2 Step Experiments
2 Step Experiments in NSW Year 10 Core maths involve two actions, such as tossing a coin and spinning a spinner, or rolling two dice (NESA outcome MA5‑PRO‑C‑01). Listing every outcome in a table, an array or a tree diagram makes it easy to count the outcomes you want.
You will count outcomes, find probabilities from tables of sums and products of two dice, and compare drawing cards with and without replacement. You will also multiply probabilities for independent events, such as a train and a bus both being on time, find relative frequencies from a simulation, and predict how many wins to expect from a game at a school fete. It is part of Year 10 Core Probability.
Every question with a fully worked solution.
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Theory
This Year 10 maths page covers two-step experiments: listing every outcome in a grid, multiplying probabilities for independent events, the difference between drawing with and without replacement, the expected number of results, and the relative frequency from a simulation.
A two-step (multistage) experiment has two parts, such as rolling two dice or drawing two cards. Each outcome is a pair, such as \((3, 5)\).
The number of outcomes is the number for step one times the number for step two: a coin and a die give \(2 \times 6 = 12\) outcomes.
Events are independent if the result of one does not change the probability of the other, such as two separate dice. They are dependent if it does, such as drawing a second card without replacement.
The expected number of results in \(n\) trials is \(P(\text{event}) \times n\).
The relative frequency (observed probability) from a simulation or experiment is the number of times the event happened divided by the number of trials.
For independent events \(A\) and \(B\), and for repeated trials:
How to find a probability in a two-step experiment
- List the outcomes in a grid (or a tree) and count them.
- Mark the outcomes in the event, counting each pair once.
- Divide by the total number of outcomes.
How to use the multiplication rule
- Check whether the steps are independent. If not (no replacement), change the second probability: one fewer item in total, and one fewer of the type taken.
- Multiply the probabilities of the steps.
- Add the results if the event can happen in more than one order.
There are \(6 \times 6 = 36\) equally likely outcomes. List the pairs whose largest number is \(4\):
| \(\text{first die } 4\) | \(:\) | \((4,1), (4,2), (4,3), (4,4)\) |
| \(\text{second die } 4\) | \(:\) | \((1,4), (2,4), (3,4)\) |
That is \(7\) outcomes (\((4,4)\) is listed only once).
| \(P\) | \(=\) | \(\boldsymbol{\dfrac{7}{36}}\) |
(i) both are on time
(ii) both are late.
Independent events: multiply.
| \((\text{i})\quad P(\text{both on time})\) | \(=\) | \(0.85 \times 0.7 = \boldsymbol{0.595}\) |
(ii) First find each probability of being late with the complement.
| \(P(\text{bus late})\) | \(=\) | \(1 - 0.85 = 0.15\) |
| \(P(\text{train late})\) | \(=\) | \(1 - 0.7 = 0.3\) |
| \(P(\text{both late})\) | \(=\) | \(0.15 \times 0.3 = \boldsymbol{0.045}\) |
(i) put back before the second draw
(ii) not put back.
The odd cards are \(1\), \(3\) and \(5\).
(i) With replacement, the second draw is the same as the first.
| \(P\) | \(=\) | \(\dfrac{3}{6} \times \dfrac{3}{6} = \dfrac{9}{36} = \boldsymbol{\dfrac{1}{4}}\) |
(ii) Without replacement, one odd card has gone: \(2\) odd cards out of \(5\) are left.
| \(P\) | \(=\) | \(\dfrac{3}{6} \times \dfrac{2}{5} = \dfrac{6}{30} = \boldsymbol{\dfrac{1}{5}}\) |
The second events are dependent in (ii): the first draw changes the second.
(i) Find the probability of winning.
(ii) How many wins would you expect in \(240\) games?
(iii) A computer simulation of \(300\) games gave \(46\) wins. Find the relative frequency and compare.
(i) The totals of \(7\) are \((1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\): \(6\) of \(36\) outcomes.
| \(P(\text{win})\) | \(=\) | \(\dfrac{6}{36} = \boldsymbol{\dfrac{1}{6}}\) |
| \((\text{ii})\quad \text{expected wins}\) | \(=\) | \(\dfrac{1}{6} \times 240 = \boldsymbol{40}\) |
| \((\text{iii})\quad \text{relative frequency}\) | \(=\) | \(\dfrac{46}{300} \approx \boldsymbol{0.153}\) |
This is close to \(\dfrac{1}{6} \approx 0.167\). More trials would usually bring it closer.
Common pitfalls
Frequently asked questions
How many outcomes does a two-step experiment have?
Multiply the number of outcomes of each step. Two six-sided dice have \(6 \times 6 = 36\) outcomes; a coin and a die have \(12\).
What is the difference between independent and dependent events?
Independent events do not affect each other, like two coin tosses. Dependent events do, like drawing a second card after the first is kept out.
When can you multiply probabilities?
To find \(P(A \text{ and } B)\), multiply \(P(A)\) by \(P(B)\) when the events are independent. For dependent events, multiply by the probability of \(B\) after \(A\) has happened.
What is relative frequency?
It is the fraction of trials in which an event actually happened: the count divided by the number of trials. It estimates the probability.
How do you find the expected number of results?
Multiply the probability by the number of trials. For a probability of \(\dfrac{1}{4}\) over \(120\) games, expect about \(30\) results.