Twostep experiments are those that incorporate two simple experiments, for example tossing a coin and rolling a die, or tossing a coin twice. Finding probabilities of twostep experiments is easier if we use a list, table, or tree diagram to show all possible outcomes.
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.
Two fair dice are thrown and then their difference is calculated.
What is the probability of a difference of $1$1?
Think: We can create an array to think about the outcomes that are possible. With the two dice rolls being along either side.
Do: Complete the array and count the total number of outcomes and the number of outcomes with a difference of $1$1.
$0$0  $1$1  $2$2  $3$3  $4$4  $5$5  
$1$1  $0$0  $1$1  $2$2  $3$3  $4$4  
$2$2  $1$1  $0$0  $1$1  $2$2  $3$3  
$3$3  $2$2  $1$1  $0$0  $1$1  $2$2  
$4$4  $3$3  $2$2  $1$1  $0$0  $1$1  
$5$5  $4$4  $3$3  $2$2  $1$1  $0$0 
There are $6\times6=36$6×6=36 different outcomes and $10$10 of those have a difference of $1$1. The probability will be $\frac{10}{36}$1036. This can be simplified to $\frac{5}{18}$518.
A tree diagram is useful in tracking twostep experiments. It is named because the diagram that results looks like a tree.
IMAGE OF TREE
It is useful if the events have different weightings or are unequal events.
The important components of the tree diagram are:
When the outcomes are not equally likely the probability will be written on the branches.
The sum of the probabilities on branches from a single node should sum to $1$1.
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 the outcomes are equally likely.



The outcomes for tossing a coin once  Outcomes for randomly selecting the top, middle, or bottom shelf  Outcomes from rolling a standard die 
Here are some examples that have probabilities on the branches, because they do not have an equal chance of occurring.



When looking at a group of branches that come from a single point, the sum of the group always adds to $1$1 (or $100%$100%). This indicates that all the outcomes are listed.
When more than one experiment is carried out, we have two (or more) columns of branches.
Here are some examples. These ones do not have the probabilities written, because the outcomes are equally likely.

Tree diagram showing outcomes and sample space for tossing a coin twice to calculate the probability of HH we would have 1/4, or the probability of just 1 head, would be 2/4 as the outcomes HT and TH both have just one head. 
Here is an example that have probabilities on the branches. This probability tree diagram shows the outcomes of playing two games of tennis where the probability of winning is $\frac{3}{10}$310 and the probability of losing is $\frac{7}{10}$710.
The probabilities of the events are multiplied along each branch, for example the probability of winning both games is $9%$9% which is found by $0.3\times0.3=0.09$0.3×0.3=0.09
To find the probability of at least $1$1 win, we could do either
a) P (win, lose) + P (lose, win) + P (win, win) = $21%+21%+9%=51%$21%+21%+9%=51%
or b) use the complementary event of losing both games and calculate: 1  P (lose, lose) = $149%=51%$1−49%=51%
Multiply along the branches to calculate the probability of individual outcomes.
Add down the list of outcomes to calculate the probability of multiple options.
The final percentage should add to $100$100, or the final fractions should add to $1$1  this is useful to see if you have calculated everything correctly.
For multistage events where the next stage is affected by the previous stage, we call these dependent events. We need to take care when drawing the tree diagram accordingly.
One type of experiment that is dependent on previous trials is an experiment without replacement. This means that the object selected (e.g. card, marble, person) is not able to be selected in the second selection.
For example, the probability of drawing a red card from a standard pack of $52$52 cards is $\frac{26}{52}=\frac{1}{2}$2652=12. If we do draw a red card and choose to select a second card without replacement there are only $25$25 red cards left, but there are still $26$26 black cards. And there are only $51$51 cards left in the entire deck. The probability of selecting a second red card is $\frac{25}{51}$2551. This can be seen in the top branches of the tree diagram above.
A player is rolling $2$2 dice and looking at their sum. They draw up a table of all the possible dice rolls for two dice and what they sum to.
$1$1  $2$2  $3$3  $4$4  $5$5  $6$6  

$1$1  $2$2  $3$3  $4$4  $5$5  $6$6  $7$7 
$2$2  $3$3  $4$4  $5$5  $6$6  $7$7  $8$8 
$3$3  $4$4  $5$5  $6$6  $7$7  $8$8  $9$9 
$4$4  $5$5  $6$6  $7$7  $8$8  $9$9  $10$10 
$5$5  $6$6  $7$7  $8$8  $9$9  $10$10  $11$11 
$6$6  $7$7  $8$8  $9$9  $10$10  $11$11  $12$12 
What is the probability the dice will sum to $8$8?
On the island of Timbuktoo the probability that a set of traffic lights shows red, yellow or green is equally likely. Christa is travelling down a road where there are two sets of traffic lights.
What is the probability that both sets of traffic lights will be red?
Han owns four green ties and three blue ties. He selects one of the ties at random for himself and then another tie at random for his friend.
Fill in the probabilities matching the edges of the probability tree:
$\editable{}$  $\editable{}$  $\editable{}$ 
$\editable{}$  $\editable{}$  $\editable{}$ 
What is the probability that Han selects a blue tie for himself?
Calculate the probability that Han selects two green ties.