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### Course: Grade 8 math (FL B.E.S.T.) > Unit 9

Lesson 4: Sample spaces & probability- Sample spaces for compound events
- Sample spaces for compound events
- Die rolling probability
- Probability of a compound event
- Compound events example with tree diagram
- Probabilities of compound events
- Theoretical and experimental probabilities
- Making predictions with probability
- Making predictions with probability

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# Compound events example with tree diagram

Sal figures out the probability of flipping three coins and getting at least two tails. Created by Sal Khan.

## Want to join the conversation?

- How would you do this without a diagram? I am competing in a math competition soon and will need to be able to do this faster than this method.(58 votes)
- There are other videos in the probability and combinatorics playlist that cover this, but I agree that this video is far more mechanical than most of Sal's other stuff. The way I think of this problem mathematically is to decide how many outcomes there are for 3 coins:
`(flip*flip*flip) = 2*2*2 = 8`

Then figure out how many ways there can be 2 tails in 3 coins:

(Sal covers this one in his videos also. The reasoning behind it is a bit confusing at first, so if you haven't seen it before don't worry too much-- you'll get there.) So you know that 3 combinations out of the 8 contain 2 tails. Next I add the one occasion where all three coins are tails:`3 choose 2 = 3!/2!(3-2)! = 6/2 = 3`

`3 (outcomes with 2 tails)+ 1 (outcome with 3 tails) = 4 (outcomes with at least 2 tails)`

Out of the possible 8, that's 4, so we wind up with 4/8 or 1/2.

Works for me, anyway.(43 votes)

- If there are three coins why are their 4 showing in the 3rd row. I'm a little confused.(8 votes)
- So, the picture you see in the video is what computer scientists and mathematicians refer to as a tree. Notice it has branches, (the black edges the lead to a Heads or Tails). You can trace these branches and you will get a "sequence" of how those the flip of the coins can turn out.

The first row represents the outcome of just one flip of a coin: {H,T}

The second row represents the outcome of two flips of a coin: {(H,T), (T,T), (T,H), (H,H)}. For example, the first flip is heads and second is tails is represented by (H,T).

The third row represents the outcome of three flips of a coin: {(H,T,H), (T,T,H), (T,H,H), (H,H,H), (H,T,T), (T,T,T), (T,H,T), (H,H,T)}. Again for example, the first flip is heads, second is tails, and third tails is represented by (H,T,T).

Each of those showings represent either a Heads or Tails. There needs to be 4 in row three because row 2 had 2 showings. Each row is doubled because of the number of options it can take, in this example it's either going to be heads or tails.

If they wanted to continue this by adding another flip row 4 would have 8 showings.(12 votes)

- Why is there no explanation without the diagram? I get the feeling there is a way to do it without using a diagram tree. Possibly using permutations and combinations and factor out what you don't want.(12 votes)
- You could multiply....for example, how many outcomes can come out of 3 shirts, 2 skirts, 5 pants, and 10 accesories. Well....I wouldn't be able to do such a tree diagram so I multiply 3x2x5x10= and that would give....300!! See! It's easy and you don't have to do that tedious tree diagram! -Green Panda-(2 votes)

- I need help solving this problem on probability, please.

A number cube is rolled in an experiment, and the frequency of each different roll is recorded.

Based on the experimental results, how many rolls of 2 would you expect to get in a new experiment with 75 trials?

Roll 1 2 3 4 5 6

Freq 9 8 12 11 7 13(3 votes)- I'm no expert on probabilities, but this looks like a problem that can be solved as a ratio. There are 60 rolls shown, and 8 of them are 2. So out of 75, 8/60 = x/75. You can expect ten 2's in 75 rolls, according to the results.(3 votes)

- Does the term node refer to a "leaf" in any tree diagram?(3 votes)
- Yes. "Node" and "Leaf" mean the same thing. Sal mentions this at0:57.(2 votes)

- I don’t get it, how do I use a tree diagram for coins? He explained it but it isn’t really making sense to me.(3 votes)
- a tree diagram is just "branches" that separate from a point depending on how many outcomes there are.

one coin flip would have two branches, one for heads then one for tails.

two coin flips would have the same two, but then each of those would split into two, so now there are four end results.

Another flip adds another two to each of those four end results.

the reason two are added each time is because there are two results. if you did something like a 6 sided dice, each roll would have 6 branches, so 6 for one roll, then all the way to 36 for two rolls.

Did that help>(2 votes)

- A spinnner is divided into 6 equal sections numbered 1 through 6. Predict how many times out of 180 spins the spinner is most likely to stop on a odd number?(2 votes)
- what is a tree diagram I heard this in a class how do I use this(2 votes)
- A tree diagram is used to show all the possible combinations. So let's say there are 3 shirts and 4 pants, and you want to know how many possible pairs there are. You can use a tree diagram, but you really could do 3x4 to get the answer.(3 votes)

- at2:00, why did you count out only the outcomes of the third coin? and not the other ones?

Also, could you maybe make a compound tree diagram for Statistics?(0 votes) - What formula can be used to find the favorable outcomes in this problem: 'If you roll two fair six-sided dice, what is the probability that the sum is 3 or lower?'?(1 vote)
- Instead of memorizing formulas it is often better to just reason through the situation. Hopefully my logic can help you understand;

So, there are (6 * 6 = 36) different possible outcomes. So that means there is a (1/(6 * 6)) chance that we will get any 1 outcome. Based off that, how many of those would result in a 3 or lower sum?

In this situation it would probably be easiest just to count it out, where D1/D2 equal Die-1/Die-2;

(D1 = 1 & D2 = 1) => 2 => TRUE!

(D1 = 2 & D2 = 1) => 2 => TRUE!

(D1 = 1 & D2 = 2) => 2 => TRUE!

Everything else would result in a greater result, so that means there are a total of 3 outcomes (out of the 36) which would result in a sum of 3 or lower.

So, the percent = (number of outcomes)/(number of possible outcomes) = (3/36) = (1/12).

I hope this helps,

- Convenient Colleague(3 votes)

## Video transcript

If you flip three
fair coins, what is the probability that
you'll get at least two tails? The tree diagram below shows
all the possible outcomes of flipping three coins. At the top of the
tree, this shows us the two outcomes
for the first coin, and then given each
of those outcomes, it shows us what's possible
for the second coins. If we got a heads
on the first coin, we could get a heads or a
tails on the second coin. Got a tails on the
first coin, well, we could get a heads or a
tails on the second coin. And then for each
of those outcomes, it shows us the different
outcomes for the third coins. So let's just think
about where on this-- how do we represent
getting at least two tails? Or what are the total outcomes,
and which of those meet our constraints of getting
at least two tails? So this node right
over here, this is getting a heads
on the third coin, a heads on the
second coin, we just have to follow up the tree,
and a heads on the first coin. So this is getting three heads,
so this is definitely not going to meet our constraint. This node right over
here, we have a head, head-- this is
often called a leaf if we're talking
about a tree diagram-- a head, head, and a tail. So that's one tail. That doesn't meet our constraint
of at least two tails. What about this one here? Heads, tails, heads. Once again, only one tail,
so that doesn't meet it. Heads, tails, tails. This one does. So let me color this in. Let me color all the ones
that meet our constraints. This is getting a
tail on the third one, a tail on the second one,
and a heads on the first. So that's at least two tails. Here we have tails,
heads, heads. That doesn't meet it. Tails, heads, tails. That does. So let's color this one in. And then tails, tails. Well, if you got a tail on
the first and the second, then either of these are
going to meet the constraints, because you already
got two tails. So that one meets it. That you've got tails, tails,
heads, and tails, tails, tails. Both of them. So, let's go back
to the question. What is the probability that
you'll get at least two tails? Well, how many equally
likely outcomes are there? We're assuming this
is a fair coin. We see that there are
1, 2, 3, 4, 5, 6, 7, 8 equally likely outcomes. And how many of these
outcomes met our constraints? 1, 2, 3, 4. 4 out of the 8. 4/8, which could also be
viewed is equivalent to 1/2. The probability that I'll get
at least two tails is 1/2.