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7th grade (Illustrative Mathematics)
Course: 7th grade (Illustrative Mathematics) > Unit 8
Lesson 6: Lesson 9: Multi-step experimentsProbability of a compound event
Learn how to use sample space diagrams to find probabilities.
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How can we solve this without a chart?(20 votes)
There is 2/3 probability of getting a vacation that is 2 days or higher. There is 2/3 probability of getting a vacation away from snow.
If you want both conditions satisfied, multiply the two probabilities. 2/3 * 2/3 = 4/9.
Sal has explanations on his compound probability videos. https://www.khanacademy.org/math/statistics-probability/probability-library/multiplication-rule-independent/v/compound-probability-of-independent-events(4 votes)
Khan Acamedy is pretty cool for people who are studying in this program(9 votes)- A jar holds 15 red pencils and 10 blue. What is the probability of picking a red pencil?(5 votes)
think of it as a fraction. There are 15 pencils out of 25 pencils so the probability of picking a red pencil would be 15/25 or 60%.(8 votes)
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?(5 votes)- 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!(6 votes)
Could there be a way to do this in our heads? Or without a chart?(5 votes)
You Could but you are more likely to miss something(4 votes)
- 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?(6 votes)
- Is there a video on the probability of compound events?(3 votes)
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.(3 votes)
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.(3 votes)
- I seriously need a vacation to one of those places. Upvote this post if any of you have younger siblings that are tiring.(4 votes)
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!(4 votes)
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.