Probability (Grade 10) — Mathematics Grades 8–11

Calculate probabilities for two events using Venn diagrams and the addition rule.

Gr 10 Term 4 - ProbabilityTerm 4 - 2 weeks (CAPS)

Practise Probability in the app →

What gets asked

Calculate the theoretical probability of an event from a sample space.
Read counts or probabilities off a Venn diagram for two events.
Use the addition rule to find a missing probability.
Decide whether two events are mutually exclusive.

You must be able to

Count favourable outcomes over the total number of outcomes in the sample space.
Fill in a Venn diagram starting with the overlap, then the rest of each circle.
Subtract the overlap once when combining P(A) and P(B) for 'or'.
Use P(not A) = 1 minus P(A) to find the probability an event does not happen.

Traps that cost marks

Multiplying P(A) by P(B) to get P(A and B). That shortcut needs independent events, which is Grade 11 work. Read P(A and B) off the diagram instead.
Writing the full total for each event in its circle, so the overlap gets counted twice. Place the overlap first. Then subtract it from each total before filling in the rest of the circle.
Giving overlapping events a special name of their own, instead of just saying they overlap. There is no separate name for this at this level. If two events can happen together, say they are not mutually exclusive.

Worked example

In a class, P(A) = 0,5, P(B) = 0,4 and P(A and B) = 0,2. Find P(A or B).

Use P(A or B) = P(A) + P(B) − P(A and B).
P(A or B) = 0,5 + 0,4 − 0,2.
P(A or B) = 0,7.

Probability and frequency

What the counting says should happen, what really happened, and how far apart the two are allowed to be.

outcome: One of the things that can happen. Drawing a red bead is one outcome.
trial: One go: one toss of a coin, or one draw from a bag.
probability: How likely an outcome is, written as a number from $0$ to $1$ .
sample space: The list of every outcome that is possible.
event: An outcome, or a group of outcomes, that you are interested in.
relative frequency: How often an outcome really happened, divided by the number of trials.

Start by writing the sample space. An event is the part of it you care about. Count the outcomes in the event, then put that count over the size of the whole sample space. That is the theoretical probability.

This counting only works when the outcomes are equally likely. A fair die gives six equally likely faces. A drawing pin also has two outcomes, but they are not equally likely, so you may not write one half.

Relative frequency comes from actually doing it. Divide the number of times the outcome happened by the number of trials. It is a fraction. The plain count of successes is not a relative frequency.

The expected number is a count of things, not a chance. Multiply the probability by the number of trials. Asked how many sixes in $60$ rolls, answer with a number of sixes.

Relative frequency wanders about near the theoretical probability. It only settles close to it after very many trials. A small gap is not proof of a loaded die, and the next few rolls do not owe you a correction.

Rules to remember

if the outcomes are equally likely, theoretical probability $= \frac{outcomes in the event}{all outcomes}$
expected number $=$ probability $\times$ number of trials

Examples

A box holds

20

airtime vouchers. Seven of them are worth R

29

and the rest are worth R

12

. One is drawn. Find the probability that it is worth R

29

Worked answer

The sample space is all $20$ vouchers.
The event holds the $7$ vouchers worth R $29$ .
So the probability is $\frac{7}{20}$ .
Written as a decimal, $\frac{7}{20} = 0, 35$ .

Answer: $\frac{7}{20}$

The bottom number is the whole sample space, not the $13$ other vouchers.

A learner rolls a die

60

times and gets a six

12

times. Find the relative frequency of a six, and the expected number of sixes.

Worked answer

Relative frequency: $\frac{12}{60} = \frac{1}{5} = 0, 2$ .
A fair die has six equally likely faces, so P(six) is $\frac{1}{6}$ .
Expected sixes: $\frac{1}{6} \times 60 = 10$ .
So $12$ turned up where $10$ was expected.

Answer: relative frequency $0, 2$ , expected $10$ sixes

The expected number counts sixes. The relative frequency is a fraction.

The same learner rolls that die

600

times and gets

108

sixes. Is the die loaded?

Worked answer

Expected sixes: $600 \div 6 = 100$ .
Relative frequency: $\frac{108}{600} = 0, 18$ .
The theoretical probability is $\frac{1}{6}$ , which is near $0, 17$ .
So $8$ extra sixes in $600$ rolls is an ordinary result.

Answer: no, a gap that size is ordinary

Even $600$ trials leave room for a wobble this small.

Traps

Putting the number of unwanted outcomes underneath instead of the sample space. The bottom of the fraction counts everything that can happen: all $20$ vouchers.
Giving the count of successes as the relative frequency. Divide that count by the number of trials: $\frac{12}{60} = 0, 2$ .
Answering with a probability when the question asked how many times. Multiply the probability by the number of trials to turn it into a count.
Expecting the next few rolls to even out the results so far. A die keeps no record of past rolls. Every roll starts fresh.

Learn and practise “Probability and frequency” in the app →

Venn diagrams and addition rule

Two overlapping circles hold everything you need for two events, once you know how to read them.

Venn diagram: A rectangle for the sample space, with a circle inside it for each event.
intersection: The overlap of the two circles. Its members are in both events at once.
union: Everything in either circle, with the overlap counted once, not twice.
addition rule: The rule that adds two probabilities and then removes the overlap once.

A Venn diagram draws a rectangle for the sample space, with a circle inside it for each event. Everything belongs somewhere in the rectangle. Where the two circles cross is the intersection, and whoever sits there belongs to both events.

Fill one in from the middle outwards. Write the intersection first. Take it off each event's total, and write what is left in the rest of that circle. Anything the circles do not account for goes in the rectangle, outside both.

Reading it takes care. "A and B" is the intersection. "A or B" is the union: either circle, with the overlap counted once. "Not A" is everything outside circle A, and part of circle B sits out there too.

To turn a count into a probability, divide by the size of the sample space. Never by the count in one circle. The whole class is the sample space, whether or not a learner falls inside a circle.

The addition rule comes straight off the picture. Adding $P (A)$ and $P (B)$ counts the intersection twice, so you take it off once. Given any three of the four probabilities, you can work back to the fourth.

Rules to remember

$P (A or B) =$ $P (A) + P (B) - P (A and B)$
$P (A and B)$ is the overlap, read off the diagram

Examples

In a class of

40

25

take a taxi to school and

16

buy lunch at the tuck shop.

7

do both. Fill in the four regions.

Worked answer

Start in the middle: $7$ sit in the intersection.
Taxi only: $25 - 7 = 18$ .
Tuck shop only: $16 - 7 = 9$ .
Inside the circles altogether: $18 + 7 + 9 = 34$ .
Outside both, still in the rectangle: $40 - 34 = 6$ .

Answer: $18$ taxi only, $7$ both, $9$ tuck shop only, $6$ neither

Writing $25$ and $16$ straight into the circles would count the $7$ twice.

One learner from that class is picked. Find the probability that the learner takes a taxi or buys at the tuck shop.

Worked answer

The union holds $18 + 7 + 9 = 34$ learners.
Over the whole class: $\frac{34}{40} = \frac{17}{20}$ .
The addition rule agrees: $\frac{25}{40} + \frac{16}{40} - \frac{7}{40} = \frac{34}{40}$ .
The $\frac{7}{40}$ came off once, because it had been added twice.

Answer: $\frac{17}{20}$

Counting the regions and using the addition rule must agree.

For two events,

P (A) = 0, 6

and

P (B) = 0, 5

. The probability of A or B is

0, 9

. Find the probability of A and B.

Worked answer

Put what you know into the addition rule.
$0, 6 + 0, 5 = 1, 1$ .
The union is only $0, 9$ , so the overlap is $1, 1 - 0, 9 = 0, 2$ .
Check: $0, 2$ lies between $0$ and $1$ , so it can be a probability.

Answer: $0, 2$

The rule takes the overlap off, so finding it means subtracting back.

Traps

Writing the full total for each event in its circle, so the overlap gets counted twice. Place the overlap first. Then subtract it from each total before filling in the rest of the circle.
Adding $P (A)$ and $P (B)$ without taking the overlap off. Subtract $P (A and B)$ once. It sits inside both events.
Multiplying $P (A)$ by $P (B)$ to get $P (A and B)$ . That shortcut needs independent events, which is Grade 11 work. Read $P (A and B)$ off the diagram instead.
Reading "or" as the overlap alone. "Or" covers either circle, overlap included. Count the whole union.

Learn and practise “Venn diagrams and addition rule” in the app →

Mutually exclusive events

Events that cannot both happen, events that leave nothing out, and the short cut each one gives you.

mutually exclusive: Two events that cannot both happen in the same trial.
complement: Everything in the sample space that is not in the event.
complementary events: Two events that cannot both happen and between them leave nothing out.
at least one: One or more. Its opposite is none at all.

Two events are mutually exclusive when they cannot both happen in one trial. Roll a die once. "The number is a two" and "the number is odd" cannot both come true. So there is no intersection at all, and $P (A and B) = 0$ .

Events that can happen together are simply not mutually exclusive. There is no other name for them at this level. "An even number" and "more than three" do overlap, because $4$ and $6$ are both.

With no overlap, the addition rule loses its last term and you just add. Check before you use that short form. Used on events that can happen together, it counts the overlap twice.

The complement of an event is everything else in the sample space. Its probability is $1$ minus the probability of the event. Subtract from $1$ , not from $100$ , unless the figures you were given are percentages.

Complementary events are mutually exclusive and they also fill the sample space. Being mutually exclusive is not enough on its own. A two and a five cannot both come up, but a four is neither of them. The complement of "at least one" is "none", and that is often the shorter route.

Rules to remember

$P (not A) = 1 - P (A)$
mutually exclusive: $P (A and B) = 0$ , so $P (A or B) = P (A) + P (B)$
complementary: mutually exclusive, and $P (A) + P (B) = 1$

Examples

A die is rolled once. Event A is "the number is a

2

". Event B is "the number is odd". Are they mutually exclusive, and what is the probability of A or B?

Worked answer

The number $2$ is not odd, so A and B cannot both happen.
They are mutually exclusive, so the overlap is $0$ .
$P (A)$ is $\frac{1}{6}$ and $P (B)$ is $\frac{3}{6}$ .
Nothing to take off, so add: $\frac{1}{6} + \frac{3}{6} = \frac{4}{6} = \frac{2}{3}$ .

Answer: yes, and $\frac{2}{3}$

Nothing was subtracted, because there was no overlap to subtract.

At a school, the probability that a learner walks to school is

0, 65

. Find the probability that a learner does not walk.

Worked answer

Walking and not walking cover every learner between them.
So they are complementary events, and one is the complement of the other.
$1 - 0, 65 = 0, 35$ .

Answer: $0, 35$

You subtract from $1$ , not from $100$ , because these are probabilities.

In a group of

50

learners,

8

play neither soccer nor netball. Find the probability that a learner plays at least one of the two.

Worked answer

The opposite of "at least one" is "neither".
$P (neither)$ is $\frac{8}{50}$ .
So take that off $1$ : $1 - \frac{8}{50} = \frac{42}{50}$ .
Simplify: $\frac{42}{50} = \frac{21}{25}$ .

Answer: $\frac{21}{25}$

Adding the three other regions works too, but this route is shorter.

Traps

Giving overlapping events a special name of their own, instead of just saying they overlap. There is no separate name for this at this level. If two events can happen together, say they are not mutually exclusive.
Subtracting a probability from $100$ instead of from $1$ . $P (not A) = 1 - P (A)$ . Use $100$ only when the figures are percentages.
Calling every pair of mutually exclusive events complementary. They must fill the sample space as well. A $2$ and a $5$ do not.
Using the short addition rule on events that can happen together. Drop the last term only once you have checked that the overlap is $0$ .

Learn and practise “Mutually exclusive events” in the app →

Grade 10 Probability

What gets asked

You must be able to

Traps that cost marks

Worked example

Probability and frequency

Rules to remember

Examples

Traps

Venn diagrams and addition rule

Rules to remember

Examples

Traps

Mutually exclusive events

Rules to remember

Examples

Traps

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