10.02 Introduction to multi-stage experiments

Multi-stage experiments are those that incorporate two or more simple experiments, for example tossing a coin and rolling a die. Finding probabilities of multistage events is easier if we use visual display such as a a list or a tree diagram to show all possible outcomes.

Before we practice creating displays, a quick note about independent events.Two events are independent if the occurrence of one does not affect the probability of the occurrence of the other. When tossing an unbiased coin repeatedly the probability of the occurrence of a tail on any individual toss is $0.5$0.5. This probability remains unchanged, regardless of whether there has been a run of heads or tails in previous tosses, since the coin has no memory. So each coin toss is an independent event.  The events of tossing a coin and then rolling a die are independent, because they use completely different objects.  The die is not affected by the coin and vice versa.

If events $A$A and $B$B are independent we are calculate the probability of both $A$A and $B$B occurring by simply multiplying the probabilities of the two events together. We can use the set notation $P\left(A\cap B\right)$P(A∩B) to describe this situation, and summarise this property as: 

Worked Example

Example 1

If two dice are rolled, what is the probability of rolling snake eyes (double ones)?

(Note: It does not matter if two separate dice were rolled or a single die was rolled twice)

The two rolls are independent events, so we can multiply the probability of rolling a one on each die.

$P\left(\text{Double Ones}\right)$P(Double Ones)	$=$=	$\frac{1}{6}\times\frac{1}{6}$16​×16​
	$=$=	$\frac{1}{36}$136​

Probability Displays

Lists

Let's say we're interested in how many Personal Identification Numbers (PINs) can be made from the digits $1$1, $2$2, $3$3 and $8$8.

To see all options in front of us we can systematically list all combinations as follows:

$1238$1238	$2138$2138	$3128$3128	$8123$8123
$1283$1283	$2183$2183	$3182$3182	$8132$8132
$1823$1823	$2318$2318	$3218$3218	$8213$8213
$1832$1832	$2381$2381	$3281$3281	$8231$8231
$1328$1328	$2813$2813	$3812$3812	$8312$8312
$1382$1382	$2831$2831	$3821$3821	$8321$8321

At a glance we can now see there are $24$24 combinations and we can easily use our list to determine probabilities.

a)  What is the probability that a PIN starts with a $3$3 and ends with a $2$2?

We look down the column where all PINs start with $3$3 and count those that also end with a $2$2. This gives us $\frac{2}{24}$224​

b)  What is the probability that a PIN starts with a $3$3 or ends with a $2$2?

This time we need to count all the PINs that either start with a $3$3, end with a $2$2 OR both. This gives us $\frac{10}{24}$1024​

Tables

A table is useful for showing all possible outcomes of two events in the rows and columns. For example, if we tossed $1$1 coin and $1$1 die we can show the outcomes for the coin along the first column and the outcomes for the die across the top row.

Each cell in the table is an outcome of rolling a die and a coin. There are $12$12 possible outcomes in the sample space and each are equally likely to occur. 

Tree Diagrams

Tree diagrams can be very useful to display the outcomes in a multi-stage events, particularly if the number of options and stages is low. A tree diagram is named because the diagram that results looks like a tree. We can summarise the information by putting the probability of each stage on the branch and the outcome at the end of the branch. Probabilities of each outcome can be found by multiplying along a branch.

When a single trial is carried out, we have just one column of branches. Here are some examples. None of these have probabilities written on the branches because they all have equal chance of happening.  

Here are some examples that have probabilities on the branches, because each outcome is not equally likely.

You should notice that when we look at a group of branches, the sum of the group should add to $1$1.  This indicates that all the outcomes are listed.  

When more than one trial is carried out, we have two (or more) columns of branches. The first example is drawn without probabilities on the branches, as each outcome has an equal chance of happening. You could, of course, write the probability of each outcome on each branch as $\frac{1}{2}$12​. 

We can see the player has a $3$3% chance of winning, so the chance that they win $2$2 games in a row is $9$9%. 

If I wanted to know what the chance is that they win at least $1$1 match, we could do either

a) add $9+21+21$9+21+21 together and get $51$51%,

or b) use the complementary event of losing both games and calculate $100-49=51$100−49=51%.

In summary:

Practice Questions

Question 1

QUestion 2

Question 3

Question 4