Alternative Format

Try This

"Best of Three" Coin Toss

Two friends, Deepak and Ty, are tossing a fair coin three times to play a best-of-three game. If the coin shows heads, then Deepak wins the toss, and if the coin shows tails, then Ty wins the toss.

Best of Three

Deepak and Ty are tossing a fair coin three times.

What is the probability that Deepak wins the game after exactly two coin tosses?

This experiment is different from those we previously studied. Instead of studying the outcomes of a single coin toss, we instead need to study the outcomes of three consecutive coin tosses. 

To do this, we must take into consideration that each time you toss the coin, the probability of tossing heads is not affected by the outcome of any previous toss.

"Each individual coin toss is independent of the other two tosses."

Lesson Goals

• Define and identify independent events.
• Use a tree diagram to list all possible outcomes of an experiment.
• Calculate the theoretical probability of two independent events.

Try This!

Two friends, Deepak and Ty, are tossing a fair coin three times to play a "best of three" game. 

1. What is the probability that Deepak wins the "best of three" after two coin tosses?
2. What is the probability that exactly \(2\) of the \(3\) tosses are heads?

Think about this problem, then move on to the next part of the lesson.

Independent Events

Independence

An experiment consists of rolling a standard die and tossing a fair coin.

One possible outcome of this experiment is to roll a \(2\) on the die and toss a head with the coin.

\(2\), heads

If we think about these as two independent experiments, then we know that 

• the probability of rolling a \(2\) is \(\dfrac{1}{6}\) and
• the probability of tossing a head is \(\dfrac{1}{2}\).

But how do we determine the probability of both events happening? Before we can answer this question, let's think about what we mean by independence.

We call these experiments independent if we can assume that one does not affect the other. 

We can think of "Rolling a \(2\)" and "Tossing a head" as two independent events if we assume that the roll of the die does not affect the flip of the coin..

In this example, independence means that rolling a \(2\) does not make you any more or less likely to toss a head.

Order of Events

Perhaps someone might try to argue that it is possible that the roll of a die could affect the toss of a coin or vise versa. In probability theory, we make reasonable assumptions about whether it's appropriate to consider two experiments to be independent of one another.

We assume that rolling a die and tossing a coin are independent, so, when calculating probabilities, order does not matter. 

We can think of

• rolling a die and then tossing a coin, or
• tossing a coin and then rolling a die, or
• rolling a die and tossing a coin at the same time.

In all cases, the probability of rolling a \(2\) and tossing a head will be the same.

You may find that it's helpful to choose one of the above situations when working with probabilities. As long as you can argue that each piece of an experiment is independent from the rest, then you can choose whichever one you would like.

Independence vs. Dependence

To gain an even clearer picture of what makes two events independent, it is helpful to think about what would make two events not independent, or in other words, dependent. 

Suppose an experiment is drawing \(2\) balls, one after the other from a bag. In the bag, there is \(1\) striped ball and \(2\) solid balls. 

Experiment: Draw two balls, one after the other, from a bag. 

Independent Events

We have independence if one ball is drawn, then replaced,
and then the second ball is drawn. 

Notice that before we make the first draw, we have \(3\) balls to choose from, \(1\) striped and \(2\) solid. Since we've replaced the first ball before making our second draw, the outcome of the first draw does not actually change the probability of choosing striped or solid on the second draw.

	Independent
Before Draw 1
Before Draw 2

In comparison, we have dependents if \(1\) ball is drawn and not replaced, and then the second ball is drawn.

Dependent Events

One ball is drawn, and not replaced,
and then the second ball is drawn. 

Notice again that before we make the first draw, we have \(3\) balls to choose from, \(1\) striped and \(2\) solid. Since we are not replacing the first ball before making our second draw, the particular type of ball drawn affects the probability of choosing striped or solid on the second draw.

• If we choose a solid ball on the first draw, then on the second draw, we are choosing between \(1\) striped and \(1\) solid ball.
• On the other hand, if the first ball chosen is striped, then on the second draw, we are choosing from \(2\) solid balls. 

	Dependent
Before Draw 1
Before Draw 2	or

Here, the outcome of the first draw does affect the probability of choosing striped or solid on the second draw.

Check Your Understanding 1

Question — Part 1

An experiment consists of rolling a standard die and then tossing a fair coin. Bay is interested in the probability of rolling an even number and then tossing heads.

Indicate whether the events are independent or dependent.

Answer — Part 1

Independent

Feedback — Part 1

Rolling an even number does not have affect the probability of tossing heads.

Therefore, the events are independent events.

Question — Part 2

An experiment consists of drawing two cards, one after the other, from a deck of cards. One card is drawn, and not replaced, and then a second card is drawn. Jem is interested in the probability of drawing a King and then drawing a Jack.

Indicate whether the events are independent or dependent.

Answer — Part 2

Dependent

Solution — Part 2

Drawing a King on the first draw affects the probability of drawing a Jack on the second draw since the first card drawn was not replaced.

Therefore, the events are dependent events.

Listing Possible Outcomes

Example 1

Solution

So here, we're not looking for probability, but just a way to list all of the outcomes. There are many possible outcomes of this experiment. One place to start might be to just start writing a few of them down. 

Listing Outcomes

• \(1\), heads
• \(2\), tails
• \(5\), tails
•    \(\vdots\)

If we continue to list outcomes in this way, then we're going to have to be extra careful that our list is complete and that we have, in fact, listed all of the possible outcomes. Instead, to make sure that we have all possible outcomes in our list, it's a good idea to list the outcomes systematically. This means that we follow a particular order or plan.

• We can start by considering all outcomes where the die shows \(1\). In this case, there are exactly two possibilities: either we have a \(1\) and heads, or a \(1\) and tails. And that's exhausting all possibilities where the die rolls a \(1\).
• Now we can move on to a different number. Let's consider all outcomes were the die rolls a \(2\). Similarly, the outcomes here are rolling a \(2\) and tossing heads, as well as rolling a \(2\) and tossing tails. Again, we've exhausted all possibilities where a \(2\) is rolled.
• We can continue in this way, listing all possibilities for when the die rolls a \(3\), a \(4\), a \(5\), or a \(6\).

Listing Outcomes Systematically

• \(1\), heads
• \(2\), heads
• \(3\), heads
• \(4\), heads
• \(5\), heads
• \(6\), heads
• \(1\), tails
• \(2\), tails
• \(3\), tails
• \(4\), tails
• \(5\), tails
• \(6\), tails

The result is the following list of the \(12\) possible outcomes. Take a look at the symmetry within the columns and the rows of our list. Because we followed a particular order and plan for listing and organized them to reflect this, we can be confident that the list of possible outcomes is complete.

Example 2

Solution

As in the previous example, let's systematically list all of the possible outfits. But we'll take a slightly different approach here and start by using a tree diagram keeping an eye on probabilities for later. We assume that choosing a shirt is independent from choosing a pair of pants.

And so we will think about choosing a shirt first and then choosing a pair of pants to add to this shirt. We will use this order when we make our tree. We start by considering all possible choices for the shirt which are white, green, or red.

To represent this using a tree, we draw three branches and indicate the color of each shirt at the end of each of the three branches. 

For each choice of shirt, there are now two choices for the pants. So beside each choice of shirt we draw two branches and write the color of each possible pair of pants at the end of each of the branches.

Now we have a completed tree. You can list the possible outcomes by following any path along the branches of the tree from left to right.

For example, we can choose a white shirt paired with brown pants:

Or we can choose a red shirt paired with black pants:

Again, from the shape and symmetry within the tree we can be sure that we have all of the possible outfits that Jason could make.

To confirm our answer another way we can think about the counting. There are \(3\) shirts. And for each shirt we have \(2\) choices for pants. So it makes sense that we would have \(3\times 2 = 6\) outfits in total on our tree.

Outcomes (Outfits)

• white shirt, black pants
• white shirt, brown pants
• green shirt, black pants
• green shirt, brown pants
• red shirt, black pants
• red shirt, brown pants 

Note that we could have chosen a different order for our tree. For example, we could have instead thought about choosing a pair of pants first and then adding a shirt. This would change the look of our tree and the look and order of the list of outcomes we write down. But convince yourself that both would represent the same complete list of outcomes in some order.

Check Your Understanding 2

Question

The letters of the word RAT are cut apart and put into a bag. One letter is drawn and then a standard die is rolled.

How many possible outcomes are there?

Answer

\(18\)

Feedback

There are \(18\) possible outcomes.

• R, 1
• R, 2
• R, 3
• R, 4
• R, 5
• R, 6
• A, 1
• A, 2
• A, 3
• A, 4
• A, 5
• A, 6
• T, 1
• T, 2
• T, 3
• T, 4
• T, 5
• T, 6

Probability of Independent Events

Example 3

Solution

Again, we assume that the roll and the toss are independent. Since both the die and the coin are fair, meaning that all outcomes of both are equally likely to occur, we know that all of the possible outcomes listed are also equally likely.

There is exactly \(1\) outcome where a \(2\) is rolled and a head is tossed.

There are \(12\) possible outcomes.

The probability of rolling a \(2\) and tossing heads is \( \dfrac{1}{12}\).

Check Your Understanding  3

Question

A fair coin is tossed and a standard die is rolled.

What is the probability of tossing tails and rolling a \(3\)?

Answer

\(\dfrac{1}{12}\)

Feedback

There are \(12\) possible outcomes.

• heads, \(1\)
• tails, \(1\)
• heads, \(2\)
• tails, \(2\)
• heads, \(3\)
• tails, \(3\)
• heads, \(4\)
• tails, \(4\)
• heads, \(5\)
• tails, \(5\)
• heads, \(6\)
• tails, \(6\)

There is \(1\) favourable outcome. It is tails, \(3\).

Therefore, the probability of tossing tails and rolling a \(3\) is \(\dfrac{1}{12}\).

Example 4

Solution

You may think there are three possible outcomes: 

• Two heads
• A head and a tail
• Two tails

This is definitely true. This might suggest that the probability of getting two heads is \(\dfrac{1}{3}\) because there's \(3\) possibilities, and \(1\) of them is two heads. Unfortunately, this answer is incorrect because these three outcomes are not actually equally likely. So let's analyze this experiment.

As usual, we will assume that these two tosses are independent of each other. Using a tree diagram, the possible outcomes for the first toss are heads or tails. 

And independent of the result of the first toss, we have two equally likely possibilities for the second toss, heads or tails.

From this tree, we see that there are actually four possible outcomes. Now, since the tosses are independent and the likelihood of tossing heads or tails on each is 50/50, these \(4\) outcomes are equally likely.

By drawing this tree diagram, we are reminded that there are actually two ways that we can get one head and one tail. Either we toss ahead first and then a tail or we toss a tail first and then a head.

Now for the probability calculation. We have \(4\) equally likely outcomes.

Outcomes

• HH
• HT
• TH
• TT

Exactly one has the outcome in our event, which is two heads.

As a result, the probability of tossing two heads must be \( \dfrac{1}{4}\).

Try This Problem Revisited

Solution

Let's start by creating a tree diagram to determine all of the possible outcomes. Here all three of the coin tosses are independent of one another. The first coin toss can be heads or tails as usual.

And independent of the result of the first toss, we have two equally likely possibilities for the second toss, heads or tails.

Independent of the result of the first two tosses, we have the same two possibilities for the third toss. So, we get the following tree diagram. 

We can list all of the possible outcomes by following each path along the branches of the tree from left to right. Here we see that there are \(8\) different possibilities.

Outcomes

• HHH
• HHT
• HTH
• HTT
• THH
• THT
• TTH
• TTT

There are \(8\) equally likely outcomes.

Now, it may seem strange to toss the coin a third time after one player has already won with two tosses. But we will assume we'll toss all three no matter what. So these are really the true outcomes of the experiment.

We can now use our list of possible outcomes to answer the probability questions. Note that since each toss has an equal chance of landing on heads or tails the \(8\) outcomes must be equally likely. 

1. For part a), we determine the probability that Deepak wins after just two coin tosses. Here we need to determine the probability that the first two tosses are heads.

   Out of the \(8\) possible outcomes we have listed, there are exactly \(2\) outcomes in which the two first tosses are heads.

   Therefore, the probability of Deepak winning after just two coin tosses is \(\dfrac{2}{8} = \dfrac{1}{4} \).

2. In part b), we want to calculate the probability that exactly two heads are tossed in any order. What this means is that we are looking for outcomes that include two heads and one tail in any order.

   Out of the \(8\) possible outcomes that we have listed there are \(3\) outcomes in which exactly two heads and one tail occurs.

   Therefore the probability of getting exactly two heads is \( \dfrac{3}{8} \).

Check Your Understanding 4

Question — Step 1

A spinner divided into equal yellow, blue, and red sections is spun and a four-sided die is rolled.

Represent the outcomes of the experiment using a tree and determine the number of outcomes of the spinner and the die.

Answer — Step 1

Number of outcomes of the spinner: \(3\)

Number of outcomes of the die: \(4\)

Feedback — Step 1

Notice that the number of branches corresponds to the number of outcomes at each level.

Created with GeoGebra. Author: University of Waterloo. CC BY-NC-SA 4.0. https://ggbm.at/typzegar

Question — Step 2

Using Step 1, identify the number of equally-likely outcomes.

Answer — Step 2

\(12\)

Feedback — Step 2

There are \(12\) possible paths in the tree. Since the events are independent these \(12\) outcomes are equally likely.

Therefore, there are \(12\) equally-likely outcomes.

Online Version

https://ggbm.at/typzegar

Take It With You

\(5\) fair coins are tossed at the same time.

1. How many possible outcomes are there?
2. Are all possible outcomes equally likely? Explain.
3. What is the probability of tossing exactly \(3\) heads?

So think about the tools we've learned in this lesson, things like systematic lists and tree diagrams and independence. For this question you might not want to use a tree diagram or a list to come up with all of the possible outcomes.

But can you use the examples from the lesson to come up with a strategy for solving larger scale problems like this one?