal to the sum of the probabilities of the events $A$ and $B.$
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 can be done about the value for members that like both classics and romances?
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 | $8$ | $11$ | $198+11=$ |
Does Not Like Romances | $20$ | $11$ | $3120+11=$ |
Total | $288+20=$ | $2211+11=$ | $50$ |
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.Rewrite $39$ as $28+19−8$
Write as a sum of fractions
Substitute values
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.
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(3or4)=P(3)+P(4)=61 +61 ⇓P(3or4)=62 =31 $ 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.
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)=63 =21 $ |
Prime | $2,$ $3,$ $5$ | $P(B)=63 =21 $ |
Even and prime | $2$ | $P(AandB)=61 $ |
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)=21 +21 −61 =65 $ This probability can be verified by accounting for the five outcomes that are even or prime: $2,$ $3,$ $4,$ $5,$ and $6.$
For mutually exclusive events, the Addition Rule of Probability is a postulate.
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:Meaning: P(A∩B)=cThe probability of eventsAandBhappening isc. $ 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 $a+b+c.$ |
Identity Property of Addition
Rewrite $0$ as $c−c$
Commutative Property of Addition
Associative Property of Addition
Substitute values
Consider the following probabilities for events $A$ and $B.$
Use the formula for the Addition Rule of Probability.
Substitute values
Add and subtract terms
Use the Addition Rule of Probability to answer each question. Write each probability rounded to two decimal places.
As shown in the following diagram, $A,$ $B,$ and $C$ are three overlapping events.
Write an expression for $P(AorBorC).$
$P(AorBorC)$ $=$ $P(A)$ $+$ $P(B)$ $+$ $P(C)−P(AandB)$ $−$ $P(BandC)$ $−$ $P(AandC)$ $+$ $P(AandBandC)$
Use a Venn diagram to construct the expression.
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 than5.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 $P(AorBorC)$ was found. This expression can be used to find the probability that Diego wins. $P(AorBorC)=P(A)+P(B)+P(C)−P(AandB)−P(BandC)−P(AandC)+P(AandBandC) $ 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 than5.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 |
---|---|---|
$A$ | $2♣,$ $4♣,$ $6♣,$ $8♣,$ $10♣,$ $2♠,$ $4♠,$ $6♠,$ $8♠,$ $10♠,$ $2♢,$ $4♢,$ $6♢,$ $8♢,$ $10♢,$ $2♡,$ $4♡,$ $6♡,$ $8♡,$ $10♡$ | $5220 $ |
$B$ | $A♣,$ $2♣,$ $3♣,$ $4♣,$ $A♠,$ $2♠,$ $3♠,$ $4♠,$ $A♢,$ $2♢,$ $3♢,$ $4♢,$ $A♡,$ $2♡,$ $3♡,$ $4♡$ | $5216 $ |
$C$ | $A♡,$ $2♡,$ $3♡,$ $4♡,$ $5♡,$ $6♡,$ $7♡,$ $8♡,$ $9♡,$ $10♡,$ $J♡,$ $Q♡,$ $K♡$ | $5213 $ |
$AandB$ | $2♣,$ $4♣,$ $2♠,$ $4♠,$ $2♢,$ $4♢,$ $2♡,$ $4♡$ | $528 $ |
$AandC$ | $2♡,$ $4♡,$ $6♡,$ $8♡,$ $10♡$ | $525 $ |
$BandC$ | $A♡,$ $2♡,$ $3♡,$ $4♡$ | $524 $ |
$AandBandC$ | $2♡,$ $4♡$ | $522 $ |
Substitute values
Add fractions
$ba =b/2a/2 $
Calculate quotient
Round to ${\textstyle 1 \, \ifnumequal{1}{1}{\text{decimal}}{\text{decimals}}}$