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Probability of a compound event

Learn how to use sample space diagrams to find probabilities.

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  • spunky sam blue style avatar for user Konrad Freymiller
    How can we solve this without a chart?
    (27 votes)
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  • blobby green style avatar for user Lois Thall
    A jar holds 15 red pencils and 10 blue. What is the probability of picking a red pencil?
    (7 votes)
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  • leaf green style avatar for user Lana Gramlich
    Example question;
    If you roll two fair six-sided dice, what is the probability that the sum is 4 or higher?

    Without drawing out a grid, what is the mathmatical formula for such a question?
    (7 votes)
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    • blobby green style avatar for user Zane Rodnick-Smith
      The hard answer is that there really isn't one catch-all plug-and-play formula for what you want. What you're asking for is really the combination of several probability events. Let's break it down:

      Let P(A) be the probability of the first die roll, and P(B) be the second die roll. A and B are discrete random variables with outcomes S = {1:6} Our sample space contains 36 possible outcomes, but not all of those outcomes are equally likely. For example, there are many ways to roll 7, but only one combination will roll 12.
      So mathematically, you want this: P(A + B) >= 4.

      This is a trickier question to answer than you might think, since from a mathematical perspective, here is what you are really asking:
      P(A + B) = P(4 OR 5 OR 6 OR 7 OR 8 OR 9 OR 10 OR 11 OR 12)

      So when you bring inequalities into the mix, you are actually asking about every single possible outcome above the one you want! Each of these probability events must be individually calculated and summed. Probability gets very complex very quickly when you start asking about probabilities beyond single events.

      An easier way would be to use the complement:

      P(A+B) = 1 - P(2 OR 3)

      This is much easier to find. There is only one combination that gives us 2, so P(2) = 1/36. There are two possiblities for 3, 1 and 2, and 2 and 1. So P(3) = 2/36. Since these events are independent (the dice do not influence each other), we can sum the probabilities. Therefore:

      P( (A+B) >= 4) = 1 - 3/36 or about 92%.

      Hope that helps!
      (9 votes)
  • primosaur sapling style avatar for user igeorge58
    Could there be a way to do this in our heads? Or without a chart?
    (7 votes)
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  • blobby green style avatar for user 1000045
    Khan Acamedy is pretty cool for people who are studying in this program
    (9 votes)
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  • blobby green style avatar for user wschof8
    there are 15 servers in a restaurant , each owns 5 identically colored shirts. each shirt chosen independently by each server each day. what is the probability that they all show up on the same day wearing the same colored shirt?
    (7 votes)
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  • hopper happy style avatar for user Veronica Mackar
    Is there a video on the probability of compound events?
    (5 votes)
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  • purple pi purple style avatar for user louisaandgreta
    What would be the mathematical way, with numbers that is, to find these probabilities, especially for the numerator, without having to draw the sample space?
    For the practice problem after “Probabilities of compound events” I don’t want to have to create these tables for the coins and dices.
    (4 votes)
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    • mr pink green style avatar for user David Severin
      I do not think any one formula will cover these examples.
      So you have to consider each case independently.
      WIth dice, there is 1 way to get 2 or 12, 2 ways to get 3 or 11, 3 ways to get 4 or 10, 4 ways to get 5 or 9, 5 ways to get 6 or 8, and 6 ways to get 7. Note 6*6=36 and 2(1+2+3+4+5)+6=36.
      For 3 coin flips, if order does not matter, having all 3 the same would be 1 (either heads or tails) and having 2 of 1 and 1 of other would be 3 (either two heads and a tail or two tails and a head) which gives 2(1+3)=8 which is 2^3=8.
      If order matters, each one is separate from the others.
      (4 votes)
  • purple pi pink style avatar for user strawberrypie2003
    I don't get how to do this subject AT ALL. HEEELLLPPP! (Those who manage give me a reasonable answer gets a home-made invisible cookie and a million katroodle dollars. :D)
    (3 votes)
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    • blobby green style avatar for user Zane Rodnick-Smith
      All the vacations seem equally fun to you, so you just decide to pick one of the three at random. That is 3 possible trips to go on.

      Instead of thinking of day length of vacations, because I think that is confusing, instead think of it like this:
      You can choose one friend to take with you on the vacation, either Amy, Ben, or Chris. Because you are fair, you decide to pick which friend gets to go with you at random. This means there are 9 possible ways to go on vacation:

      Island with Amy, Ben, or Chris (3 possibilities): IA, IB, IC
      Skiing with Amy, Ben, or Chris (3 possibilities): SA, SB, SC
      Camping with Amy, Ben, or Chris (3 possibilities): CA, CB, CC

      So our SAMPLE SPACE consists of these 9 total outcomes.

      Here's where it gets trickier: we can set other conditions. Say that Ben does not like to ski. This means that one possible vacation, Skiing with Ben (SB), has been eliminated. How many possible vacations are left? 8, since we removed one possibility from our total SAMPLE SPACE of 9. Therefore, the probability of picking a "good" vacation (because Ben would ruin our vacation by complaining a lot) would be 8/9.

      Or another example: You don't want to go camping because it's too hot. If you pick a vacation at random, what is the probability you will enjoy your vacation? Well, there are three possible outcomes that result in us going camping, so if we want to have fun, we would want to pick outcomes that are NOT camping. How many of those are there? Well, there are 9 total outcomes, minus 3 outcomes for camping, giving us 6 outcomes out of a total of 9, or 6/9.

      Let's make it a little more complicated and put these two conditions together. Ben is a wimp and doesn't want to go skiing, but it's too hot to go camping. How many outcomes will satisfy us for a vacation? Now in this case, it's easier to count the situations where we would not like the vacation, so 1/9 (Skiing with Ben) and 3/9 (Camping with Anybody), so 4/9 outcomes will not satisfy us. That leaves 5 out of 9 that will.

      It gets harder when the conditions overlap. Say we do not want to go skiing, OR we do not want to go with Ben. There are 3 conditions were we go with Ben, and 3 conditions where we go skiing. So the probability should be 6/9, right?

      Well, no. The reason is that we are counting some outcomes more than once. We have to take out all the outcomes where they overlap. In this case, we are counting Skiing with Ben twice. So we have to take out that one condition and only count it once, leaving us with 5/9 chances.

      Hope that helps! Message me if you need any more help!
      (6 votes)
  • blobby green style avatar for user Nathaniel
    How can you solve this type of question without setting up a table. What is the equation?
    (4 votes)
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Video transcript

- [Voiceover] Let's say that you're on some type of a game show and you've been doing quite well. And you're now at the round where you get to pick your fantabulous vacation. And so there are three possible places that you could go. You could go on an island beach vacation. Island beach vacation. You could go skiing on a ski vacation. Or, you could go camping. Now those aren't the only possibilities because for each of those vacations there's different amount of time that you could go on them. So you could go for one day. You could go for two days. Two days. Or you could go for three days. Three. Put that in a different color. You could go for three days. You could go for three days. So the first question I'd wanna know is, well what is the-- They're gonna randomly pick either a one day ski vacation or a two day island vacation. The first question I wanna know is, what are all of the possible outcomes here? What is the sample space? What is the space from which we are going to pick your particular vacation package? Well for the sample space, we can construct a grid. Which you can see that I've essentially been constructing while I wrote down all of the possibilities. So let me draw out the sample space with these uneven looking grid lines. All right, I think you get the picture. And I'll just abbreviate it. You can go on a one day, a one day island trip. This one I, this is one day island trip. You can go on a two day, two day-- Actually, let me just write it this way. All of these are going to be one day. Right? Because of the one day column. All of these are going to be two days. Two days. Two days. And all of these are going to be three days because it's on the three day column. And all of the ones in this row are going to be island trips. So it's one day island trip, two day island trip, three day island trip. This second row, it's all ski trips. One day ski trip, two day ski trip, three day ski trip. And then finally everything in this third row, they're campng trips. One day camping trip, two day camping trip, three day camping trip. So just like that, we have constructed the sample space right over here. You see that there's one, two, three, four, five, six, seven, eight, nine outcomes. And let's say that each of these outcomes are on a little piece of paper and they put it in a barrel and they roll it up. And for our purposes we can assume that they are all equally likely outcomes. So we're gonna assume equally likely outcomes. So if we do assume equally likely outcomes, we can figure out a probability. Maybe you live in someplace that's cold and you're really not in the mood to go skiing. In fact, you'd like to spend several days away from the snow. So let's ask ourselves a question. What is the probability that you're going to win something at least two days on a vacation without snow. Two days on vacation without snow. You're gonna randomly pick one of these nine outcomes. What's the probability that it's going to be at least-- It's gonna give you a vacation that gives you at least two days without snow. Well, let's just think a little bit about it. We know the sample space and we know each of the outcomes are equally likely. There are nine equal outcomes here. So let's write that down. We got nine equal outcomes. Now how many of the outcomes satisfy this? This event, this constraint. At least 2 days of vacation-- Let me write this. Two days. Without snow. Whether it falls or touching it or whatever. So you're essentially avoiding skiing. You want at least two days on something other than skiing. And we're assuming you're not gonna go camping in some type of alpine. You're camping in some place that's warm. Well, let's think about these outcomes. So this one is no snow but it's only one day. This is two days without snow so we can circle that one. This is three days without snow so we can circle that one. All of these have snow. This is one day without snow so we're not gonna do this one. This is two days without snow and this is three days without snow. And so four of the equally likely outcomes satisfy this constraint. So you have a 4/9 probability of getting a vacation that keeps you away from snow for at least two days. Hopefully you found that fun and useful for the next time that you are some type of strange game show.