ACARA v9 CONTENT DESCRIPTION “list all outcomes for compound events both with and without replacement, using lists, tree diagrams, tables or arrays; assign probabilities to outcomes”
Builds on: the Probability strand. This unit builds on the probability of single events and on systematic counting. Listing outcomes for compound events is the foundation for estimating probabilities from data and running simulations in the rest of this strand.
Listing the possibilities first
A compound event is one made of two or more simple events happening together or in sequence, like tossing two coins, spinning a spinner twice, or drawing two cards. To work out the chance of such an event, the first job is almost always to list every possible outcome carefully and completely, because a probability is only as reliable as the list of possibilities it rests on. This unit is about generating those outcomes systematically and then assigning probabilities to them.
A two-stage compound event
Two coins combine into one event with four possible outcomes.
Tossing two coins is one compound event. Coin 1 can be H or T, and for each result coin 2 can be H or T, giving the four outcomes HH, HT, TH and TT.
With replacement
The single most important distinction is whether the experiment is carried out with replacement or without replacement. With replacement means whatever was selected is put back before the next stage, so every stage faces exactly the same options. Spinning a three-colour spinner twice is a with-replacement situation: the first spin does not use up a colour, so the second spin again has all three. The number of outcomes is three times three, which is nine.
With replacement: same options
Spin a three-colour spinner twice; each spin still has all three colours.
With replacement, the chosen colour goes back, so each spin again has all three colours. The grid lists 3 × 3 = 9 equally likely outcomes.
Without replacement
Without replacement means the selected item is not returned, so later stages have fewer options to choose from. Drawing two letters from the set A, B and C without replacement gives three choices for the first letter but only two for the second, making three times two, or six, outcomes. A crucial consequence is that repeats become impossible: you cannot draw A twice, so outcomes like AA simply do not exist. Recognising which kind of experiment you are dealing with decides how many outcomes there are and which ones are even possible.
Without replacement: fewer, no repeats
Draw two letters from A, B, C; the second draw cannot repeat the first.
Without replacement the first letter is not returned, so the second draw has only two letters left. That gives 3 × 2 = 6 outcomes, and the repeats AA, BB and CC are impossible.
Lists, trees, tables and arrays
Several tools help list outcomes without missing any. The simplest is an organised list: for two coins, writing HH, HT, TH and TT captures all four possibilities. A tree diagram draws a branch for each option at every stage, so two coins give two first branches, each splitting into two, and reading along the branches recovers the same four outcomes. A table or array is ideal for two-stage experiments with several options each, setting the first stage as rows and the second as columns.
A tree diagram for two coins
Each branch is one half; multiply along a path to get the probability of each outcome.
Each branch of the tree carries probability one half. Multiplying along a path gives 1/2 × 1/2 = 1/4, so every one of the four outcomes has probability 1/4.
Two dice: the power of a table
Two dice show how powerful a table can be. Listing them as a six-by-six array gives thirty-six equally likely outcomes, far too many to track in your head but easy to read from a grid. The table also makes patterns visible: if you record the sum of the two dice in each cell, you can see that a sum of seven appears six times while a sum of two appears only once, which immediately tells you that seven is the most likely total.
Two dice as a table
A six-by-six table holds all 36 outcomes; the cells summing to seven stand out.
A six-by-six table lists all 6 × 6 = 36 outcomes for two dice. Six of them, shown highlighted, add to seven, so P(sum 7) = 6/36 = 1/6.
Assigning probabilities
Once the outcomes are listed, assigning probabilities is straightforward when they are equally likely. The probability of any single outcome is one divided by the total number of outcomes, so each of the four coin outcomes has probability one quarter, and each of the thirty-six dice outcomes has probability one thirty-sixth. The probability of an event is then the number of outcomes that satisfy it divided by the total. From the two dice, the probability of a sum of seven is six out of thirty-six, which simplifies to one sixth.
Probabilities add to one
Four equally likely outcomes each take one quarter; together they make a whole.
The four coin outcomes are equally likely, so each takes one quarter of the whole. Four quarters make one, which is why the probabilities of all the outcomes add to 1.
Multiplying along a tree
Tree diagrams also let you find probabilities by multiplying along the branches, and this is where the with or without replacement choice reappears. For two spins of the spinner with replacement, the probability of red then green is one third times one third, which is one ninth, matching the nine-outcome list. For the letters drawn without replacement, the probability of A then B is one third times one half, since only two letters remain for the second draw, giving one sixth and again matching the listed outcomes. Always check that the probabilities of all the outcomes add to one.
Quick self-check
1. Two coins are tossed. How many equally likely outcomes are there?
2. Two coins are tossed. What is the probability of getting two heads (HH)?
3. A 3-colour spinner is spun twice WITH replacement. How many outcomes are there?
4. Two letters are drawn from {A, B, C} WITHOUT replacement. How many outcomes are there?
5. Two dice are rolled (a 6 × 6 table of 36 outcomes). What is the probability the sum is 7?