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8.2: Probabilites of Multi-step Events - Mathematics


8.2: Probabilites of Multi-step Events - Mathematics

8.2: Probabilites of Multi-step Events - Mathematics

The graph shows the number of combinations assigned to each of the 14 teams in the NBA Draft Lottery. In Example 2, you determined the probability that the team with the worst record wins the first pick is 25%. The graph supports this conclusion because the height of the bar for the team with the worst record is greater than the heights of all the other bars.

Is it more likely that the team with the worst record will not win the first pick? Using the graph, imagine stacking all of the bars for the other 13 teams, as shown in the figure below. The height of this stack is greater than the height of the bar for the worst team. So, it is more likely that the team with the worst record will not win the first pick.

If you're a fan of the NBA you might want to check out the Sports Authority, where you can get accessories and apparel for all your favorite teams.

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8.2: Probabilites of Multi-step Events - Mathematics

The table shows the probabilities of a winter storm in New York for several months. Complete the table. Then describe the likelihood of each event in words.

Remember that a probability is measured from 0 (impossible) to 1 (certain). Each of the probabilities in the table are less than 50%. So, each event is more unlikely than it is likely.

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Describe the likelihood that at least 1 inch of snow will accumulate in Billings.

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Describe the likelihood that at least 1 inch of snow will accumulate in Helena.

The map indicates that the probability that at least 1 inch of snow will accumulate in Helena is about 80%. So, this event is likely.

Helena is the capital of Montana.

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Describe the likelihood that at least 1 inch of snow will accumulate in Sidney.

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Describe the likelihood that at least 1 inch of snow will accumulate in Opheim.

The map indicates that the probability that at least 1 inch of snow will accumulate in Opheim is only about 1%. So, this event is nearly impossible.

Opheim is a small town near the Canadian border in Montana. Its population is about 100 people. Although it is cold in the winter, it is also dry. That is the reason the probability of snow is so low.

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8.2: Probabilites of Multi-step Events - Mathematics

You randomly choose a card from a standard deck of playing cards. Find the theoretical probability of choosing a card of each suit.

There are 52 cards in the deck and 13 of each suit. So, the probability of drawing any particular suit is

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You have a standard deck of cards. The bar graph shows the results of randomly choosing 1 card, recording its suit, and placing it back in the deck for 50 trials. Find the experimental probability of choosing a card of each suit.

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Compare the probabilities found in Exercises 23 and 24.

The theoretical and experimental probabilities for drawing each suit are as follows.

Anyone who has played cards recognizes that a theoretical probability is the result you expect over hundreds or even thousands of card hands. For a small number of hands (like 50), the results can vary. This is the whole point of card games, isn't it? You are hoping that you will have good "luck" and that your "experimental probability" will be better than the average.

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Sophie&rsquos Tree Diagram

Sophie conducted a multistage experiment in which she tossed a coin two times. She drew this tree diagram to represent the sample space.

  • How many outcomes does the tree diagram show?
  • What is the probability that Sophie will toss heads both times?
    P(heads, heads)
  • What is the probability that Sophie will toss heads and tails in any order?
    P(heads and tails)
  • How many outcomes are in the sample space?
  • How many times does “heads, heads” appear in the sample space?

Probability Models

The sample space S for a probability model is the set of all possible outcomes.

For example, suppose there are 5 marbles in a bowl. One is red , one is blue , one is yellow , one is green , and one is purple . If one marble is to be picked at random from the bowl, the sample space possible outcomes S = < red , blue , yellow , green , purple >. If 3 of the marbles are red and 2 are blue , then the sample space S = < red , blue >, since only two possible color outcomes be possible. If, instead, two marbles are picked from a bowl with 3 red marbles and 2 blue marbles, then the sample space S = <(2 red ), (2 blue ), (1 red and 1 blue )>, the set of all possible outcomes.

An event A is a subset of the sample space S.

Suppose there are 3 red marbles and 2 blue marbles in a bowl. If an individual picks three marbles, one at a time, from the bowl, the event "pick 2 red marbles" can be achieved in 3 ways, so the set of outcomes A = <( red , red , blue ),( red , blue , red ), ( blue , red , red )>. The sample space for picking three marbles, one at a time, is all of the possible ordered combinations of three marbles, S = <( red , red , red ), ( red , red , blue ), ( red , blue , red ), ( blue , red , red ), ( blue , blue , red ), ( blue , red , blue ), ( red , blue , blue )>. Since there are only 2 blue marbles, it is impossible to achieve the event < blue , blue , blue >.

A probability is a numerical value assigned to a given event A . The probability of an event is written P(A) , and describes the long-run relative frequency of the event. The first two basic rules of probability are the following:

Rule 1: Any probability P(A) is a number between 0 and 1 (0 < P(A) < 1).
Rule 2: The probability of the sample space S is equal to 1 (P(S) = 1).

Suppose five marbles, each of a different color, are placed in a bowl. The sample space for choosing one marble, from above, is S = < red , blue , yellow , green , purple >. Since one of these must be selected, the probability of choosing any marble is equal to the probability of the sample space S = 1. Suppose the event of interest is choosing the purple marble, A = < purple >. If it is equally likely that any one marble will be selected, then the probability of choosing the purple marble, P(A) = 1/5. In general, the following formula describes the calculation of probabilities for equally likely outcomes:

If there are k possible outcomes for a phenomenon and each is equally likely, then each individual outcome has probability 1/ k . The probability of any event A is

If two events have no outcomes in common, then they are called disjoint . For example, the possible outcomes of picking a single marble are disjoint: only one color is possible on each pick. The addition of probabilities for disjoint events is the third basic rule of probability:

Rule 3: If two events A and B are disjoint, then the probability of either event is the sum of the probabilities of the two events:
P(A or B) = P(A) + P(B).

The chance of any (one or more) of two or more events occurring is called the union of the events. The probability of the union of disjoint events is the sum of their individual probabilities.

For example, the probability of drawing either a purple , red , or green marble from a bowl of five differently colored marbles is the sum of the probabilities of drawing any of these marbles: 1/5 + 1/5 + 1/5 = 3/5.

If there are three red marbles and two blue marbles, then the probability of drawing any red marble is the number of outcomes in the event " red ," which is equal to three, divided by the total number of outcomes, 5, or 3/5 = 0.6. The sample space is this case is < red , blue >, which must have total probability equal to 1, so the probability of drawing a blue marble is equal to 2/5 = 0.4. The event of drawing a blue marble does not occur if a red marble is chosen, so the event A blue">blue " is called the complement A c of the event " red ." Since an event and its complement together form the entire sample space S , the probability of an event A is equal to the probability of the sample space S , minus the probability of A c , as follows:

Rule 4: The probability that any event A does not occur is P(A c ) = 1 - P(A).

Venn Diagrams

In the Venn diagram to the right, events A and B are disjoint. Suppose, for example, event A is drawing a red marble from a bowl of five differently colored marbles, and event B is drawing a blue marble. These events cannot both occur, so there is no overlapping area.

In the Venn diagram on the left, events A and B are not disjoint. This means that it is possible for both events to occur, and the overlapping area represents this possibility. Suppose, for instance, there are 3 red marbles and two blue marbles in a bowl. Two marbles are to be drawn from the bowl, one after the other. After the first draw, the marble drawn is returned to the bowl. Define event A to be drawing a red marble from the bowl on the first draw, and define event B to be drawing a blue marble on the second draw. The occurence of event A is represented by the red area, the occurence of event B is represented by the blue area, the occurence of both events is represented by the overlapping area (also known as the intersection of the two events), and the occurence of either event is represented by the entire colored area (also known as the union of the two events).

Independence

Rule 5: If two events A and B are independent, then the probability of both events is the product of the probabilities for each event:
P(A and B) = P(A)P(B).

The chance of all of two or more events occurring is called the intersection of events. For independent events, the probability of the intersection of two or more events is the product of the probabilities.

In the case of two coin flips, for example, the probability of observing two heads is 1/2*1/2 = 1/4. Similarly, the probability of observing four heads on four coin flips is 1/2*1/2*1/2*1/2 = 1/16.

If two events A and B are not disjoint, then the probability of their union (the event that A or B occurs) is equal to the sum of their probabilities minus the sum of their intersection.

In the example corresponding to the second Venn diagram above, we know that the probability of drawing a red marble on the first draw (event A ) is equal to 3/5, and the probability of drawing a blue marble on the second draw (event B ) is equal to 2/5. Since events A and B are independent, the probability of the intersection of A and B, P(A and B) , equals the product P(A)*P(B) = 3/5*2/5 = 6/25. The probability of the union of A and B , P(A or B) , is equal to
P(A) + P(B) - P(A and B) = 3/5 + 2/5 - 6/25 = 1 - 6/25 = 19/25 = 0.76.

For another example, consider tossing two coins. The probability of a head on any toss is equal to 1/2. Since the tosses are independent, the probability of a head on both tosses (the intersection) is equal to 1/2*1/2 = 1/4. The probability of a head on either toss (the union) is equal to the sum of the probabilities of a head on each toss minus the probability of the intersection, 1/2 + 1/2 - 1/4 = 3/4.

Note: Disjoint events are not independent. In the marble example, consider drawing one marble from the bowl of five, where each marble is a different color. Suppose the event of interest, event A , is drawing a blue marble. The probability of drawing this marble is 1/5. Suppose event B is drawing a green marble. These events are disjoint, since event B cannot occur if event A occurs. Obviously, they are not independent, since the outcome of event A directly affects the outcome of event B . If, instead, two marbles were to be drawn from the bowl, with the first marble replaced before the second marble was drawn, then the event of drawing a blue marble on the first draw would not affect the outcome of the second draw. The event of drawing a green marble on the second draw would be independent of the event of drawing a blue marble on the first draw, so the probability of both events occurring would be the product of the probabilities of each event, 1/5*1/5 = 1/25.

The probability of the intersection of two or events which are not independent is determined using conditional probabilities.

For some more definitions and examples, see the probability index in Valerie J. Easton and John H. McColl's Statistics Glossary v1.1 .


Lesson 8

Noah will select 1 card at random from a standard deck of cards. Find the probabilities. Explain or show your reasoning.

Expand Image

8.2: Under One Condition

Kiran notices that the probabilities from the warm-up can be arranged into at least two equations.

Kiran wonders if it is always true that (P( ext) = P( ext) oldcdot P(B)) for events A and B. He wants to check additional examples from drawing a card from a deck.

  1. If Event A is “the card is black” and Event B is “the card is a king,” does the equation hold? Explain or show your reasoning.
  2. If Event A is “the card is a face card” and Event B is “the card is a spade,” does the equation hold? Explain or show your reasoning.

8.3: Coin and Cube

A coin is flipped, then a standard number cube is rolled. Let A represent the event “the coin lands showing heads” and B represent “the standard number cube lands showing 4.”

Expand Image

Attribution: US Nickel, by Lordnikon. Public Domain. Wikimedia Commons. Source.

  1. Are events A and B independent or dependent? Explain your reasoning.
  2. Find the probabilities
    1. (P( ext))
    2. (P( ext))
    3. (P( ext))
    4. (P( ext))
    1. (P( ext))
    2. (P( ext))

    Students are writing programs in their robotics class using two different programming languages that are very similar. 70% of the programs are written in the first programming language and the other 30% percent are written in the second programming language. The newer robots only accept programs written in the first programming language and the older robots only accept programs written in the second programming language.

    In this robotics class, Andre’s job is to determine which programming language each program is written in. He determines the programming language correctly 90% of the time for each of the programs. If Andre determines that a program is written in the second programming language, what is the probability that the program is actually written in the first programming language? Explain your reasoning.

    Summary

    A conditional probability is the probability that one event occurs under the condition that another event occurs.

    For example, we will remove two marbles from a jar that contains 3 green marbles, 2 blue marbles, 1 white marble, and 1 black marble. We might consider the conditional probability that the second marble we remove is green given that the first marble removed was green. The notation for this probability is (P( ext)) where the vertical line can be read as “under the condition that the next event occurs” or “given that the next event occurs.” In this example, (P( ext) = frac<2><6>) since we assume the condition that the first marble drawn was green has happened, so the second draw only has 2 possible green marbles left to draw out of 6 marbles still in the jar.

    To find the probability of two events happening together, we can use a multiplication rule:

    For example, to find the probability that we draw two green marbles from the jar, we could write out the entire sample space and find the probability from that or we could use this rule.

    Since the probability of getting green on the first draw is (frac<3><7>) and the conditional probability was considered previously, we can find the probability that both events occur using the multiplication rule.

    This tells us that the probability of getting green marbles in both draws is (frac<1><7>) (since this is equivalent to (frac<6><42>) ).

    For example, when flipping a coin and rolling a standard number cube, the events “getting a tails for the coin” and “getting 5 for the number cube” are independent. That means we can find the probability of both events occurring to be (frac<1><12>) by using the multiplication rule.

    Glossary Entries

    The probability that one event occurs under the condition that another event occurs.

    Dependent events are two events from the same experiment for which the probability of one event depends on whether the other event happens.

    Independent events are two events from the same experiment for which the probability of one event is not affected by whether the other event occurs or not.

    The Illustrative Mathematics name and logo are not subject to the Creative Commons license and may not be used without the prior and express written consent of Illustrative Mathematics.

    This book includes public domain images or openly licensed images that are copyrighted by their respective owners. Openly licensed images remain under the terms of their respective licenses. See the image attribution section for more information.


    Adding Probabilities - Not Mutually Exclusive

    In the previous section, we showed you a formula for calculating the the probability of two (or more) mutually exclusive outcomes. The formula was pretty simple:

    But this only works if the outcomes are mutually exclusive. What happens if they aren't mutually exclusive?

    Well, the formula gets just a little bit more complicated. Now it looks like this:

    P(A or B) = P(A) + P(B) - P(A and B)

    That's right - now we don't just need to figure out the probability of A and the probability of B - we need to figure out the probability that both happen. Why? Well, suppose I asked you to find the probability that a card was either a spade or a king. You can add the probability of a spade, and the probability of a king, but what have you done? You've counted one desired outcome twice! You counted the king of spades because it's a king, and you also counted it because it's a spade. So we need to subtract that one outcome back out of the result. That's what the last term of our formula is: subtract out the probability of it being both a king and a spade.

    So let's go ahead and calculate that probability. We'll use S for spade, and K for king:


    7.8 Probability and Sampling

    IM 6–8 Math was originally developed by Open Up Resources and authored by Illustrative Mathematics®, and is copyright 2017-2019 by Open Up Resources. It is licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). OUR's 6–8 Math Curriculum is available at https://openupresources.org/math-curriculum/.

    Adaptations and updates to IM 6–8 Math are copyright 2019 by Illustrative Mathematics, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0).

    Adaptations to add additional English language learner supports are copyright 2019 by Open Up Resources, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0).

    The second set of English assessments (marked as set "B") are copyright 2019 by Open Up Resources, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0).

    Spanish translation of the "B" assessments are copyright 2020 by Illustrative Mathematics, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0).

    The Illustrative Mathematics name and logo are not subject to the Creative Commons license and may not be used without the prior and express written consent of Illustrative Mathematics.

    This site includes public domain images or openly licensed images that are copyrighted by their respective owners. Openly licensed images remain under the terms of their respective licenses. See the image attribution section for more information.


    Lesson 2

    When reaching into a dark closet and pulling out one shoe from a pile of 20 pairs of shoes, you pull out a left shoe.

    When listening to a playlist—which has 5 songs on it—in shuffle mode, the first song on the playlist plays first.

    2.2: How Likely Is It?

    Label each event with one of these options:

    impossible, unlikely, equally likely as not, likely, certain

    1. You will win grand prize in a raffle if you purchased 2 out of the 100 tickets.
    2. You will wait less than 10 minutes before ordering at a fast food restaurant.
    3. You will get an even number when you roll a standard number cube.
    4. A four-year-old child is over 6 feet tall.
    5. No one in your class will be late to class next week.
    6. The next baby born at a hospital will be a boy.

    g. It will snow at our school on July 1.

    h. The Sun will set today before 11:00 p.m.

    i. Spinning this spinner will result in green.

    j. Spinning this spinner will result in red.

    Expand Image

    Description: <p>A circular spinner divided into 3 parts. The top half of the spinner is divided into two equal parts, a yellow section, labeled “Y” and a blue section, labeled “B.” The bottom half of the spinner is one green section labeled “G.” The spinner dial points to the section labeled "G."</p>

    Invent another situation for each label, for a total of 5 more events.

    2.3: Take a Chance

    This applet displays a random number from 1 to 6, like a number cube. With a partner, you will play a game of chance.

    In the first round, one of you will score on an even roll and one of you will score on an odd roll. You decide that first.

    In the second round, the winner of round 1 will score on numbers (1 - 4) , and the other player will score on numbers (5 - 6) .

    Each round is 10 rolls. Be sure to turn on "History" after your first roll and wait for it to update before rolling again.

    1. When each player had three numbers, did one of them usually win?
    2. When one player had four numbers, did you expect them to win? Explain your reasoning.

    On a game show, there are 3 closed doors. One door has a prize behind it. The contestant chooses one of the doors. The host of the game show, who knows where the prize is located, opens one of the other doors which does not have the prize. The contestant can choose to stay with their first choice or switch to the remaining closed door.

    Practice playing the game with your partner and record your results. Whoever is the host starts each round by secretly deciding which door has the prize.

    1. Play 20 rounds where the contestant always stays with their first choice.
    2. Play 20 more rounds where the contestant always switches doors.

    2.4: Card Sort: Likelihood

    Your teacher will give you some cards that describe events. Order the events from least likely to most likely.

    After ordering the first set of cards, pause here so your teacher can review your work. Then, your teacher will give you a second set of cards.

    Add the new set of cards to the first set so that all of the cards are ordered from least likely to most likely.

    Summary

    A chance experiment is something that happens where the outcome is unknown. For example, if we flip a coin, we don’t know if the result will be a head or a tail. An outcome of a chance experiment is something that can happen when you do a chance experiment. For example, when you flip a coin, one possible outcome is that you will get a head. An event is a set of one or more outcomes.

    We can describe events using these phrases:

    For example, if you flip a coin:

    • It is impossible that the coin will turn into a bottle of ketchup.
    • It is unlikely the coin will land on its edge.
    • It is equally likely as not that you will get a tail.
    • It is likely that you will get a head or a tail.
    • It is certain that the coin will land somewhere.

    The probability of an event is a measure of the likelihood that an event will occur. We will learn more about probabilities in the lessons to come.

    Glossary Entries

    A chance experiment is something you can do over and over again, and you don’t know what will happen each time.

    For example, each time you spin the spinner, it could land on red, yellow, blue, or green.

    Expand Image

    An event is a set of one or more outcomes in a chance experiment. For example, if we roll a number cube, there are six possible outcomes.

    Expand Image

    Examples of events are “rolling a number less than 3,” “rolling an even number,” or “rolling a 5.”

    An outcome of a chance experiment is one of the things that can happen when you do the experiment. For example, the possible outcomes of tossing a coin are heads and tails.

    IM 6–8 Math was originally developed by Open Up Resources and authored by Illustrative Mathematics®, and is copyright 2017-2019 by Open Up Resources. It is licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). OUR's 6–8 Math Curriculum is available at https://openupresources.org/math-curriculum/.

    Adaptations and updates to IM 6–8 Math are copyright 2019 by Illustrative Mathematics, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0).

    Adaptations to add additional English language learner supports are copyright 2019 by Open Up Resources, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0).

    The second set of English assessments (marked as set "B") are copyright 2019 by Open Up Resources, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0).

    Spanish translation of the "B" assessments are copyright 2020 by Illustrative Mathematics, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0).

    The Illustrative Mathematics name and logo are not subject to the Creative Commons license and may not be used without the prior and express written consent of Illustrative Mathematics.

    This site includes public domain images or openly licensed images that are copyrighted by their respective owners. Openly licensed images remain under the terms of their respective licenses. See the image attribution section for more information.


    Watch the video: - Multi-Step Calculations and Constants (November 2021).