Learning Outcomes

Describe a sample an are and an easy and compound events in it using typical notationCalculate the probability of an occasion using conventional notationCalculate the probability of two independent events using conventional notationRecognize when two occasions are mutually exclusiveCalculate a conditional probability using standard notation

Probability is the likelihood the a details outcome or event happening. Statisticians and also actuaries use probability to make predictions about events. An actuary that functions for a vehicle insurance firm would, because that example, be interested in exactly how likely a 17 year old male would certainly be to obtain in a automobile accident. Castle would use data from past occasions to make predictions about future events using the characteristics of probabilities, then usage this information to calculate an insurance allowance rate.

You are watching: If the probability of an event is 1 then

In this section, us will discover the meaning of an event, and learn just how to calculation the probability that it’s occurance. Us will likewise practice using typical mathematical notation to calculate and also describe different kinds that probabilities.



Basic Concepts

If you roll a die, pick a map from deck of playing cards, or randomly pick a person and also observe your hair color, we room executing an experiment or procedure. In probability, us look at the likelihood of different outcomes.

We begin with part terminology.


Events and Outcomes

The an outcome of one experiment is referred to as an outcome.An event is any specific outcome or team of outcomes.A simple occasion is an occasion that can not be broken down furtherThe sample space is the set of all feasible simple events.

example

If we roll a traditional 6-sided die, explain the sample an are and some an easy events.


Show Solution

The sample space is the set of all feasible simple events: 1,2,3,4,5,6

Some examples of simple events:

We roll a 1We role a 5

Some link events:

We role a number bigger than 4We role an also number

Basic Probability

Given that all outcomes room equally likely, we can compute the probability of an occasion E utilizing this formula:

P(E)=\frac\textNumber of outcomes equivalent to the occasion E\textTotal number of equally-likely outcomes


examples

If we roll a 6-sided die, calculate

P(rolling a 1)P(rolling a number bigger 보다 4)
Show Solution

Recall that the sample an are is 1,2,3,4,5,6

There is one outcome matching to “rolling a 1,” therefore the probability is \frac16There are two outcomes bigger 보다 a 4, therefore the probability is \frac26=\frac13

Probabilities are basically fractions, and also can be lessened to reduced terms choose fractions.


This video describes this example and also the ahead one in detail.

Let’s speak you have a bag v 20 cherries, 14 sweet and also 6 sour. If you choose a cherry at random, what is the probability the it will certainly be sweet?


Show Solution

There are 20 feasible cherries that can be picked, so the number of possible outcomes is 20. Of these 20 feasible outcomes, 14 space favorable (sweet), for this reason the probability that the cherry will certainly be sweet is \frac1420=\frac710.There is one potential complication to this example, however. It should be assumed that the probability of picking any type of of the cherries is the exact same as the probability that picking any kind of other. This wouldn’t it is in true if (let united state imagine) the sweet cherries are smaller than the tart ones. (The cake cherries would involved hand an ext readily once you sampled from the bag.) allow us save in mind, therefore, that once we assess probabilities in terms of the proportion of favorable to all potential cases, we rely greatly on the presumption of same probability for every outcomes.


Try It

At part random moment, you look at her clock and note the minutes reading.

a. What is probability the minutes analysis is 15?

b. What is the probability the minutes reading is 15 or less?


Cards

A traditional deck of 52 playing cards is composed of 4 suits (hearts, spades, diamonds and clubs). Spades and also clubs room black while hearts and diamonds are red. Each suit contains 13 cards, each of a different rank: one Ace (which in countless games features as both a low card and a high card), cards numbered 2 with 10, a Jack, a Queen and a King.


example

Compute the probability the randomly illustration one map from a deck and getting an Ace.


Show Solution

There are 52 cards in the deck and 4 Aces so P(Ace)=\frac452=\frac113\approx 0.0769

We can additionally think of probabilities as percents: over there is a 7.69% opportunity that a randomly selected card will be one Ace.

Notice the the smallest feasible probability is 0 – if there are no outcomes the correspond v the event. The largest feasible probability is 1 – if all feasible outcomes correspond v the event.


This video clip demonstrates both this example and the ahead cherry example on the page.


Certain and also Impossible events

An impossible event has a probability that 0.A certain event has a probability of 1.The probability of any event need to be 0\le P(E)\le 1

Try It


In the course of this section, if you compute a probability and get an answer that is an unfavorable or higher than 1, you have actually made a mistake and should check your work.

Types that Events

Complementary Events

Now let us research the probability that an occasion does not happen. As in the ahead section, take into consideration the situation of roll a six-sided dice and very first compute the probability of rojo a six: the answer is P(six) =1/6. Now consider the probability that we carry out not role a six: there are 5 outcomes that room not a six, therefore the answer is P(not a six) = \frac56. Notice that

P(\textsix)+P(\textnot a six)=\frac16+\frac56=\frac66=1

This is no a coincidence. Take into consideration a generic situation with n possible outcomes and an occasion E that synchronizes to m of this outcomes. Climate the continuing to be nm outcomes correspond to E not happening, thus

P(\textnotE)=\fracn-mn=\fracnn-\fracmn=1-\fracmn=1-P(E)



Complement of an Event

The complement of an occasion is the occasion “E no happen”

The notation \barE is used for the enhance of occasion E.We can compute the probability that the complement using P\left(\barE\right)=1-P(E)Notice likewise that P(E)=1-P\left(\barE\right)

example

If you traction a random map from a deck of playing cards, what is the probability the is no a heart?


Show Solution

There space 13 mind in the deck, therefore P(\textheart)=\frac1352=\frac14.

The probability that not drawing a heart is the complement: P(\textnot heart)=1-P(\textheart)=1-\frac14=\frac34


This case is explained in the adhering to video.


Try It


Probability of 2 independent events


example

Suppose we flipped a coin and also rolled a die, and also wanted to recognize the probability of gaining a head on the coin and a 6 top top the die.


Show Solution

We can list all feasible outcomes: H1,H2,h2,H4,H5,H6,T1,T2,T3,T4,T5,T6.

Notice there space 2\cdot6=12 full outcomes. The end of these, only 1 is the wanted outcome, so the probability is \frac112.


The prior instance contained two independent events. Obtaining a particular outcome from rolling a die had actually no influence on the outcome from flipping the coin.


Independent Events

Events A and B room independent events if the probability of event B occurring is the exact same whether or not occasion A occurs.


example

Are these events independent?

A fair coin is tossed two times. The two occasions are (1) first toss is a head and also (2) 2nd toss is a head.The two events (1) “It will certainly rain tomorrow in Houston” and (2) “It will certainly rain morning in Galveston” (a city near Houston).You attract a map from a deck, then attract a 2nd card there is no replacing the first.
Show Solution
The probability that a head come up on the second toss is 1/2 nevertheless of whether or no a head come up on the first toss, therefore these events are independent.These occasions are not independent because it is more likely the it will rain in Galveston on days it rain in Houston than on days the does not.The probability that the 2nd card being red counts on whether the an initial card is red or not, for this reason these events are no independent.

When two events are independent, the probability the both emerging is the product that the probabilities that the separation, personal, instance events.


P(A and B) because that independent events

If events A and also B are independent, climate the probability the both A and also B occurring is

P\left(A\text and B\right)=P\left(A\right)\cdotP\left(B\right)

where P(A and also B) is the probability of occasions A and B both occurring, P(A) is the probability of event A occurring, and also P(B) is the probability of occasion B occurring


If girlfriend look back at the coin and die instance from earlier, you have the right to see just how the variety of outcomes of the very first event multiplied by the variety of outcomes in the 2nd event multiply to same the total variety of possible outcomes in the merged event.


example

In your drawer you have actually 10 pairs of socks, 6 that which space white, and 7 tee shirts, 3 that which space white. If girlfriend randomly reach in and pull the end a pair of socks and also a tee shirt, what is the probability both space white?


Show Solution

The probability of picking a white pair the socks is \frac610.

The probability of choosing a white tee shirt is \frac37.

The probability the both gift white is \frac610\cdot\frac37=\frac1870=\frac935


Examples of joint probabilities are questioned in this video.


Try It


The previous instances looked in ~ the probability the both events occurring. Currently we will certainly look in ~ the probability the either event occurring.


example

Suppose us flipped a coin and rolled a die, and wanted to understand the probability of obtaining a head on the coin or a 6 ~ above 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 have the right to see that 7 of the outcomes have actually a head on the coin or a 6 ~ above the dice or both – we usage or inclusively below (these 7 outcomes are H1, H2, h2, H4, H5, H6, T6), for this reason the probability is \frac712. How could we have found this indigenous the individual probabilities?

As we would expect, \frac12 of these outcomes have actually a head, and \frac16 of these outcomes have a 6 top top the die. If we add these, \frac12+\frac16=\frac612+\frac212=\frac812, i m sorry is not the exactly probability. Looking at the outcomes we deserve to see why: the outcome H6 would have actually been count twice, because it includes both a head and also a 6; the probability of both a head and roll a 6 is \frac112.

If us subtract the end this double count, we have actually the exactly probability: \frac812-\frac112=\frac712.


P(A or B)

The probability of one of two people A or B occurring (or both) is

P(A\text or B)=P(A)+P(B)–P(A\text and B)


example

Suppose we draw one map from a typical deck. What is the probability that we obtain a Queen or a King?


Show Solution

There are 4 Queens and 4 kings in the deck, for this reason 8 outcomes corresponding to a Queen or King out of 52 feasible outcomes. For this reason the probability of illustration a Queen or a King is:

P(\textKing or Queen)=\frac852

Note the in this case, there are no cards that space both a Queen and also a King, so P(\textKing and also Queen)=0. Using our probability rule, we can have said:

P(\textKing or Queen)=P(\textKing)+P(\textQueen)-P(\textKing and also Queen)=\frac452+\frac452-0=\frac852


See much more about this example and the vault one in the adhering to video.


In the last example, the occasions were mutually exclusive, for this reason P(A or B) = P(A) + P(B).


Try It


example

Suppose we attract one map from a conventional deck. What is the probability that we acquire a red map or a King?


Show Solution

Half the cards room red, therefore P(\textred)=\frac2652

There are four kings, therefore P(\textKing)=\frac452

There are two red kings, therefore P(\textRed and King)=\frac252

We can then calculate

P(\textRed or King)=P(\textRed)+P(\textKing)-P(\textRed and King)=\frac2652+\frac452-\frac252=\frac2852


Try It

In her drawer you have actually 10 bag of socks, 6 that which space white, and 7 tee shirts, 3 that which room white. If you reach in and also randomly grab a pair that socks and also a tee shirt, what the probability at least one is white?


Example

The table below shows the number of survey topics who have actually received and not got a speeding ticket in the last year, and the shade of their car. Find the probability the a randomly chosen person:

Has a red auto and acquired a speeding ticketHas a red car or got a speeding ticket.
Speeding ticketNo speeding ticketTotal
Red car15135150
Not red car45470515
Total60605665

Show Solution

We have the right to see that 15 human being of the 665 surveyed had actually both a red car and also got a speeding ticket, so the probability is \frac15665\approx0.0226.

Notice that having a red car and getting a speeding ticket space not live independence events, therefore the probability the both the them occurring is not merely the product of probabilities of every one occurring.

We could answer this question by simply adding up the numbers: 15 people with red cars and speeding tickets + 135 through red cars yet no ticket + 45 v a ticket yet no red auto = 195 people. Therefore the probability is \frac195665\approx0.2932.

We likewise could have found this probability by:

P(had a red car) + P(got a speeding ticket) – P(had a red car and also got a speeding ticket)

= \frac150665+\frac60665-\frac15665=\frac195665.


This table example is in-depth in the adhering to explanatory video.


Try It


Conditional Probability

In the previous ar we computed the probabilities of occasions that were independent of every other. We saw that getting a details outcome native rolling a die had actually no affect on the outcome from flipping a coin, even though us were computer a probability based on doing them in ~ the same time.

In this section, we will consider events that are dependent on each other, referred to as conditional probabilities.


Conditional Probability

The probability the occasion B occurs, provided that event A has happened, is represented as

P(B | A)

This is check out as “the probability of B offered A


For example, if you draw a card from a deck, climate the sample space for the next card drawn has changed, since you are now working v a deck of 51 cards. In the following instance we will present you just how the computations for events like this are various from the computations we did in the last section.


example

What is the probability that 2 cards drawn at random from a deck of playing cards will both be aces?


Show Solution

It could seem that you could use the formula because that the probability of 2 independent events and also simply multiply \frac452\cdot\frac452=\frac1169. This would certainly be incorrect, however, due to the fact that the two events are no independent. If the first card attracted is an ace, climate the probability the the 2nd card is additionally an ace would be lower due to the fact that there would only be three aces left in the deck.

Once the very first card liked is an ace, the probability the the second card favored is likewise an ace is dubbed the conditional probability of drawing an ace. In this instance the “condition” is that the an initial card is one ace. Symbolically, we compose this as:

P(ace on second draw | one ace ~ above the an initial draw).

The upright bar “|” is review as “given,” therefore the above expression is quick for “The probability the an ace is attracted on the second draw provided that one ace was attracted on the first draw.” What is this probability? after an ace is attracted on the very first draw, there room 3 aces the end of 51 complete cards left. This way that the conditional probability of illustration an ace ~ one ace has currently been attracted is \frac351=\frac117.

Thus, the probability that both cards gift aces is \frac452\cdot\frac351=\frac122652=\frac1221.


Since we recognize the person has actually a red car, we are only considering the 150 people in the very first row that the table. Of those, 15 have actually a speeding ticket, for this reason P(ticket | red car) = \frac15150=\frac110=0.1Since we understand the person has a speeding ticket, we are just considering the 60 world in the very first column that the table. That those, 15 have a red car, for this reason P(red automobile | ticket) = \frac1560=\frac14=0.25.

Notice native the last instance that P(B | A) is not same to P(A | B).


These kinds of conditional probabilities are what insurance companies use to determine your insurance allowance rates. Lock look in ~ the conditional probability of you having actually accident, given your age, her car, your vehicle color, your driving history, etc., and price your policy based on that likelihood.

View much more about conditional probability in the following video.


Example

If you attract two cards native a deck, what is the probability the you will acquire the Ace that Diamonds and also a black card?


Show Solution

You can meet this condition by having instance A or situation B, together follows:

Case A) girlfriend can get the Ace that Diamonds an initial and climate a black color card or

Case B) friend can get a black color card very first and then the Ace that Diamonds.

Let’s calculate the probability of situation A. The probability that the an initial card is the Ace the Diamonds is \frac152. The probability that the 2nd card is black offered that the an initial card is the Ace the Diamonds is \frac2651 since 26 that the remaining 51 cards room black. The probability is as such \frac152\cdot\frac2651=\frac1102.

Now for case B: the probability that the an initial card is black is \frac2652=\frac12. The probability the the second card is the Ace the Diamonds given that the an initial card is black color is \frac151. The probability of situation B is therefore \frac12\cdot\frac151=\frac1102, the very same as the probability of instance 1.

Recall the the probability that A or B is P(A) + P(B) – P(A and also B). In this problem, P(A and B) = 0 because the an initial card cannot be the Ace that Diamonds and be a black card. Therefore, the probability of case A or situation B is \frac1101+\frac1101=\frac2101. The probability that you will gain the Ace the Diamonds and also a black color card when illustration two cards from a deck is \frac2101.


These 2 playing card scenarios are debated further in the complying with video.

See more: 100 Ways To Say “ I Love You In Aztec Language, Useful Phrases In Nahuatl


Try It


Since we recognize the test result was positive, we’re restricted to the 75 females in the an initial column, of i m sorry 5 were no pregnant. P(not pregnant | positive test result) = \frac575\approx0.067.Since we know the woman is no pregnant, us are limited to the 19 ladies in the second row, of i beg your pardon 5 had a hopeful test. P(positive test result | not pregnant) = \frac519\approx0.263

The second result is what is usually called a false positive: A positive an outcome when the mrs is no actually pregnant.