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2011-02-01T00:00:01
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I was in the Bibliotèque the other day and I heard a couple chatting. At first I thought, "Who are these people, it's a bibliotèque, don't they know you're supposed to be quiet?". Then I realised they were my good friends Elle and Joe.<br></br>
Elle: "Joe, what's the deal with tree diagrams?"<br></br>
Joe: "What, those things we use for probability?"<br></br>
Elle: "Yeah the ones that look like trees."<br></br>
Joe: "What do you mean what's the deal with tree diagrams? They're easy, you just multiply probabilities across and add probabilities vertically."<br></br>
Elle: "I know that, but why? It's all pretty abstract to me."<br></br>
<br></br>
That got me thinking, is this a common problem? Do people mindlessly draw tree diagrams without really understanding the concepts behind them?<br></br>
<br></br>
Let's take a couple of examples back to first principles and see if we can gain a deeper insight into tree diagrams and their use for calculating probabilities.<br></br>
<br></br>
<span style="font-weight: bold;">What is a Tree Diagram?</span><br></br>
<br></br>
A tree diagram is simply a way of representing a sequence of events. Tree diagrams are particularly useful in probability since they record all possible outcomes in a clear and uncomplicated manner.<br></br>
<br></br>
<span style="font-weight: bold;">Example:</span><br></br>
Let's take a look at a simple example, flipping a coin and then rolling a die. We might want to know the probability of getting a Head and a 4.<br></br>
If we wanted, we could list all the possible outcomes:<br></br>
<br></br>
<div style="text-align: center; font-style: italic;">(H,1) (H,2) (H,3) (H,4) (H,5) (H,6)</div>
<div style="text-align: center;"><span style="font-style: italic;">(T,1) (T,2) (T,3) (T,4) (T,5) (T,6)</span></div>
<div style="text-align: center;"> </div>
<div style="text-align: center; font-style: italic;">Probability of getting a Head and a 4:</div>
<div style="text-align: center; font-style: italic;">P(H,4) = $\frac{1}{12}$</div>
<div style="text-align: center;"> </div>
<div style="text-align: left;">Here is one way of representing the situation using a tree diagram. To save time, I have chosen not to list every possible die throw (1, 2, 3, 4, 5, 6) separately, so I have just listed the outcomes "4" and "not 4":</div>
<div style="text-align: left;"> </div>
<div style="text-align: left;"> <mdo:image alt="tree diagram" height="280" src="diag1.png" width="385"></mdo:image></div>
<div style="text-align: left;">Each path represents a possible outcome, and the fractions indicate the probability of travelling along that branch. For each pair of branches the sum of the probabilities adds to 1.</div>
<div style="text-align: left;"> </div>
<div style="text-align: left;"><span style="font-weight: bold;">"And" Means Multiply</span></div>
<div style="text-align: left;"> </div>
<div style="text-align: left;">So how might we work out P(H,4) from the tree diagram? We could word this as the probability of getting a Head <span style="font-style: italic;">and then</span> a 4. This is the green path.</div>
<div style="text-align: left;"> </div>
<div style="text-align: left;">Half the time, I expect to travel along the first green branch.</div>
<div style="text-align: left;"> </div>
<div style="text-align: left;"> <mdo:image alt="halved rectangle" height="215" src="diag2.png" width="279"></mdo:image></div>
<div style="text-align: left;">Then, on one sixth of those occasions, I will also travel along the second green branch. We can think of this as $\frac{1}{6} \text{ of } \frac{1}{2}$.</div>
<div style="text-align: left;"> </div>
<div style="text-align: left;"> <mdo:image alt="rectangle in twelve pieces" height="215" src="diag3.png" width="279"></mdo:image></div>
<div style="text-align: left;"> </div>
<div style="text-align: left;">$\frac{1}{6} \text{ of } \frac{1}{2} = \frac{1}{6}$ x $\frac{1}{2} = \frac{1}{12}$</div>
<div style="text-align: left;"> </div>
<div style="text-align: left;">So this is why Joe said that you multiply across the branches of the tree diagram.</div>
<div style="text-align: left;"> </div>
<div style="text-align: left;"> <mdo:image alt="Tree diagram showing probability calculations" height="276" src="diag5.png" width="501"></mdo:image></div>
<div style="text-align: left;"> </div>
<div>"And" only means multiply if events are independent, that is, the outcome of one event does not affect the outcome of another. This is certainly true for our example, since flipping the coin has no impact on the outcome of the die throw. </div>
<br></br>
<span style="font-weight: bold;">"Or" Means Add</span><br></br>
<br></br>
Now let's consider the probability of getting a Head <span style="font-style: italic;">or</span> a 4.<br></br>
We are using the word "or" in its mathematical sense to mean "Head or 4 or both", as opposed to the common usage which often means "either a Head or a 4":<br></br>
<br></br>
<div style="text-align: center;"><span style="font-style: italic;">(H,1) (H,2) (H,3) (H,4) (H,5) (H,6) (T,4)</span></div>
<br></br>
So P(H or 4) is $\frac{7}{12}$<br></br>
<br></br>
Again, we can work this out from the tree diagram, by selecting every branch which includes a Head or a 4:<br></br>
<br></br>
<mdo:image alt="Tree diagram: heads and fours selected" height="276" src="diag4.png" width="477"></mdo:image><br></br>
<br></br>
Each of the ticked branches shows a way of achieving the desired outcome. So P(H or 4) is the sum of these probabilities:<br></br>
<br></br>
$P(H\text{ or }4) = P(H,4) + P(H, \text{ not }4 ) + P(T, 4) = \frac{1}{12} + \frac{5}{12} + \frac{1}{12} = \frac{7}{12}$.<br></br>
<br></br>
So this is why Joe said that you add down the ends of the branches.<br></br>
<br></br>
<span style="font-weight: bold;">Picturing the Probabilities</span><br></br>
<br></br>
Imagine I roll an ordinary die three times, and I'm interested in the probability of getting one, two or three sixes.<br></br>
I might draw a tree diagram like this:<br></br>
<br></br>
<mdo:image alt="6 and not a 6 tree diagram" height="487" src="diag6.png" width="556"></mdo:image><br></br>
Check that you agree with the probabilities at the end of each branch before reading on.<br></br>
<br></br>
We can now work out:<br></br>
<div style="margin-left: 40px;">P(three sixes) = $\frac{1}{216}$<br></br>
P(exactly two sixes) = $\frac{15}{216}$<br></br>
P(exactly one six) = $\frac{75}{216}$<br></br>
P(no sixes) = $\frac{125}{216}$</div>
<div> </div>
<div>Again, check that you understand where these probabilities have come from before reading on.</div>
<div> </div>
<div>To really check your understanding, think about the outcomes that contribute to each of the probabilities on the tree diagram. For example, P(6, not 6, 6) is $\frac{5}{216}$, because out of the 216 total outcomes there are five outcomes which satisfy (6, not 6, 6):</div>
<div style="text-align: center;">6, 1, 6</div>
<div style="text-align: center;">6, 2, 6</div>
<div style="text-align: center;">6, 3, 6</div>
<div style="text-align: center;">6, 4, 6</div>
<div style="text-align: center;">6, 5, 6</div>
<div style="text-align: left;"> </div>
<div style="text-align: left;">Can you explain why there are 25 outcomes that satisfy (not 6, not 6, 6)?</div>
<div style="text-align: left;">What about the other probabilities on the tree diagram? </div>
<br></br>
I hope this article helps you to understand what's happening next time you come across a tree diagram, and that it helps you to construct your own tree diagrams to solve problems.<br></br>
<br></br>
<br></br>
<span style="font-style: italic;">Click</span> <a href="http://nrich.maths.org/public/leg.php?group_id=40&code=206#results" style="font-style: italic;">here</a> <span style="font-style: italic;">for a selection of NRICH problems where tree diagrams can be used.</span> <br></br></mdoxml>
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An Introduction to Tree Diagrams
This article explains how tree diagrams are constructed and helps
you to understand how they can be used to calculate probabilities.
Probability
Tree diagrams
Probability
Probability