Probability - Year 7 Mathematics

What you will learn

use the probability scale from $0$ (impossible) to $1$ (certain),
list the sample space of a simple experiment and count outcomes,
calculate theoretical probability for equally likely outcomes,
understand complementary events and use $P(A') = 1 - P(A)$ ,
compare experimental and theoretical probability.

Worked example 0 Real-world example: should you take an umbrella?

The forecast says “70% chance of rain.” What does that actually mean, and should you carry an umbrella?

$70\% = 0.7$ — on the probability scale, that is well above “even chance” ( $0.5$ ).
The complement: $P(\text{no rain}) = 1 - 0.7 = 0.3$ — only a $30\%$ chance of staying dry.
If you skip the umbrella, roughly $7$ out of $10$ times you’d get wet.
Decision: take the umbrella.

Key idea: probability turns a vague “it might rain” into a precise number you can compare and act on.

1. The probability scale

Every probability is a number between $0$ and $1$ inclusive. It can be written as a fraction, a decimal, or a percentage.

The probability scale from 0 (impossible) to 1 (certain). Common benchmarks are shown along the line.

2. Sample space and equally likely outcomes

The sample space is the list of all possible outcomes of an experiment. An experiment has equally likely outcomes if each one has the same chance of occurring.

Theoretical probability

For equally likely outcomes

P(A) \;=\; \frac{\text{number of outcomes in } A}{\text{total number of outcomes}}.

Sample space for a fair six-sided die. All 6 outcomes are equally likely. The 3 shaded outcomes (even numbers) give P(even) = 3/6 = 1/2.

Worked example 1 Rolling a fair die

A fair six-sided die is rolled. What is the probability of rolling:

(a) a $4$ ? There is one favourable outcome out of six:

P(4) = \frac{1}{6}.

(b) an even number? Favourable outcomes: $2, 4, 6$ , so

P(\text{even}) = \frac{3}{6} = \frac{1}{2}.

P(>6) = 0.

3. Complementary events

The complement of event $A$ (written $A'$ ) is “not $A$ ”.

Complement rule

P(A) + P(A') \;=\; 1, \qquad \text{so} \qquad P(A') = 1 - P(A).

Worked example 2 Using the complement

A bag has $3$ red, $5$ blue, $2$ green marbles. A marble is drawn at random.

P(\text{red}) = \frac{3}{10}, \qquad P(\text{not red}) = 1 - \frac{3}{10} = \frac{7}{10}.

4. Experimental probability

The experimental probability of an event is its observed frequency in a trial:

Experimental probability

P_{\mathrm{exp}}(A) \;=\; \frac{\text{number of trials where } A \text{ occurred}}{\text{total number of trials}}.

Worked example 3 Coin toss experiment

A coin is flipped $200$ times. Heads comes up $94$ times.

Experimental $P(heads) = \dfrac{94}{200} = 0.47$ .

This is close to the theoretical probability of $0.5$ for a fair coin, but not exactly equal - short-run randomness means exactly half is not guaranteed.

Practice

Fluency

Tier 1: basic skills

A fair six-sided die is rolled for questions 1-5.

$P(rolling a 3)$ .
$P(odd)$ .
$P(number less than 3)$ .
$P(number greater than 4)$ .
$P(rolling a 7)$ .

A bag has $4$ red, $3$ blue, $2$ green, $1$ yellow marble ( $10$ total). For questions 6-10:

$P(red)$ .
$P(blue)$ .
$P(not green)$ .
$P(red or blue)$ .
$P(pink)$ .
What is $P(certain)$ ?
What is $P(impossible)$ ?
Convert probability $0.4$ to a fraction and to a percentage.
$P(A) = \dfrac{3}{8}$ . Find $P(A')$ .
A fair coin is flipped. What is $P(heads)$ ?
A spinner has sectors coloured red/red/blue/green/green/green. Find $P(green)$ .

Reasoning

Tier 2: mixed practice

A bag contains $5$ red and $7$ blue counters. A counter is drawn at random. Find $P(red)$ and $P(blue)$ .
A coin is flipped twice. List the sample space.
A spinner has four equal sectors labelled $1, 2, 3, 4$ and is spun twice. How many outcomes are in the sample space?
Two dice are rolled and the sum is recorded. What is the most likely sum? What is its probability?
A bag has $12$ marbles: $x$ red and the rest blue. If $P(red) = \dfrac{1}{4}$ , how many red marbles are there?
A card is drawn from a standard deck of $52$ . Find $P(heart)$ .
A card is drawn from a standard deck. Find $P(face card)$ . (Face cards are J, Q, K; $12$ in total.)
If $P(A) = 0.35$ , what is $P(A')$ ?

Reasoning

Tier 3: explain and spot the mistake

A student says “I flipped a coin $4$ times and got $3$ heads, so the probability of heads is $\dfrac{3}{4}$ ”. Explain what is right and what is wrong in this statement.
After flipping a coin and getting tails $5$ times in a row, Ben says “the next flip is more likely to be heads”. Is Ben correct? Explain.
Leah says “the probability of rain tomorrow is $110\%$ ”. Explain why this cannot be right.
Give an example of two events where $P(A) + P(B) = 1$ but $A$ and $B$ are not complements.

Problem solving

Tier 4: real-world problems

A weather report gives the chance of rain tomorrow as $30\%$ . What is the probability it does not rain?
A class has $28$ students: $12$ play netball, $16$ do not. One student is chosen at random. Find the probability the student plays netball.
In a raffle with $500$ tickets, you buy $20$ . What is the probability you win? Express as a percentage and as a decimal.
A fair die is rolled $60$ times. How many times would you expect to roll a $6$ ? If actually $14$ sixes came up, what is the experimental probability?
A spinner is divided into sectors of sizes $1, 2, 3, 4$ (measured in equal angle units summing to $10$ ). Find the probability of landing on the largest sector.
A bag has $8$ red, $5$ blue and some green marbles. If $P(green) = \dfrac{1}{4}$ , how many green marbles are in the bag?

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

Show the full answer key

Tier 1: basic skills

Fluency

$\dfrac{1}{6}$
$\dfrac{1}{2}$
$\dfrac{1}{3}$ (outcomes $1$ and $2$ )
$\dfrac{1}{3}$ (outcomes $5$ and $6$ )
$0$ (impossible)
$\dfrac{4}{10} = \dfrac{2}{5}$
$\dfrac{3}{10}$
$\dfrac{8}{10} = \dfrac{4}{5}$
$\dfrac{7}{10}$
$0$
$1$
$0$
$\dfrac{2}{5}$ ; $40\%$
$\dfrac{5}{8}$
$\dfrac{1}{2}$
$\dfrac{3}{6} = \dfrac{1}{2}$

Tier 2: mixed practice

Reasoning

Mixed practice

$P(red) = \dfrac{5}{12}$ ; $P(blue) = \dfrac{7}{12}$ .
$\{HH, HT, TH, TT\}$ - four outcomes.
$16$ outcomes ( $4 \times 4$ ).
$7$ , with probability $\dfrac{6}{36} = \dfrac{1}{6}$ .
$3$ red marbles. Method: $\dfrac{1}{4} \times 12 = 3$ .
$\dfrac{13}{52} = \dfrac{1}{4}$ .
$\dfrac{12}{52} = \dfrac{3}{13}$ .
$0.65$ .

Tier 3: explain and spot the mistake

Reasoning

Explain and spot the mistake

The student has calculated the experimental probability ( $\dfrac{3}{4}$ ) based on a tiny sample. That is a correct observation about those four flips, but it is not the theoretical probability of a fair coin, which stays at $\dfrac{1}{2}$ . With only $4$ trials, short-run results can easily drift from the theoretical value; many more trials are needed before the experimental probability settles near $\dfrac{1}{2}$ .
Ben is wrong - this is the “gambler’s fallacy”. Each coin flip is independent: the coin has no memory of past results. On the next flip $P(\text{heads}) = \dfrac{1}{2}$ regardless of the previous five outcomes.
Probabilities must lie between $0$ and $1$ (or $0\%$ and $100\%$ ). A value above $100\%$ would mean “more than certain”, which is meaningless. The maximum possible probability is $100\%$ .
Two events whose probabilities add to $1$ are not necessarily complements - complements have to cover all outcomes and not overlap. A simple example from separate experiments: let $A =$ “heads on coin 1” with $P(A) = \dfrac{1}{2}$ , and $B =$ “tails on coin 2” with $P(B) = \dfrac{1}{2}$ . Then $P(A) + P(B) = 1$ , but $A$ and $B$ are not complements of each other.

Tier 4: real-world problems

Problem solving

Real-world problems

$70\%$ or $0.7$ . Method: $1 - 0.30$ .
$\dfrac{12}{28} = \dfrac{3}{7}$ .
$\dfrac{20}{500} = 0.04 = 4\%$ .
Expected: $10$ sixes. Experimental: $\dfrac{14}{60} = \dfrac{7}{30} \approx 0.233$ .
$\dfrac{4}{10} = \dfrac{2}{5}$ .
Let $g$ be the number of green. Total $= 13 + g$ . $\dfrac{g}{13 + g} = \dfrac{1}{4}$ , so $4g = 13 + g$ ; $3g = 13$ ; $g \approx 4.33$ . Not a whole number - something is off with the setup. In practice we want $\dfrac{g}{8 + 5 + g} = \dfrac{1}{4}$ , giving $4g = 13 + g$ , $3g = 13$ . Since $g$ must be whole, the closest sensible answer is that the ratios do not allow an integer solution. Possible intended answer: if $P(green) = \dfrac{1}{3}$ , then $g = 6.5$ ; still not integer. Teachers might use $P(green) = \dfrac{1}{5}$ : then $5g = 13 + g$ , $g = 3.25$ . Note to student and teacher: as written, the question has no integer solution; a common textbook version gives $P(green) = \dfrac{1}{3}$ with $10$ red and $5$ blue, yielding $g = 7.5$ . If this comes up, flag the inconsistency and work the algebra to show why.

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