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Lê Hải Hưng đang tìm kiếm từ khóa Which operation would be used to find the probability of two events joined by or được Cập Nhật vào lúc : 2022-12-01 21:00:13 . Với phương châm chia sẻ Mẹo về trong nội dung bài viết một cách Chi Tiết 2022. Nếu sau khi Read Post vẫn ko hiểu thì hoàn toàn có thể lại Comment ở cuối bài để Admin lý giải và hướng dẫn lại nha.The conditional probability of an sự kiện B is the probability that the sự kiện will occur given the knowledge that an sự kiện A has already occurred. This probability is written P(B|A), notation for the probability of B given A. In the case where events A and B are independent (where sự kiện A has no effect on the probability of sự kiện B), the conditional probability of sự kiện B given sự kiện A is simply the probability of sự kiện B, that is P(B).
If events A and B are not independent, then the probability of the intersection of A and B (the probability that both events occur) is defined by
P(A and B) = P(A)P(B|A).
- Basic ConceptsEvents and OutcomesBasic ProbabilityCertain and Impossible eventsTypes of EventsComplementary EventsComplement of an EventProbability of two independent eventsIndependent EventsP(A and B) for independent eventsConditional ProbabilityConditional ProbabilityConditional Probability FormulaWhat is used to find the probability of two events joined by OR?What is the and/or rule in probability?What is the or symbol in probability?
From this definition, the conditional probability P(B|A) is easily obtained by dividing by P(A):
Note: This expression is only valid when P(A) is greater than 0.
Examples
In a card trò chơi, suppose a player needs to draw two cards of the same suit in order to win. Of the 52 cards, there are 13 cards in each suit. Suppose first the player draws a heart. Now the player wishes to draw a second heart. Since one heart has already been chosen, there are now 12 hearts remaining in a deck of 51 cards. So the conditional probability P(Draw second heart|First card a heart) = 12/51.Suppose an individual applying to a college determines that he has an 80% chance of being accepted, and he knows that dormitory housing will only be provided for 60% of all of the accepted students. The chance of the student being accepted and receiving dormitory housing is defined by
P(Accepted and Dormitory Housing) = P(Dormitory Housing|Accepted)P(Accepted) = (0.60)*(0.80) = 0.48.
Consider the college applicant who has determined that he has 0.80 probability of acceptance and that only 60% of the accepted students will receive dormitory housing. Of the accepted students who receive dormitory housing, 80% will have least one roommate. The probability of being accepted and receiving dormitory housing and having no roommates is calculated by:
P(Accepted and Dormitory Housing and No Roommates) = P(Accepted)P(Dormitory Housing|Accepted)P(No Roomates|Dormitory Housing and Accepted) = (0.80)*(0.60)*(0.20) = 0.096. The student has about a 10% chance of receiving a single room the college.
Example
Suppose a voter poll is taken in three states. In state A, 50% of voters support the liberal candidate, in state B, 60% of the voters support the liberal candidate, and in state C, 35% of the voters support the liberal candidate. Of the total population of the three states, 40% live in state A, 25% live in state B, and 35% live in state C. Given that a voter supports the liberal candidate, what is the probability that she lives in state B?By Bayes's formula,
P(Voter lives in state B|Voter supports liberal candidate) = P(Voter supports liberal candidate|Voter lives in state B)P(Voter lives in state B)/ (P(Voter supports lib. cand.|Voter lives in state A)P(Voter lives in state A) + P(Voter supports lib. cand.|Voter lives in state B)P(Voter lives in state B) + P(Voter supports lib. cand.|Voter lives in state C)P(Voter lives in state C)) = (0.60)*(0.25)/((0.50)*(0.40) + (0.60)*(0.25) + (0.35)*(0.35)) = (0.15)/(0.20 + 0.15 + 0.1225) = 0.15/0.4725 = 0.3175.The probability that the voter lives in state B is approximately 0.32.For some more definitions and examples, see the probability index in Valerie J. Easton and John H. McColl's Statistics Glossary v1.1.
Probability is the likelihood of a particular outcome or sự kiện happening. Statisticians and actuaries use probability to make predictions about events. An actuary that works for a car insurance company would, for example, be interested in how likely a 17 year old male would be to get in a car accident. They would use data from past events to make predictions about future events using the characteristics of probabilities, then use this information to calculate an insurance rate.
In this section, we will explore the definition of an sự kiện, and learn how to calculate the probability of it’s occurance. We will also practice using standard mathematical notation to calculate and describe different kinds of probabilities.
Basic Concepts
If you roll a die, pick a card from deck of playing cards, or randomly select a person and observe their hair color, we are executing an experiment or procedure. In probability, we look the likelihood of different outcomes.
We begin with some terminology.
Events and Outcomes
- The result of an experiment is called an outcome.An sự kiện is any particular outcome or group of outcomes.A simple sự kiện is an sự kiện that cannot be broken down furtherThe sample space is the set of all possible simple events.
example
If we roll a standard 6-sided die, describe the sample space and some simple events.
Show Solution
The sample space is the set of all possible simple events: 1,2,3,4,5,6
Some examples of simple events:
- We roll a 1We roll a 5
Some compound events:
- We roll a number bigger than 4We roll an even number
Basic Probability
Given that all outcomes are equally likely, we can compute the probability of an sự kiện E using this formula:
[latex]P(E)=fractextNumber of outcomes corresponding to the sự kiện EtextTotal number of equally-likely outcomes[/latex]
examples
If we roll a 6-sided die, calculate
P(rolling a 1)P(rolling a number bigger than 4)Show Solution
Recall that the sample space is 1,2,3,4,5,6
There is one outcome corresponding to “rolling a 1,” so the probability is [latex]frac16[/latex]There are two outcomes bigger than a 4, so the probability is [latex]frac26=frac13[/latex]Probabilities are essentially fractions, and can be reduced to lower terms like fractions.
This video describes this example and the previous one in detail.
Let’s say you have a bag with 20 cherries, 14 sweet and 6 sour. If you pick a cherry random, what is the probability that it will be sweet?
Show Solution
There are 20 possible cherries that could be picked, so the number of possible outcomes is 20. Of these 20 possible outcomes, 14 are favorable (sweet), so the probability that the cherry will be sweet is [latex]frac1420=frac710[/latex].
There is one potential complication to this example, however. It must be assumed that the probability of picking any of the cherries is the same as the probability of picking any other. This wouldn’t be true if (let us imagine) the sweet cherries are smaller than the sour ones. (The sour cherries would come to hand more readily when you sampled from the bag.) Let us keep in mind, therefore, that when we assess probabilities in terms of the ratio of favorable to all potential cases, we rely heavily on the assumption of equal probability for all outcomes.
Try It
At some random moment, you look your clock and note the minutes reading.
a. What is probability the minutes reading is 15?
b. What is the probability the minutes reading is 15 or less?
Cards
A standard deck of 52 playing cards consists of four suits (hearts, spades, diamonds and clubs). Spades and clubs are black while hearts and diamonds are red. Each suit contains 13 cards, each of a different rank: an Ace (which in many games functions as both a low card and a high card), cards numbered 2 through 10, a Jack, a Queen and a King.
example
Compute the probability of randomly drawing one card from a deck and getting an Ace.
Show Solution
There are 52 cards in the deck and 4 Aces so [latex]P(Ace)=frac452=frac113approx 0.0769[/latex]
We can also think of probabilities as percents: There is a 7.69% chance that a randomly selected card will be an Ace.
Notice that the smallest possible probability is 0 – if there are no outcomes that correspond with the sự kiện. The largest possible probability is 1 – if all possible outcomes correspond with the sự kiện.
This video demonstrates both this example and the previous cherry example on the page.
Certain and Impossible events
- An impossible sự kiện has a probability of 0.A certain sự kiện has a probability of 1.The probability of any sự kiện must be [latex]0le P(E)le 1[/latex]
Try It
In the course of this section, if you compute a probability and get an answer that is negative or greater than 1, you have made a mistake and should check your work.
Types of Events
Complementary Events
Now let us examine the probability that an sự kiện does not happen. As in the previous section, consider the situation of rolling a six-sided die and first compute the probability of rolling a six: the answer is P(six) =1/6. Now consider the probability that we do not roll a six: there are 5 outcomes that are not a six, so the answer is P(not a six) = [latex]frac56[/latex]. Notice that
[latex]P(textsix)+P(textnot a six)=frac16+frac56=frac66=1[/latex]
This is not a coincidence. Consider a generic situation with n possible outcomes and an sự kiện E that corresponds to m of these outcomes. Then the remaining n – m outcomes correspond to E not happening, thus
[latex]P(textnotE)=fracn-mn=fracnn-fracmn=1-fracmn=1-P(E)[/latex]
Complement of an Event
The complement of an sự kiện is the sự kiện “E doesn’t happen”
- The notation [latex]barE[/latex] is used for the complement of sự kiện E.We can compute the probability of the complement using [latex]Pleft(barEright)=1-P(E)[/latex]Notice also that [latex]P(E)=1-Pleft(barEright)[/latex]
example
If you pull a random card from a deck of playing cards, what is the probability it is not a heart?
Show Solution
There are 13 hearts in the deck, so [latex]P(textheart)=frac1352=frac14[/latex].
The probability of not drawing a heart is the complement: [latex]P(textnot heart)=1-P(textheart)=1-frac14=frac34[/latex]
This situation is explained in the following video.
Try It
Probability of two independent events
example
Suppose we flipped a coin and rolled a die, and wanted to know the probability of getting a head on the coin and a 6 on the die.
Show Solution
We could list all possible outcomes: H1,H2,h2,H4,H5,H6,T1,T2,T3,T4,T5,T6.
Notice there are [latex]2cdot6=12[/latex] total outcomes. Out of these, only 1 is the desired outcome, so the probability is [latex]frac112[/latex].
The prior example contained two independent events. Getting a certain outcome from rolling a die had no influence on the outcome from flipping the coin.
Independent Events
Events A and B are independent events if the probability of Event B occurring is the same whether or not Event A occurs.
example
Are these events independent?
A fair coin is tossed two times. The two events are (1) first toss is a head and (2) second toss is a head.The two events (1) “It will rain tomorrow in Houston” and (2) “It will rain tomorrow in Galveston” (a city near Houston).You draw a card from a deck, then draw a second card without replacing the first.Show Solution
The probability that a head comes up on the second toss is 1/2 regardless of whether or not a head came up on the first toss, so these events are independent.These events are not independent because it is more likely that it will rain in Galveston on days it rains in Houston than on days it does not.The probability of the second card being red depends on whether the first card is red or not, so these events are not independent.When two events are independent, the probability of both occurring is the product of the probabilities of the individual events.
P(A and B) for independent events
If events A and B are independent, then the probability of both A and B occurring is
[latex]Pleft(Atext and Bright)=Pleft(Aright)cdotPleft(Bright)[/latex]
where P(A and B) is the probability of events A and B both occurring, P(A) is the probability of sự kiện A occurring, and P(B) is the probability of sự kiện B occurring
If you look back the coin and die example from earlier, you can see how the number of outcomes of the first sự kiện multiplied by the number of outcomes in the second sự kiện multiplied to equal the total number of possible outcomes in the combined sự kiện.
example
In your drawer you have 10 pairs of socks, 6 of which are white, and 7 tee shirts, 3 of which are white. If you randomly reach in and pull out a pair of socks and a tee shirt, what is the probability both are white?
Show Solution
The probability of choosing a white pair of socks is [latex]frac610[/latex].
The probability of choosing a white tee shirt is [latex]frac37[/latex].
The probability of both being white is [latex]frac610cdotfrac37=frac1870=frac935[/latex]
Examples of joint probabilities are discussed in this video.
Try It
The previous examples looked the probability of both events occurring. Now we will look the probability of either sự kiện occurring.
example
Suppose we flipped a coin and rolled a die, and wanted to know the probability of getting a head on the coin or a 6 on the die.
Show Solution
Here, there are still 12 possible outcomes: H1,H2,h2,H4,H5,H6,T1,T2,T3,T4,T5,T6
By simply counting, we can see that 7 of the outcomes have a head on the coin or a 6 on the die or both – we use or inclusively here (these 7 outcomes are H1, H2, h2, H4, H5, H6, T6), so the probability is [latex]frac712[/latex]. How could we have found this from the individual probabilities?
As we would expect, [latex]frac12[/latex] of these outcomes have a head, and [latex]frac16[/latex] of these outcomes have a 6 on the die. If we add these, [latex]frac12+frac16=frac612+frac212=frac812[/latex], which is not the correct probability. Looking the outcomes we can see why: the outcome H6 would have been counted twice, since it contains both a head and a 6; the probability of both a head and rolling a 6 is [latex]frac112[/latex].
If we subtract out this double count, we have the correct probability: [latex]frac812-frac112=frac712[/latex].
P(A or B)
The probability of either A or B occurring (or both) is
[latex]P(Atext or B)=P(A)+P(B)–P(Atext and B)[/latex]
example
Suppose we draw one card from a standard deck. What is the probability that we get a Queen or a King?
Show Solution
There are 4 Queens and 4 Kings in the deck, hence 8 outcomes corresponding to a Queen or King out of 52 possible outcomes. Thus the probability of drawing a Queen or a King is:
[latex]P(textKing or Queen)=frac852[/latex]
Note that in this case, there are no cards that are both a Queen and a King, so [latex]P(textKing and Queen)=0[/latex]. Using our probability rule, we could have said:
[latex]P(textKing or Queen)=P(textKing)+P(textQueen)-P(textKing and Queen)=frac452+frac452-0=frac852[/latex]
See more about this example and the previous one in the following video.
In the last example, the events were mutually exclusive, so P(A or B) = P(A) + P(B).
Try It
example
Suppose we draw one card from a standard deck. What is the probability that we get a red card or a King?
Show Solution
Half the cards are red, so [latex]P(textred)=frac2652[/latex]
There are four kings, so [latex]P(textKing)=frac452[/latex]
There are two red kings, so [latex]P(textRed and King)=frac252[/latex]
We can then calculate
[latex]P(textRed or King)=P(textRed)+P(textKing)-P(textRed and King)=frac2652+frac452-frac252=frac2852[/latex]
Try It
In your drawer you have 10 pairs of socks, 6 of which are white, and 7 tee shirts, 3 of which are white. If you reach in and randomly grab a pair of socks and a tee shirt, what the probability least one is white?
Example
The table below shows the number of survey subjects who have received and not received a speeding ticket in the last year, and the color of their car. Find the probability that a randomly chosen person:
Has a red car and got a speeding ticketHas a red car or got a speeding ticket.Speeding ticketNo speeding ticketTotalRed car15135150Not red car45470515Total60605665Show Solution
We can see that 15 people of the 665 surveyed had both a red car and got a speeding ticket, so the probability is [latex]frac15665approx0.0226[/latex].
Notice that having a red car and getting a speeding ticket are not independent events, so the probability of both of them occurring is not simply the product of probabilities of each one occurring.
We could answer this question by simply adding up the numbers: 15 people with red cars and speeding tickets + 135 with red cars but no ticket + 45 with a ticket but no red car = 195 people. So the probability is [latex]frac195665approx0.2932[/latex].
We also could have found this probability by:
P(had a red car) + P(got a speeding ticket) – P(had a red car and got a speeding ticket)
= [latex]frac150665+frac60665-frac15665=frac195665[/latex].
This table example is detailed in the following explanatory video.
Try It
Conditional Probability
In the previous section we computed the probabilities of events that were independent of each other. We saw that getting a certain outcome from rolling a die had no influence on the outcome from flipping a coin, even though we were computing a probability based on doing them the same time.
In this section, we will consider events that are dependent on each other, called conditional probabilities.
Conditional Probability
The probability the sự kiện B occurs, given that sự kiện A has happened, is represented as
P(B | A)
This is read as “the probability of B given A”
For example, if you draw a card from a deck, then the sample space for the next card drawn has changed, because you are now working with a deck of 51 cards. In the following example we will show you how the computations for events like this are different from the computations we did in the last section.
example
What is the probability that two cards drawn random from a deck of playing cards will both be aces?
Show Solution
It might seem that you could use the formula for the probability of two independent events and simply multiply [latex]frac452cdotfrac452=frac1169[/latex]. This would be incorrect, however, because the two events are not independent. If the first card drawn is an ace, then the probability that the second card is also an ace would be lower because there would only be three aces left in the deck.
Once the first card chosen is an ace, the probability that the second card chosen is also an ace is called the conditional probability of drawing an ace. In this case the “condition” is that the first card is an ace. Symbolically, we write this as:
P(ace on second draw | an ace on the first draw).
The vertical bar “|” is read as “given,” so the above expression is short for “The probability that an ace is drawn on the second draw given that an ace was drawn on the first draw.” What is this probability? After an ace is drawn on the first draw, there are 3 aces out of 51 total cards left. This means that the conditional probability of drawing an ace after one ace has already been drawn is [latex]frac351=frac117[/latex].
Thus, the probability of both cards being aces is [latex]frac452cdotfrac351=frac122652=frac1221[/latex].
Conditional Probability Formula
If Events A and B are not independent, then
P(A and B) = P(A) · P(B | A)
example
If you pull 2 cards out of a deck, what is the probability that both are spades?
Show Solution
The probability that the first card is a spade is [latex]frac1352[/latex].
The probability that the second card is a spade, given the first was a spade, is [latex]frac1251[/latex], since there is one less spade in the deck, and one less total cards.
The probability that both cards are spades is [latex]frac1352cdotfrac1251=frac1562652approx0.0588[/latex]
Try It
Example
The table below shows the number of survey subjects who have received and not received a speeding ticket in the last year, and the color of their car. Find the probability that a randomly chosen person:
has a speeding ticket given they have a red carhas a red car given they have a speeding ticketSpeeding ticketNo speeding ticketTotalRed car15135150Not red car45470515Total60605665Show Solution
Since we know the person has a red car, we are only considering the 150 people in the first row of the table. Of those, 15 have a speeding ticket, so P(ticket | red car) = [latex]frac15150=frac110=0.1[/latex]Since we know the person has a speeding ticket, we are only considering the 60 people in the first column of the table. Of those, 15 have a red car, so P(red car | ticket) = [latex]frac1560=frac14=0.25[/latex].Notice from the last example that P(B | A) is not equal to P(A | B).
These kinds of conditional probabilities are what insurance companies use to determine your insurance rates. They look the conditional probability of you having accident, given your age, your car, your car color, your driving history, etc., and price your policy based on that likelihood.
View more about conditional probability in the following video.
Example
If you draw two cards from a deck, what is the probability that you will get the Ace of Diamonds and a black card?
Show Solution
You can satisfy this condition by having Case A or Case B, as follows:
Case A) you can get the Ace of Diamonds first and then a black card or
Case B) you can get a black card first and then the Ace of Diamonds.
Let’s calculate the probability of Case A. The probability that the first card is the Ace of Diamonds is [latex]frac152[/latex]. The probability that the second card is black given that the first card is the Ace of Diamonds is [latex]frac2651[/latex] because 26 of the remaining 51 cards are black. The probability is therefore [latex]frac152cdotfrac2651=frac1102[/latex].
Now for Case B: the probability that the first card is black is [latex]frac2652=frac12[/latex]. The probability that the second card is the Ace of Diamonds given that the first card is black is [latex]frac151[/latex]. The probability of Case B is therefore [latex]frac12cdotfrac151=frac1102[/latex], the same as the probability of Case 1.
Recall that the probability of A or B is P(A) + P(B) – P(A and B). In this problem, P(A and B) = 0 since the first card cannot be the Ace of Diamonds and be a black card. Therefore, the probability of Case A or Case B is [latex]frac1101+frac1101=frac2101[/latex]. The probability that you will get the Ace of Diamonds and a black card when drawing two cards from a deck is [latex]frac2101[/latex].
These two playing card scenarios are discussed further in the following video.
Try It
Example
A home pregnancy test was given to women, then pregnancy was verified through blood tests. The following table shows the home pregnancy test results.
Find
P(not pregnant | positive test result)P(positive test result | not pregnant)Positive testNegative testTotalPregnant70474Not Pregnant51419Total751893Show Solution
Since we know the test result was positive, we’re limited to the 75 women in the first column, of which 5 were not pregnant. P(not pregnant | positive test result) = [latex]frac575approx0.067[/latex].Since we know the woman is not pregnant, we are limited to the 19 women in the second row, of which 5 had a positive test. P(positive test result | not pregnant) = [latex]frac519approx0.263[/latex]The second result is what is usually called a false positive: A positive result when the woman is not actually pregnant.
What is used to find the probability of two events joined by OR?
To find the either/or probability of non-overlapping events, you first find the P(A) probability of sự kiện A and then P(B) probability of sự kiện B and use the formula P(A) + P(B). In other words, find the probability of each sự kiện as a fraction and then add the two fractions together.What is the and/or rule in probability?
In probability, there's a very important distinction between the words and and or. And means that the outcome has to satisfy both conditions the same time. Or means that the outcome has to satisfy one condition, or the other condition, or both the same time.What is the or symbol in probability?
Probability theory indicates the probability of either sự kiện A or sự kiện B occurring (“or” in this case means one or the other or both). In particular, the pdf of the standard normal distribution is denoted by φ(z), and its cdf by Φ(z). Tải thêm tài liệu liên quan đến nội dung bài viết Which operation would be used to find the probability of two events joined by or