Two-step Experiments Using Tables
Every question with a fully worked solution.
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Theory
A two-step experiment, such as rolling two dice or drawing two counters, can be shown in an array (a table of every outcome) or a tree diagram. This Year 10A maths page covers arrays of sums, products, differences and the larger number, draws with and without replacement, "at least one", trees that stop early, expected numbers and fair games.
A multistage experiment is done in steps, such as rolling a die then spinning a spinner. If step one has \(m\) outcomes and step two has \(n\), there are \(m\times n\) outcomes altogether.
An array (or table of outcomes) shows the first step down the side and the second across the top. Each cell is one outcome, and the cells are equally likely when each step's outcomes are. The cells can hold pairs, sums, products, differences or the larger number.
A tree diagram shows each step as a set of branches labelled with probabilities. The branches from any point add to \(1\). Trees are best when the outcomes are not equally likely, or when there are three or more stages.
With replacement, the item is put back, so the second step has the same probabilities as the first. Without replacement, the item is kept out, so the second step has one fewer item and its probabilities change.
| Two dice: the sum | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) | \(8\) | \(9\) | \(10\) | \(11\) | \(12\) |
|---|---|---|---|---|---|---|---|---|---|---|---|
| number of ways | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(5\) | \(4\) | \(3\) | \(2\) | \(1\) |
How to solve a two-step problem
- Decide: array or tree? With or without replacement?
- Draw it: fill the array, or label every branch. Check the branches at each point add to \(1\).
- Mark the outcomes or paths in the event. "One of each" usually means two paths, such as \(RW\) and \(WR\).
- Calculate: count cells over the total, or multiply along each path and add the paths.
- Use the complement for "at least one".
Other question types
- Different spinner sections: a spinner with a red half, a blue quarter and a green quarter has unequal branches \(\dfrac{1}{2}, \dfrac{1}{4}, \dfrac{1}{4}\). Use a tree, not a list of colours.
- Stop when successful: "succeeds on the third try" is fail, fail, succeed. "Needs a third try" is just fail, fail.
- Working backwards: with replacement, \(P(\text{both red})=p^2\), so \(p=\sqrt{P}\). Without replacement, write the path with a pronumeral and solve the equation, which may be quadratic.
- Raffles and prizes: drawing winning tickets is without replacement; the second draw has one fewer ticket.
- Repeated rounds: if a tie means the round is replayed, only the rounds with a winner matter, so compare the players' chances within a single round.
(i) \(P(\text{product is a multiple of } 4)\)
(ii) \(P(\text{product}\gt 12)\)
The array (Diagrams tab) has \(6\times 4=24\) equally likely cells.
(i) Multiples of \(4\): all \(6\) in the spinner-\(4\) column, \(2\times 2\), \(4\times 2\), \(6\times 2\) in the spinner-\(2\) column, and \(4\times 1\), \(4\times 3\). That is \(6+3+1+1=11\).
| \(P\) | \(=\) | \(\boldsymbol{\dfrac{11}{24}}\) |
(ii) Products over \(12\): \(15, 18\) (spinner \(3\)) and \(16, 20, 24\) (spinner \(4\)). \(12\) itself does not count.
| \(P\) | \(=\) | \(\boldsymbol{\dfrac{5}{24}}\) |
(i) \(P(\text{one of each colour})\)
(ii) \(P(\text{at least one white})\)
With replacement the bag is the same for both draws: \(P(R)=\dfrac{5}{8}\), \(P(W)=\dfrac{3}{8}\) each time.
(i) One of each is \(RW\) or \(WR\). Multiply along each path, then add.
| \(P\) | \(=\) | \(\dfrac{5}{8}\times\dfrac{3}{8}+\dfrac{3}{8}\times\dfrac{5}{8}\) |
| \(\) | \(=\) | \(\dfrac{30}{64}=\boldsymbol{\dfrac{15}{32}}\) |
(ii) "At least one white" is every path except \(RR\). Use the complement.
| \(P\) | \(=\) | \(1-\dfrac{5}{8}\times\dfrac{5}{8}=1-\dfrac{25}{64}=\boldsymbol{\dfrac{39}{64}}\) |
(i) \(P(\text{both the same colour})\)
(ii) \(P(\text{at least one white})\)
After the first draw, \(8\) counters are left and there is one fewer of the colour drawn.
| \(P(RR)\) | \(=\) | \(\dfrac{6}{9}\times\dfrac{5}{8}=\dfrac{30}{72}\) |
| \(P(WW)\) | \(=\) | \(\dfrac{3}{9}\times\dfrac{2}{8}=\dfrac{6}{72}\) |
(i) \(P(\text{same})=\dfrac{30}{72}+\dfrac{6}{72}=\dfrac{36}{72}=\boldsymbol{\dfrac{1}{2}}\).
(ii) Use the complement of \(RR\).
| \(P(\text{at least one } W)\) | \(=\) | \(1-P(RR)\) |
| \(\) | \(=\) | \(1-\dfrac{30}{72}=\boldsymbol{\dfrac{7}{12}}\) |
(i) Find the probability that each player wins.
(ii) Is the game fair?
(iii) In \(60\) games, how many wins would you expect for each player?
(i) The array has \(36\) equally likely cells. Count by difference: \(0\) appears \(6\) times, \(1\) appears \(10\), \(2\) appears \(8\), and \(3, 4, 5\) appear \(6, 4, 2\) times.
| \(P(\text{Mia})\) | \(=\) | \(\dfrac{6+4+2}{36}=\boldsymbol{\dfrac{1}{3}}\) |
| \(P(\text{Jo})\) | \(=\) | \(\dfrac{10+8}{36}=\boldsymbol{\dfrac{1}{2}}\) |
(ii) \(\dfrac{1}{2}\ne\dfrac{1}{3}\), so the game is not fair: Jo is more likely to win. (A draw has probability \(\dfrac{6}{36}=\dfrac{1}{6}\); check \(\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{1}{6}=1\).)
| \(\text{Mia}\) | \(=\) | \(60\times\dfrac{1}{3}=\boldsymbol{20}\) |
| \(\text{Jo}\) | \(=\) | \(60\times\dfrac{1}{2}=\boldsymbol{30}\) |
Common pitfalls
Frequently asked questions
When should I use an array instead of a tree diagram?
Use an array when there are two steps with equally likely outcomes, such as two dice, especially if you combine the numbers by adding, multiplying or subtracting. Use a tree for unequal probabilities, without replacement, or three or more steps.
How do you calculate probabilities from a tree diagram?
Multiply the probabilities along each branch of a path to get the probability of that path. If the event includes several paths, add their probabilities.
What is the difference between with and without replacement?
With replacement the item goes back, so every draw has the same probabilities. Without replacement it stays out, so the next draw has one fewer item and the probabilities change.
How do you find the probability of at least one?
Find the probability of none, then subtract it from \(1\). For at least one six in two rolls: \(1-\dfrac{5}{6}\times\dfrac{5}{6}=\dfrac{11}{36}\).
How many outcomes are there when two dice are rolled?
There are \(6\times 6=36\) equally likely ordered outcomes. \((2, 5)\) and \((5, 2)\) are different outcomes, so count both.
How do you decide if a game is fair?
Find each player's probability of winning. The game is fair only if they are equal. Expected wins in \(n\) games are \(n\) times each probability.