A fair die is rolled. P(an even number) is:

1/2. Three of the six faces (2, 4, 6) are even, so P = 3/6 = 1/2.

If P(E) = 0.28, then P(E′) is:

0.72. The complement rule gives 1 − 0.28 = 0.72. An event and its complement always account for the whole sample space, so they must total 1.

Two coins are tossed. P(exactly one head) is:

1/2. The sample space is {HH, HT, TH, TT}. Two of the four outcomes have exactly one head, so P = 2/4 = 1/2. Treating HT and TH as the same outcome gives the wrong answer 1/3.

A card is drawn from 52. P(a red king) is:

2/52. There are exactly two red kings, hearts and diamonds, so P = 2/52 = 1/26. Option B counts all four kings, including the black ones.

Which value could NOT be a probability?

5/4. 5/4 = 1.25 exceeds 1, which is impossible: an event cannot be more than certain. Zero and one are both perfectly valid, representing impossible and certain events.

A bag has 4 red and 6 blue balls. Two are drawn with replacement. P(both red) is:

16/100. With replacement the bag is restored, so both draws have P(red) = 4/10 and the events are independent: (4/10)(4/10) = 16/100 = 4/25. Option B would be the answer without replacement.

Events A and B are mutually exclusive with P(A) = 0.3 and P(B) = 0.45. P(A or B) is:

0.75. Mutually exclusive means the overlap is zero, so the general rule reduces to simple addition: 0.3 + 0.45 = 0.75. Option A multiplies instead of adding.

P(at least one head in three tosses of a fair coin) is:

7/8. Use the complement: P(no heads) = (1/2)³ = 1/8, so P(at least one) = 1 − 1/8 = 7/8. Listing all the favourable cases would take seven lines to reach the same answer.

Probability — Mathematics Notes, Worked Examples and Quiz

A number for how likely something is

An experiment is any process with an uncertain result — rolling a die, drawing a card, checking whether a bulb is faulty. The set of everything that could happen is the sample space S, and an event is any subset of it.

When every outcome is equally likely, the probability of an event is simply the proportion of the sample space it occupies.

P(E) = n(E) / n(S)0 ≤ P(E) ≤ 1P(E′) = 1 − P(E)(the complement)P = 0 means impossible, P = 1 means certain; nothing can fall outside that range

n(E): the number of favourable outcomes
n(S): the total number of possible outcomesall equally likely
E′: the event "E does not happen"

Theoretical and experimental probability

Theoretical probability is calculated from the structure of the experiment: a fair die has six faces, so P(4) = 1/6. Experimental probability, or relative frequency, comes from actually performing trials: if a 4 appeared 47 times in 300 rolls, the experimental probability is 47/300. The more trials you run, the closer the two tend to get.

Listing the sample space properly

Most lost marks in this chapter come from an incomplete sample space rather than from any error in the arithmetic. For two dice there are 36 outcomes, not 12 and not 21 — (2, 5) and (5, 2) are different outcomes and both must be counted.

For two-stage experiments, a table or a grid is the safest way to list them. For sequences of events with different probabilities, a tree diagram is better.

The complement is usually the short cut

Questions asking for "at least one" are almost always faster the other way round. P(at least one six in two rolls) requires you to count three separate cases; 1 − P(no six at all) = 1 − (5/6)(5/6) = 11/36 takes one line. Whenever you see "at least", consider the complement first.

Combining events: and, or

Two rules cover every combined-event question at this level. Getting them the right way round is a matter of reading whether the question says and or or.

Events are mutually exclusive when they cannot both happen — a single card cannot be both a king and a queen. Events are independent when one happening does not change the probability of the other — two separate coin tosses.

P(A or B) = P(A) + P(B) − P(A and B)P(A or B) = P(A) + P(B)(mutually exclusive only)P(A and B) = P(A) × P(B)(independent only)add for OR, multiply for AND — but check the condition before dropping the correction term

Worked example

A card is drawn from a standard pack of 52. Find the probability that it is a king or a heart.

P(king) = 4/52 and P(heart) = 13/52.Four kings, thirteen hearts.
These are not mutually exclusive: the king of hearts is both.Check for overlap before adding. This is the whole question.
P(king and heart) = 1/52.There is exactly one card that is both.
P(king or heart) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13.Without subtracting the overlap you would count the king of hearts twice and get 17/52.

4/13

Tree diagrams: with and without replacement

A tree diagram handles experiments in stages. Each branch carries a probability; you multiply along a path to get the probability of that complete outcome, and add the paths that satisfy the question.

The one thing to watch is whether the first item is replaced. With replacement, the second set of branches is identical to the first and the events are independent. Without replacement, both the numerator and the denominator change, and the events are dependent.

Switch to without replacement. The second-stage branches change, because there is one ball fewer in the bag — and possibly one fewer of the colour just taken. The first-stage branches never change.

The things that are always true

Use these as a check on every answer you write. If any of them is violated, the arithmetic is wrong somewhere, and finding out before the examiner does is free marks.

Every probability lies between 0 and 1 inclusive. A negative answer, or 7/5, means an arithmetic error.
The probabilities of all the outcomes in a sample space add to exactly 1.
P(E) + P(E′) = 1, always.
An impossible event has probability 0; a certain event has probability 1.
Probability says nothing about a single trial. A coin that has landed heads eight times running is still 50/50 on the ninth — coins have no memory.

Before you leave this chapter

P(E) = favourable outcomes ÷ total equally likely outcomes.
List the sample space completely — two dice give 36 outcomes, not 21.
"At least one" is nearly always faster as 1 − P(none).
Add for OR (subtracting the overlap unless the events are mutually exclusive); multiply for AND when the events are independent.
On a tree diagram, multiply along the branches and add the paths.

Relative frequency, and testing whether something is fair

Theoretical probability assumes the die is fair. Relative frequency makes no such assumption: it simply records what happened. Roll the die 600 times, count the sixes, divide by 600.

The two are linked by the law of large numbers: as the number of trials grows, the relative frequency settles towards the theoretical probability. That is what makes relative frequency useful as evidence — and what makes a small number of trials useless as evidence.

Worked example

A die is rolled 600 times and a six appears 137 times. Is there evidence that the die is biased?

Relative frequency of a six = 137/600 = 0.228.This is the experimental probability, taken straight from the results.
Theoretical probability for a fair die = 1/6 = 0.167.
Expected number of sixes = 600 × 1/6 = 100, against 137 observed.Comparing counts is easier to talk about than comparing decimals.
137 is well above 100 over a large number of trials, so there is evidence of bias towards six.With only 6 rolls, getting 2 sixes would be no evidence at all — the sample size is what makes the conclusion worth stating.

Yes — 0.228 against an expected 0.167 over 600 trials suggests bias.

Expected frequency

The expected number of times an event occurs in n trials is n × P(event). It is the most common one-mark question in this part of the chapter, and it is worth noting that the answer need not be a whole number — an expected 16.7 sixes in 100 rolls is a perfectly good answer.

Probability