Compound probability of independent events using diagrams
Compound events example with tree diagram
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.