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{{ printedBook.courseTrack.name }} {{ printedBook.name }} Probability measures the likelihood that something will occur. It can take any value from $0$ to $1,$ inclusive. When it is certain that the situation **will not** occur, the probability is $0.$ Further, when it is certain that the situation **will** occur, the probability is $1.$

For example, suppose a standard six-sided die is thrown. This action can be called a *probability experiment* because it has various possible *outcomes.* Specifically, a
$1,2,3,4,5,or6$
can be rolled. The set of all possible outcomes is called the *sample space.* One way to represent the sample space of an experiment is with a tree diagram, which is especially useful when an event contains more than one outcome. The sample space of the stated experiment is shown.

Sometimes a specific outcome — or *event* — is ideal or favorable. The probability of an event can be determined using the following ratio. $number of all possible outcomesnumber of favorable outcomes $
Suppose the roller wishes that a $4$ is rolled. Because one of the six sides of a standard die is a $4,$ and because there are six possible sides that can be rolled, the probability is

In a computer game, a player is assigned a character and a vehicle at random. The character can either be a human, an owl, a zombie or an alien. The possible vehicles are a bike, a car, a rocket and a submarine. Determine the sample space of all possible assignments of characters and vehicles. When assigned a character and a vehicle, what's the sample space and possible outcomes?

Show Solution

Because a sample space is the set of all possible outcomes of an event, we should determine the possibilities of characters and then vehicles. First, there are $4$ possible characters. Let's draw a tree diagram with the first letter of each character.

The diagram represents the characters human, owl, zombie and alien. Now, each character is combined with each possible vehicle.

Each character can either have a bike, car, rocker or a submarine. Looking at the tree, we can see that there are $16$ possible outcomes. They can be listed as the first letter of the character followed by the first letter of each vehicle. $ HBOBZBAB HCOCZCAC HRORZRAR HSOSZSAS $

Suppose a six-sided die is rolled. Let $A$ be the event that an $odd$ number is rolled.

$P(A)$ is $63 $ because there are three ways to roll an odd number — by rolling a $1,$ $3,$ or $5.$ The possible outcomes that do *not* satisfy the desired outcome form a noteworthy set — the $complement$.

Here, the complement of $A$ (written $A_{′}$) consists of rolling a $2,$ $4,$ or $6.$ Thus, $P(A_{′})=63 .$

The probabilities of all possible outcomes of an experiment add to equal $1.$ It follows that since the outcomes can be grouped into an event $A$ and its complement, $A_{′}.$ Thus, $P(A)+P(A_{′})=1.$ Therefore, one way of calculating $P(A_{′}),$ is to calculate $P(A)$ and subtract it from $1.$

$P(A_{′})=1−P(A)$

Test your luck on the wheel of fortune! Choose any four numbers and spin the wheel. If the marker lands on any of the chosen numbers, a prize is won.

Allister chooses the numbers $4,$ $6,$ $13$ and $19.$ What is the probability she does not win?

Show Solution

To begin, let $W$ be the event that Allister wins. Notice that this will occur if she spins a $4,6,13or19.$ This means, she'll lose if the spinner lands on any other number. Thus, $W_{′}$ is the event that Allister does not win. There are two different ways we can determine $P(W_{′}).$ We could count all of the possible outcomes that would cause her to lose, or we could determine $P(W)$ and subtract that from $1.$ We'll choose the second way. $P(W)=204 =51 $ There are $4$ outcomes that result in a win for Allister and $20$ total possible outcomes. We'll subtract $P(W)$ from $1.$ $P(W_{′}) =1−P(W)=1−51 =54 $ Therefore, the probability that Allister does not win, $P(W_{′}),$ is $54 .$

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