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{{ printedBook.courseTrack.name }} {{ printedBook.name }} In this lesson, some basics concepts of probability will be introduced and connected to real-life situations.
### Catch-Up and Review

**Here are a few recommended readings before getting started with this lesson.**

Magdalena is in a target shooting competition. For her next test, the target has three zones and she can shoot three arrows.
*all* the possible results that Magdalena can get on the test.

Assuming that Magdalena always hit the target, list

On a late-night game show, there are three closed doors. Behind one of the doors is a car, and there are sheep behind the other two. The host invites Tearrik to win the car by choosing only one door.

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b What are the chances of choosing a door with a sheep? Round the answer to two decimal places.

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Below, some basic definitions of probability are examined.

In probability, an experiment is an action that can be repeated infinitely many times and has various results called outcomes. In other words, an experiment is a process used to determine the probability of an event occurring in the future. For example, rolling a die can be considered an experiment.
*trial*. For example, if a die is thrown $100$ times, each throw can be considered a trial. An experiment is usually performed to estimate the theoretical probability of a particular outcome.

Experiments are repeated several times to collect some data, and each repetition is called a

An outcome is a possible result of a probability experiment. For example, when rolling a six-sided die, getting a $3$ is one possible outcome.

Note that when performing an experiment, each possible outcome is unique — that is, on each trial, only one outcome will occur.An event is a combination of one or more specific outcomes. For example, when playing cards, an event might be drawing a spade or a heart. For this event, one possible outcome is drawing the $A♠$ or drawing the $7♡.$

However, these are not the only outcomes of this event. All the possible outcomes that satisfy the event are listed below.

$Outcomes:A♠,2♠,3♠,4♠,5♠,6♠,7♠8♠,9♠,10♠,J♠,Q♠,K♠A♡,2♡,3♡,4♡,5♡,6♡,7♡8♡,9♡,10♡,J♡,Q♡,K♡ $The sample space of an experiment is the set of all possible outcomes. For example, when flipping a coin, there are two possible outcomes: heads, H, or tails, T. Therefore, the sample space is ${H,T}.$

Here, the sample space is shown in a tree diagram. Each row represents the possible outcomes of a toss. When the coin is flipped another time, the tree diagram gets another row with the possible outcomes.

In this case, the sample space has $4$ possible outcomes.For each of the following experiments, list the possible outcomes in the sample space and count the total number of outcomes.

a Roll one die.

b Roll two dice at the same time.

a **Number of Outcomes: $6$**

**Outcomes:**
${1,2,3,4,5,6} $

b **Number of Outcomes: $36$**

**Outcomes:**

a What are the possible outcomes when a die is rolled?

b For each outcome of the first die, there are six different outcomes for the second die.

a When a die is rolled, there are $6$ different possible outcomes.

Consequently, the sample space is the set ${1,2,3,4,5,6}.$

b As was found in Part A, each possible outcome of rolling one die is just one number. However, when rolling two dice, one possible outcome is getting a $2$ on the first die and a $5$ on the second die.

${2,5} $

Therefore, each outcome in the sample space will consist of two numbers — one for each die.To list all the possible outcomes, create all possible combinations by determining one outcome for the first die and then varying the outcome of the second die. Then, change the outcome of the first die and repeat the process.

Since each die has $6$ possible outcomes, the total number of possible outcomes for this experiment is $6⋅6=36.$

Paulina bought two white and three black marbles, all of different sizes, and put them in a bag. When she got home, her little brother Diego and sister Emily asked her to give them two marbles. Paulina agreed but told them to draw one marble each without looking inside the bag. Diego drew the first marble, then Emily.

a Assuming *all* the marbles are distinguishable, write down the sample space for the event of Diego and Emily randomly drawing two marbles from the bag. What is the total number of outcomes?

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b How many of the outcomes in the sample space represent the case where Diego and Emily both drew white marbles?

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c How many of the outcomes in the sample space represent the case where Diego and Emily both drew black marbles?

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d How many of the outcomes in the sample space represent the case where Diego and Emily drew marbles of different colors?

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e In a few words, write a different event for the situation presented.

a **Number of Outcomes:** $20$

**Outcomes:**
${W_{1},W_{2}},{W_{1},B_{1}},{W_{1},B_{2}},{W_{1},B_{3}}{W_{2},W_{1}},{W_{2},B_{1}},{W_{2},B_{2}},{W_{2},B_{3}}{B_{1},W_{1}},{B_{1},W_{2}},{B_{1},B_{2}},{B_{1},B_{3}}{B_{2},W_{1}},{B_{2},W_{2}},{B_{2},B_{1}},{B_{2},B_{3}}{B_{3},W_{1}},{B_{3},W_{2}},{B_{3},B_{1}},{B_{3},B_{2}} $
Here, $W_{1}$ and $W_{2}$ represent the white marbles, and $B_{1},$ $B_{2},$ and $B_{3}$ represent the black marbles.

b **Number of Outcomes:** $2$

**Outcomes:** ${W_{1},W_{2}}$ and ${W_{2},W_{1}}$

c **Number of Outcomes:** $6$

**Outcomes:**

${B_{1},B_{2}},{B_{1},B_{3}},{B_{2},B_{1}},{B_{2},B_{3}},{B_{3},B_{1}},{B_{3},B_{2}}, $

d **Number of Outcomes:** $12$

**Outcomes:**

${W_{1},B_{1}},{W_{1},B_{2}},{W_{1},B_{3}},{W_{2},B_{1}},{W_{2},B_{2}},{W_{2},B_{3}},{B_{1},W_{1}},{B_{1},W_{2}},{B_{2},W_{1}},{B_{2},W_{2}},{B_{3},W_{1}},{B_{3},W_{2}}, $

e **Example Event:** Diego and Emily ask Paulina to give them two marbles each.

a Give each marble a label. How many outcomes does the first drawn marble have? How many outcomes does the second marble have?

b Look for the outcomes containing only the labels assigned to the white marbles.

c Look for the outcomes containing only the labels assigned to the black marbles.

d Look for the outcomes containing one label assigned to the white marbles and one label assigned to the black marbles.

e If Diego and Emily ask for a different number of marbles, this is a different event.

a Despite having only two colors, all five marbles are said to be distinguishable. That is why starting by labeling each of the marbles is a good option.

Since both marbles are randomly drawn, each outcome of the event will consist of two labels. For the first marble, there are $5$ possible outcomes. For the second marble, there are $4$ possible outcomes because one marble has already been removed from the bag.

Since the Diego draws a marble first and Emily draws a marble after him, the order in which the marbles are drawn matters. Therefore, the outcomes ${W_{1},W_{2}}$ and ${W_{2},W_{1}}$ are different. The following table lists all the possible outcomes for the event of drawing two marbles from the bag. ${W_{1},W_{2}},{W_{1},B_{1}},{W_{1},B_{2}},{W_{1},B_{3}}{W_{2},W_{1}},{W_{2},B_{1}},{W_{2},B_{2}},{W_{2},B_{3}}{B_{1},W_{1}},{B_{1},W_{2}},{B_{1},B_{2}},{B_{1},B_{3}}{B_{2},W_{1}},{B_{2},W_{2}},{B_{2},B_{1}},{B_{2},B_{3}}{B_{3},W_{1}},{B_{3},W_{2}},{B_{3},B_{1}},{B_{3},B_{2}} $ Consequently, there are a total of $20$ possible outcomes in the sample space.

b The case in which two white marbles are drawn is represented by the outcomes containing only $W_{1}$ and $W_{2}.$ From Part A, it can be seen that there are only two such outcomes — ${W_{1},W_{2}}$ and ${W_{2},W_{1}}.$

c The case in which two black marbles are drawn is represented by the outcomes containing neither $W_{1}$ nor $W_{2}.$ Analyzing the sample space from Part A, six such outcomes can be found.

$ {B_{1},B_{2}},{B_{1},B_{3}},{B_{2},B_{1}},{B_{2},B_{3}},{B_{3},B_{1}},{B_{3},B_{2}}, $

d The case in which one white and one black marble are drawn is represented by the outcomes containing one $W-$label and one $B-$label. From Part A, there are twelve such outcomes.

${W_{1},B_{1}},{W_{1},B_{2}},{W_{1},B_{3}},{W_{2},B_{1}},{W_{2},B_{2}},{W_{2},B_{3}},{B_{1},W_{1}},{B_{1},W_{2}},{B_{2},W_{1}},{B_{2},W_{2}},{B_{3},W_{1}},{B_{3},W_{2}}, $

e There are an infinite number events that can be considered. For example, Diego and Emily could ask Paulina to give them two marbles each. In that case, the event would be the drawing of four marbles.

Sometimes more than one event can be involved in an experiment. In such cases, it is important to know how to determine the union or intersection of the events.

The union of two events $A$ and $B$ is the set of all the outcomes that are in $A$ or are in $B$ or in both $A$ and $B.$ The union of $A$ and $B$ is usually written as **or** event $B$ will occur. It can be found using the Addition Rule of Probability.

$AorB$or

$A∪B.$

The probability of the union of $A$ and $B$ is the probability that event $A$

The intersection of two events $A$ and $B$ is the set of all the outcomes that satisfy both events $A$ and $B$ simultaneously. The intersection of $A$ and $B$ is usually written as *and* $B$ will occur and can be found using the Multiplication Rule of Probability.

$AandB$or

$A∩B.$

The probability of the intersection of $A$ and $B$ is the probability that $A$

Also, there may be situations where it is easier to determine which outcomes do *not* satisfy an event rather than determining which outcomes do. For such cases, the following concept will be useful.

The complement of an event $A$ is the set of all the possible outcomes in the sample space that are **not** in event $A.$ The complement of event $A$ is usually written as $A_{′},$ $A_{c},$ or $A.$ Notice that the union of an event and its complement is the entire sample space.

For example, let $A$ be the event of getting an odd number when a six-sided die is rolled. The complement of $A$ is the event of getting an even number when the die is rolled.

Let $U$ be the set of all integers from $1$ to $9.$
### Answer

### Hint

### Solution

Consider the experiment of randomly picking a number from $U.$

- Let $A$ be the event of picking a prime number.
- Let $B$ be the event of picking an odd number.
- Let $C$ be the event of picking a multiple of $3.$

a Write the complement of each event in words.

b List the outcomes of each event and its complement.

c Write the event $A∩B$ in words and list its outcomes.

d Write the event $B∪C_{′}$ in words and list its outcomes.

e Make a Venn diagram illustrating the set $U$ and the events $A,$ $B,$ and $C.$

a **Complements:**

- $A_{′}$ is the event of picking a non-prime number.
- $B_{′}$ is the event of picking an even number.
- $C_{′}$ is the event of picking a number that is not a multiple of $3.$

b **Outcomes of the Events:**

$ABC ={2,3,5,7}={1,3,5,7,9}={3,6,9} $
**Outcomes of the Complements:**
$A_{′}B_{′}C_{′} ={1,4,6,8,9}={2,4,6,8}={1,2,4,5,7,8} $

c **Intersection of Events:** $A∩B$ is the event of picking a prime, odd number.

**Outcomes:** $A∩B={3,5,7}$

d **Union of Events:** $B∪C_{′}$ is the event of picking either an odd number or a number that is not a multiple of $3.$

**Outcomes:** $B∪C_{′}={1,2,3,4,5,7,8,9}$

e **Venn Diagram:**

a To write the complement of an event, consider the opposite event.

b The complement of an event is the set of all the possible outcomes that are **not** in that event.

c Remember how the intersection of events is defined.

d Use the definition of a union of events.

e First, look for the numbers that are common for the three events. Then, look for the numbers that are in two events but not in the third one. Finally, look for the numbers from the sample space that do not occur in any of the events.

a By definition, the complement of an event is the set of all possible outcomes that are not in that event. Therefore, the complement of picking a prime number is picking a non-prime number. The complement of events $B$ and $C$ can be similarly written.

Event | Complement |
---|---|

$A:$ picking a prime number | $A_{′}:$ picking a non-prime number |

$B:$ picking an odd number | $B_{′}:$ picking an even number |

$C:$ picking a multiple of $3$ | $C_{′}:$ picking a number that is not a multiple of $3$ |

b The outcomes of the event $A$ are all numbers in $U$ that are prime.

$A={2,3,5,7} $ In a similar way, the outcomes of the remaining events can be written. $BC ={1,3,5,7,9}={3,6,9} $ Finally, the outcomes of $A_{′}$ are the outcomes in $U$ that are not in $A.$ Similarly for the outcomes of $B_{′}$ and $C_{′}.$ $A_{′}B_{′}C_{′} ={1,4,6,8,9}={2,4,6,8}={1,2,4,5,7,8} $

c By definition, the intersection of two events is the set of all outcomes that satisfy both events simultaneously. To write $A∩B,$ start by recalling what events $A$ and $B$ represent.

$A$ - event of picking a prime number

$B$ - event of picking an odd number

Therefore, $A∩B$ is the event of picking a prime **and** odd number.

Number | Is it prime? | Is it odd? |
---|---|---|

$1$ | No | Yes |

$2$ | Yes | No |

$3$ | Yes | Yes |

$4$ | No | No |

$5$ | Yes | Yes |

$6$ | No | No |

$7$ | Yes | Yes |

$8$ | No | No |

$9$ | No | Yes |

Consequently, $A∩B={3,5,7}.$

d By definition, the union of two events is the set of all outcomes that satisfy either of the events. Before writing what $B∪C_{′}$ is, begin by remembering what each of these events represents.

$B$ - event of picking an odd number

$C_{′}$ - event of picking a number that is not a multiple of $3$

Therefore, $B∪C_{′}$ is the event of picking either an odd number **or** a number that is not multiple of $3.$

Number | Is it odd? | Is not a multiple of $3$ |
---|---|---|

$1$ | Yes | Yes |

$2$ | No | Yes |

$3$ | Yes | No |

$4$ | No | Yes |

$5$ | Yes | Yes |

$6$ | No | No |

$7$ | Yes | Yes |

$8$ | No | Yes |

$9$ | Yes | No |

Consequently, $B∪C_{′}={1,2,3,4,5,7,8,9}.$ Notice that $6$ is the only element of $U$ that is not in $B∪C_{′}$ since it satisfies neither $B$ nor $C_{′}.$

e In order to make a Venn diagram, start by considering the sample space $U,$ which is a set of integers from $1$ to $9.$

Next, draw three sets representing the events $A,$ $B,$ and $C$ and write down the outcomes of each event inside the corresponding set.

Comparing the outcomes of the three events, some conclusions can be drawn.

- The number $3$ is common for the three events. Thus, the three sets intersect each other.
- The numbers $5$ and $7$ are common for events $A$ and $B,$ but they are not in $C.$
- The number $9$ is common for events $B$ and $C,$ but it is not in $A.$
- The events $A$ and $C$ have nothing in common other than the number $3.$
- The numbers $4$ and $8$ do not belong to any of the events.

With this information, the Venn diagram can be drawn.

In the following diagram, the events and their complements can be appreciated separately.Since an event is a combination of possible outcomes of an experiment, in some cases the event happens rarely, while in others it happens frequently. This frequency depends on the experiment and the event itself.

Probability measures the likelihood that something will occur. It can be any value from $0$ or $0%$ to $1$ or $100%,$ 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 $1.$

The probability of an event occurring can be determined both theoretically and experimentally. Theoretical probability shows the expected probability when all outcomes in a sample space are equally likely, whereas experimental probability is based on data collected from repeated trials of an experiment.

When all outcomes in a sample space are equally likely, the theoretical probability of an event is the ratio of the number of favorable outcomes to the number of possible outcomes.

$P(event)=Number of possible outcomesNumber of favorable outcomes $

Therefore, the probability of event $A$ is found as follows.
$P(A)=Number of possible outcomesNumber of favorable outcomes =63 $ After simplification, it is obtained that $P(A)$ is equal to $21 .$ Consequently, there is a $50%$ chance an even number is rolled.

When an experiment is performed, the results may be a little different from what was expected. In other words, slightly different results may be obtained from what the theoretical probability predicted.

Experimental probability is the probability of an event occurring based on data collected from repeated *trials* of a probability experiment. For each trial, the outcome is noted. When all trials are performed, the experimental probability of an event is calculated by dividing the number of times the event occurs by the number of trials.

$P(event)=Number of trialsNumber of times event occurs $

It can be seen that the experimental probability of flipping tail is close to the theoretical probability, which is $0.5.$

Ramsha and Mark conducted an experiment consisting of rolling two dice and adding their results. The following diagram shows the numbers obtained in each roll.

Consider the event of getting a result greater than or equal to $8.$

a What is the theoretical probability of this event? Round the answer to two decimal places.

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b Find the experimental probability of the given event according to Ramsha's data. Round the answer to two decimal places.

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c Find the experimental probability of the given event according to Mark's data. Round the answer to two decimal places.

a Determine all the possible outputs of rolling two dice. Then, calculate the sum of the outputs and count how many of them are greater than or equal to $8.$

b Start by finding the number of successes, the number of times the event was fulfilled.

c Divide the number of successes by the number of trials.

a To find the theoretical probability, the number of favorable outcomes should be divided by the total number of possible outcomes.

$P=Number of possible outcomesNumber of favorable outcomes $ Since the experiment consists of rolling two dice and each die has $6$ possible outcomes, there is a total of $36$ possible combinations. Calculate the sum of the outputs for each combination.

The $5$ in the second row and third column represents the event of rolling a $2$ on the first die and a $3$ on the second die. The other outcomes can be calculated in the same fashion. Next, highlight the outcomes satisfying the given event, that the sum of the dice is greater than or equal to $8.$

There is a total of $15$ favorable outcomes. With this information, the theoretical probability can be calculated.$P(Sum≥8)=Total outcomesFavorable outcomes $

Substitute values and simplify

SubstituteII

$Favorable outcomes=15$, $Total outcomes=36$

$P(Sum≥8)=3615 $

ReduceFrac

$ba =b/3a/3 $

$P(Sum≥8)=125 $

CalcQuot

Calculate quotient

$P(Sum≥8)=0.416666…$

RoundDec

Round to ${\textstyle 2 \, \ifnumequal{2}{1}{\text{decimal}}{\text{decimals}}}$

$P(Sum≥8)≈0.42$

b The experimental probability of an event equals the number of successes obtained divided by the number of trials.

$P(Event)=Number of TrialsNumber of Successes $ Therefore, to find the experimental probability obtained by Ramsha, divide the number of successes she got by the number of trials she conducted. These two numbers can be deduced from the given diagram.

As can be seen, Ramsha got $2$ successes in $7$ trials. With this information, the experimental probability can be found.$P(Sum≥8)=Number of TrialsNumber of Successes $

Substitute values and simplify

SubstituteII

$Number of Successes=2$, $Number of Trials=7$

$P(Sum≥8)=72 $

CalcQuot

Calculate quotient

$P(Sum≥8)=0.285714…$

RoundDec

Round to ${\textstyle 2 \, \ifnumequal{2}{1}{\text{decimal}}{\text{decimals}}}$

$P(Sum≥8)≈0.29$

c Similar to Part B, to find the experimental probability obtained by Mark, divide the number of successes he got by the number of trials he conducted. As before, this information can be found in the given diagram.

$P(Sum≥8)=Number of TrialsNumber of Successes $

Substitute values and simplify

SubstituteII

$Number of Successes=6$, $Number of Trials=9$

$P(Sum≥8)=96 $

ReduceFrac

$ba =b/3a/3 $

$P(Sum≥8)=32 $

CalcQuot

Calculate quotient

$P(Sum≥8)=0.66666…$

RoundDec

Round to ${\textstyle 2 \, \ifnumequal{2}{1}{\text{decimal}}{\text{decimals}}}$

$P(Sum≥8)≈0.67$

The probability of drawing a club from a standard deck of cards is $0.25.$ Knowing this, what is the probability of drawing a spade, heart, or diamond if a card is drawn randomly?

To figure it out, instead of counting the favorable outcomes, the complement rule can be used.

The sum of the probability of an event and the probability of its complement is equal to $1.$

$P(A)+P(A_{′})=1$

This formula is useful when calculating the probability of the complement of an event is easier than calculating the probability of the event itself. Then, the probability of the event is calculated as follows. $P(A)=1−P(A_{′}) $

Let $A$ be an event, $A_{′}$ be its complement, and $S$ be the sample space. By the definition of the complement, the union of an event and its complement is equal to the entire sample space.
$A∪A_{′}=S $
Because $A∪A_{′}$ and $S$ represent the same event, their probabilities are equal.
$P(A∪A_{′})=P(S) $
Since the complement of $A$ consists of the outcomes that are *not* in $A,$ events $A$ and $A_{′}$ are disjoint. By the Addition Rule of Probability, the probability of the union is the sum of the individual probabilities of each of the events.
$P(A∪A_{′})=P(A)+P(A_{′}) $
Now the Transitive Property of Equality can be applied to the equalities.
${P(A∪A_{′})=P(A)+P(A_{′})P(A∪A_{′})=P(S) ⇓P(A)+P(A_{′})=P(S) $
Additionally, the probability of the entire sample space $P(S)$ is equal to $1.$ By applying the Transitive Property of Equality once more, the rule is proven.

$P(A)+P(A_{′})=1$

Using the Subtraction Property of Equality, the formula for the probability of $A$ is obtained. $P(A)+P(A_{′})=1⇕P(A)=1−P(A_{′}) $

Applying this formula, the probability of drawing a spade, heart, or diamond can be computed.

$P(♠,♡,or♢) =1−P(♣)=1−0.25=0.75 $Dylan cut out $100$ squares of paper and wrote a number from $1$ to $100$ on each square. He then put the papers in a bag and asked his dad to choose a paper at random.

What is the probability that Dylan's father picks a number that is not a multiple of $5?${"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["0.8"]}}

What is easier to count, the numbers that are not multiples of $5$ or the numbers that *are* multiples of $5?$ Note that these sets of numbers are the complements of each other.

Start by remembering what the probability of an event is. $P=Number of possible outcomesNumber of favorable outcomes $ Let $A$ be the given event, picking a number that is not a multiple of $5.$ However, counting the numbers from $1$ to $100$ that are not multiples of $5$ can be tedious. In this case, it is worth considering the complement of $A.$

Event | Complement |
---|---|

Picking a number that is not a multiple of $5.$ | Picking a number that is a multiple of $5.$ |

At the beginning of this lesson, Tearrik was invited by the host of a late-night game show to pick one door out of thee to win a car. However, behind two of the doors are sheep.
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### Hint

### Solution

Since there are $20$ prizes, there are $50$ doors with sheep. With this information, the required probability can be found.

b What are the chances of choosing a door with a sheep? Round the answer to two decimal places.

c If there were $70$ doors and $20$ prizes, what is the chance that Tearrik picks a door with a sheep behind it? Round the answer to two decimal places.

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a Divide the number of favorable outcomes by the number of possible outcomes.

b Use the Complement Rule.

c How many of the doors have sheep behind them?

a The probability of an event is the quotient of the number of favorable outcomes to the number of possible outcomes.

$P=Number of possible outcomesNumber of favorable outcomes $ Since there are three doors, there are $3$ total possible outcomes, of which only $1$ is favorable. Knowing this, the chance that Tearrik will choose the wining door can be calculated. $P(winning door)=31 ≈0.33 $

b Note that the event of picking a door with a sheep is the complement of picking the winning door. Therefore, the Complement Rule can be used.

$P(A_{′})=1−P(A) $ In Part A, the probability of picking the winning door was found to be about $0.33.$ Substituting this value into the previous equation, the probability of picking a door with a sheep will be obtained. $P(A_{′})=1−0.33≈0.67 $

c In this case, there are $70$ doors and $20$ prizes. This time finding a sheep is the favorable outcome.

$P(Sheep)=Total outcomesFavorable outcomes $

SubstituteII

$Favorable outcomes=50$, $Total outcomes=70$

$P(Sheep)=7050 $

ReduceFrac

$ba =b/10a/10 $

$P(Sheep)=75 $

CalcQuot

Calculate quotient

$P(Sheep)=0.71428…$

RoundDec

Round to ${\textstyle 2 \, \ifnumequal{2}{1}{\text{decimal}}{\text{decimals}}}$

$P(Sheep)≈0.71$

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