Probability: Grade 12 Mathematics

By the end of this lesson you should be able to

Calculate the probability of a single event
Use the addition rule for "or" and the multiplication rule for "and"
Draw and use tree diagrams, with and without replacement
Use a Venn diagram to find the probability of combined events

What is probability?

Probability measures how likely an event is. It is a number from $0$ (impossible) to $1$ (certain), written as a fraction, decimal or percentage.

P(E) = \frac{\text{number of favourable outcomes}}{\text{total number of equally likely outcomes}}

P(\text{not } E) = 1 - P(E)

Example. When a fair six-sided die is rolled, the prime numbers are $2$ , $3$ and $5$ . So $P(\text{prime}) = \dfrac{3}{6} = \dfrac{1}{2}$ .

Combining events

Two words decide which rule to use.

Word	Rule	When it applies
or	Add: $P(A \text{ or } B) = P(A) + P(B)$	Events that cannot happen together (mutually exclusive)
and	Multiply: $P(A \text{ and } B) = P(A) \times P(B)$	Independent events, where one does not affect the other

If events can happen together, use the general addition rule so that you do not count the overlap twice:

P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)

Worked examples

Example 1. A bag holds 5 red, 3 blue and 2 green counters. One counter is taken at random. Find the probability that it is not red.

There are $10$ counters in total and $5$ are not red.

P(\text{not red}) = \frac{5}{10} = \frac{1}{2}

Example 2: independent events. A fair coin is tossed and a fair die is rolled. Find $P(\text{head and a 6})$ .

The coin does not affect the die, so multiply:

P = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}

Tree diagrams

A tree diagram lists every outcome of two or more stages. Multiply along the branches to find the probability of one path, and add the paths that give the outcome you want.

Without replacement means an item is not put back, so the second probabilities change.

Example 3. A bag contains 4 red and 6 blue counters. Two counters are taken one after the other without replacement.

Tree diagram for two counters drawn without replacement. R is red and B is blue.

Multiply along each path to find the probability of each outcome.

Outcome	Working	Probability
RR	$\dfrac{4}{10} \times \dfrac{3}{9}$	$\dfrac{12}{90}$
RB	$\dfrac{4}{10} \times \dfrac{6}{9}$	$\dfrac{24}{90}$
BR	$\dfrac{6}{10} \times \dfrac{4}{9}$	$\dfrac{24}{90}$
BB	$\dfrac{6}{10} \times \dfrac{5}{9}$	$\dfrac{30}{90}$

(a) $P(\text{both red}) = \dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90} = \dfrac{2}{15}$

(b) $P(\text{one of each colour})$ is the sum of two paths, red then blue and blue then red:

\frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15}

Check that all four outcomes add to $1$ : $\dfrac{12 + 24 + 24 + 30}{90} = 1$ . ✓

Venn diagrams and overlapping events

Example 4. In a class of 40 learners, 25 study Mathematics, 18 study Science and 8 study both. Find the probability that a learner chosen at random studies neither subject.

Using the overlap rule, the number who study at least one subject is $25 + 18 - 8 = 35$ .

So $40 - 35 = 5$ learners study neither, and

P(\text{neither}) = \frac{5}{40} = \frac{1}{8}

The number who study Mathematics only is $25 - 8 = 17$ , so $P(\text{Mathematics only}) = \dfrac{17}{40}$ .

Expected frequency

If the probability of an event is $p$ and the situation is repeated $n$ times, the expected number of times it happens is $n \times p$ . For example, if $P(\text{rain}) = 0.3$ on any day, the expected number of rainy days in a 30-day month is $30 \times 0.3 = 9$ . This is an average, not a guarantee.

Common mistakes

Adding when you should multiply. “And” for independent events means multiply. Adding probabilities of separate stages can give an answer above $1$ , which is impossible.

Not changing the denominator without replacement. After one counter is removed, both the number of favourable outcomes and the total change. The second probability for $10$ counters becomes a fraction out of $9$ .

Forgetting the second path. “One of each colour” can happen in two orders. Add both paths.

Giving an answer above 1 or below 0. Always check that your probability lies between $0$ and $1$ .

Practice questions

A card is chosen at random from cards numbered $1$ to $20$ . Find the probability that it is a multiple of $5$ .
A bag holds 3 red and 5 white counters. Two counters are taken with replacement. Find the probability that both are white.
Using the same bag, two counters are taken without replacement. Find the probability that both are white.
For independent events, $P(A) = 0.4$ and $P(B) = 0.5$ . Find $P(A \text{ and } B)$ and $P(A \text{ or } B)$ .
Two fair dice are thrown. Find the probability that the sum is $8$ .
Two fair dice are thrown. Find the probability that the sum is at least $10$ .
In a group of 30 learners, 18 like football, 12 like netball and 5 like both. Find the probability that a learner chosen at random likes neither.
The probability that a learner passes a test is $0.7$ . Out of $200$ learners, how many are expected to pass?

Show answers

The multiples of $5$ are $5, 10, 15, 20$ , so $P = \dfrac{4}{20} = \dfrac{1}{5}$
$\dfrac{5}{8} \times \dfrac{5}{8} = \dfrac{25}{64}$
$\dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}$
$P(A \text{ and } B) = 0.4 \times 0.5 = 0.2$ and $P(A \text{ or } B) = 0.4 + 0.5 - 0.2 = 0.7$
The outcomes are $(2,6), (3,5), (4,4), (5,3), (6,2)$ , so $P = \dfrac{5}{36}$
A sum of $10$ has $3$ outcomes, $11$ has $2$ and $12$ has $1$ , so $P = \dfrac{6}{36} = \dfrac{1}{6}$
At least one: $18 + 12 - 5 = 25$ . Neither: $30 - 25 = 5$ , so $P = \dfrac{5}{30} = \dfrac{1}{6}$
$200 \times 0.7 = 140$ learners

Summary

$P(E) = \dfrac{\text{favourable}}{\text{total}}$ , and probabilities always lie between $0$ and $1$ .
“Or” usually means add, and “and” usually means multiply.
On a tree diagram, multiply along branches and add the paths that fit the event.
Without replacement, the second-stage probabilities change because the total has changed.

Now practise this topic

Real questions show you what still needs work. Try a past paper under timed conditions, or take a short quiz.

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