Probability — Mathslive.ie

Section 1 of 5

The Basics

Probability measures how likely an event $E$ is, on a scale from $0$ (impossible) to $1$ (certain).

The formula & two words

$P(E) = \dfrac{\text{number of desired outcomes}}{\text{total number of outcomes}}$. AND means multiply; OR means add.

Worked example — a single event

A bag has $5$ blue and $3$ green marbles. Find $P(\text{blue})$.

$P(\text{blue}) = \dfrac{5}{5+3}$

$\dfrac{5}{8}$

Section 2 of 5

Sample Space

List every equally-likely outcome, then count the ones you want.

Worked example — three coin flips

A fair coin is flipped $3$ times. Find the probability of two heads and one tail.

$8$ equally-likely outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT

Two heads & one tail: HHT, HTH, THH — that is $3$

$P = \dfrac{3}{8}$. (Note $P(\text{HHH}) = \dfrac{1}{8}$, so they are not equal.)

$\dfrac{3}{8}$

For two dice added, build a sample-space grid of the $36$ sums:

	1	2	3	4	5	6
1	2	3	4	5	6	7
2	3	4	5	6	7	8
3	4	5	6	7	8	9
4	5	6	7	8	9	10
5	6	7	8	9	10	11
6	7	8	9	10	11	12

Worked example — two dice

Two dice are thrown and the scores added. Find $P(\text{a total of }10\text{ or more})$.

Totals $\ge 10$: $10, 10, 10, 11, 11, 12$ — that is $6$ cells

$P = \dfrac{6}{36}$

$\dfrac{6}{36} = \dfrac{1}{6}$

Section 3 of 5

The Counting Principle

To count how many ways choices combine, multiply the number of options at each stage.

Worked example — John’s outfits

John packs $3$ pairs of jeans, $4$ shirts, $2$ jumpers and $3$ pairs of shoes. How many different outfits can he wear?

$3 \times 4 \times 2 \times 3$

$72$ outfits

Now you try

You try

A car comes in $4$ makes, $2$ door options, $4$ colours and $3$ fuel types. (i) How many combinations? (ii) Only four-door? (iii) $P(\text{electric, black, four-door BMW})$?

Pen and paper out — try it before you reveal.

(i) $4\times 2\times 4\times 3 = 96$

(ii) four-door fixes the doors to $1$ option: $4\times 1\times 4\times 3 = 48$

(iii) exactly one such car out of $96$

$96$; $48$; $P = \dfrac{1}{96}$

Section 4 of 5

Tree Diagrams

A tree diagram shows two stages. Multiply along the branches; add between them.

You try

A bag has $5$ blue and $3$ green marbles. One is drawn, its colour noted, and replaced; then a second is drawn. Find (i) $P(\text{both the same colour})$ and (ii) $P(\text{at least one blue})$.

Pen and paper out — try it before you reveal.

$P(BB) = \tfrac{5}{8}\times\tfrac{5}{8} = \tfrac{25}{64}$, $P(GG) = \tfrac{3}{8}\times\tfrac{3}{8} = \tfrac{9}{64}$

(i) both same $= \tfrac{25}{64} + \tfrac{9}{64} = \tfrac{34}{64} = \tfrac{17}{32}$

(ii) at least one blue $= BB + BG + GB = \tfrac{25}{64}+\tfrac{15}{64}+\tfrac{15}{64} = \tfrac{55}{64}$

(i) $\dfrac{17}{32}$ (ii) $\dfrac{55}{64}$

Section 5 of 5

Relative Frequency

Relative frequency (experimental probability) $=\dfrac{\text{result}}{\text{total number of trials}}$. It estimates probability from actual results.

You try

A basket’s fruit has relative frequencies: strawberry $0.2$, blueberry $y$, grape $0.37$, kiwi $0.23$. (i) Find $y$. (ii) In a basket of $20$ pieces, how many strawberries?

Pen and paper out — try it before you reveal.

(i) they add to $1$: $0.2 + y + 0.37 + 0.23 = 1 \Rightarrow y = 0.2$

(ii) $20 \times 0.2 = 4$

$y = 0.2$; $4$ strawberries

You try

A biased die is rolled $600$ times; the relative frequency of a $5$ is $0.25$. (i) How many $5$s came up? (ii) For a fair die rolled $600$ times, how many $5$s would you expect?

Pen and paper out — try it before you reveal.

(i) $0.25 \times 600 = 150$

(ii) fair die: $P(5) = \tfrac{1}{6}$, so $\tfrac{1}{6}\times 600 = 100$

(i) $150$ (ii) $100$

That’s Probability.

The formula, sample spaces, the counting principle, tree diagrams and relative frequency — the whole of Junior Cert probability.