Probability When Outcomes Are Not Equally Likely

Spinners with sectors of different sizes, where the probability is the angle of the sector divided by 360°, and a point chosen at random on a board of equal squares.

What You Should Already Know

From lesson 13.2: $P(E) =$ number of favourable outcomes $\div$ total number of possible outcomes, but only when the outcomes are equally likely.
The angles at the centre of a circle add up to $360^\circ$ . A sector of $90^\circ$ is $\frac{90}{360} = \frac{1}{4}$ of the circle.

Equal and Unequal Sectors

Spinner A has 5 equal sectors of $72^\circ$ . The pointer is as likely to stop on one as on another, so the 5 colours are equally likely and each has probability $\frac{1}{5}$ .

Spinner B has 4 colours, but its sectors are not equal. Counting colours would give $P(\text{Green}) = \frac{1}{4}$ , which is wrong: the green sector is half the circle, and the pointer stops in it about half the time.

The probability of a sector

The pointer of a spinner is equally likely to stop in any direction. So the chance of stopping in a sector is the fraction of the full turn that the sector takes up:

$P(\text{sector}) = \dfrac{\text{angle of the sector}}{360^\circ}$

The same fraction gives the probability that a point chosen at random inside the circle lies in the sector, since the sector takes up that fraction of the area.

Example 1: Spinner B

Spinner B above is spun once. Find the probability that it lands on each colour, and check that the probabilities add up to 1.

Solution

$P(\text{Green}) = \frac{180^\circ}{360^\circ} = \frac{1}{2}\text{,}$ $P(\text{Red}) = \frac{90^\circ}{360^\circ} = \frac{1}{4}\text{.}$

$P(\text{Blue}) = \frac{45^\circ}{360^\circ} = \frac{1}{8}\text{,}$ $P(\text{Pink}) = \frac{45^\circ}{360^\circ} = \frac{1}{8}\text{.}$

Check: $\frac{4}{8} + \frac{2}{8} + \frac{1}{8} + \frac{1}{8} = \frac{8}{8} = 1\text{.}$ The pointer always stops on one of the four colours.

$P(\text{Green}) = \frac{1}{2}$ , $P(\text{Red}) = \frac{1}{4}$ , $P(\text{Blue}) = \frac{1}{8}$ , $P(\text{Pink}) = \frac{1}{8}$ .

Example 2: Three angles given

The spinner shown has a green sector of $160^\circ$ , a red sector of $90^\circ$ , a pink sector of $50^\circ$ and a blue sector. It is spun once. Find the probability that it lands on (a) green; (b) blue; (c) black.

Solution

(a) $P(\text{green}) = \frac{160^\circ}{360^\circ} = \frac{4}{9}$ (dividing both by 40).

(b) The angles add up to $360^\circ$ , so the blue angle $= 360^\circ - 160^\circ - 90^\circ - 50^\circ = 60^\circ\text{.}$

$P(\text{blue}) = \frac{60^\circ}{360^\circ} = \frac{1}{6}\text{.}$

(a) $\frac{4}{9}$ ; (b) $\frac{1}{6}$ ; (c) $0$ .

Sectors Given in Other Ways

Example 3: Fractions of the circle

A spinner is made of a red, a blue and a green sector. The red sector takes up $\frac{1}{3}$ of the circle and the blue sector $\frac{1}{8}$ . Find (a) the probability of landing on green; (b) the angle of the green sector.

Solution

The fraction of the circle a sector takes up is its probability, and the fractions add up to 1.

(a) $P(\text{green}) = 1 - \frac{1}{3} - \frac{1}{8} = \frac{24}{24} - \frac{8}{24} - \frac{3}{24} = \frac{13}{24}\text{.}$

(b) Angle $= \frac{13}{24} \times 360^\circ = 13 \times 15^\circ = 195^\circ\text{.}$

Check: red $\frac{1}{3} \times 360^\circ = 120^\circ\text{,}$ blue $\frac{1}{8} \times 360^\circ = 45^\circ\text{,}$ and $120^\circ + 45^\circ + 195^\circ = 360^\circ\text{.}$

(a) $\frac{13}{24}$ ; (b) $195^\circ$ .

Example 4: Equal parts with colours repeated

A spinner has 8 equal sectors: 4 red, 2 blue, 1 green and 1 yellow, as shown. Find the probability of landing on each colour.

Solution

The 8 sectors are equal, so they are 8 equally likely outcomes (lesson 13.1: tell the red ones apart as $R_1$ to $R_4$ ). The colours are not equally likely.

$P(\text{red}) = \frac{4}{8} = \frac{1}{2}\text{,}$ $P(\text{blue}) = \frac{2}{8} = \frac{1}{4}\text{,}$ $P(\text{green}) = \frac{1}{8}$ , $P(\text{yellow}) = \frac{1}{8}\text{.}$

The same by angles: red takes $4 \times 45^\circ = 180^\circ$ , and $\frac{180^\circ}{360^\circ} = \frac{1}{2}\text{.}$

Red $\frac{1}{2}$ , blue $\frac{1}{4}$ , green $\frac{1}{8}$ , yellow $\frac{1}{8}$ .

Example 5: Angles in algebra

A spinner has three sectors A, B and C with angles $x^\circ$ , $2x^\circ$ and $3x^\circ$ . Find $x$ , and the probability of landing on each sector.

Solution

The angles add up to $360^\circ$ : $x + 2x + 3x = 360$ , so $6x = 360$ and $x = 60$ .

The angles are $60^\circ$ , $120^\circ$ and $180^\circ$ .

$P(A) = \frac{60}{360} = \frac{1}{6}\text{,}$ $P(B) = \frac{120}{360} = \frac{1}{3}\text{,}$ $P(C) = \frac{180}{360} = \frac{1}{2}\text{.}$

Check: $\frac{1}{6} + \frac{2}{6} + \frac{3}{6} = 1\text{.}$

$x = 60$ ; $P(A) = \frac{1}{6}$ , $P(B) = \frac{1}{3}$ , $P(C) = \frac{1}{2}$ .

Numbered Spinners

When an event is made of several sectors, add their angles first, then divide by $360^\circ$ .

Example 6: An even number

A spinner has sectors numbered 1, 2, 3 and 4 with angles $120^\circ$ , $90^\circ$ , $60^\circ$ and $90^\circ$ , as shown. It is spun once. Find the probability of getting (a) an even number; (b) a number greater than 2; (c) a 5.

Solution

(a) Even: sectors 2 and 4, $90^\circ + 90^\circ = 180^\circ\text{.}$ $P(\text{even}) = \frac{180}{360} = \frac{1}{2}\text{.}$

(b) Greater than 2: sectors 3 and 4, $60^\circ + 90^\circ = 150^\circ\text{.}$ $P = \frac{150}{360} = \frac{5}{12}\text{.}$

Note that "2 of the 4 numbers are even, so $\frac{2}{4}$ " happens to give $\frac{1}{2}$ here, but the method is wrong: (b) by counting would give $\frac{2}{4} = \frac{1}{2}$ , not $\frac{5}{12}$ .

(a) $\frac{1}{2}$ ; (b) $\frac{5}{12}$ ; (c) $0$ .

A Point Chosen at Random on a Board

A board of equal squares

A point chosen at random on a board is equally likely to be anywhere on it. When the board is made of equal small squares, each square is equally likely to contain the point, so

$P(\text{shaded}) = \dfrac{\text{number of shaded squares}}{\text{total number of squares}}$

Example 7: A shaded board

A board is made of 24 equal squares, 9 of them shaded, as shown. A point on the board is chosen at random. Find the probability that it lies (a) in a shaded square; (b) in an unshaded square.

Solution

(a) $P(\text{shaded}) = \frac{9}{24} = \frac{3}{8}\text{.}$

(b) $24 - 9 = 15$ unshaded squares: $\frac{15}{24} = \frac{5}{8}$ (or $1 - \frac{3}{8} = \frac{5}{8}$ ).

(a) $\frac{3}{8}$ ; (b) $\frac{5}{8}$ .

Common Mistakes

Counting colours on a spinner with unequal sectors: 4 colours does not mean $\frac{1}{4}$ each.
Forgetting to find a missing angle from $360^\circ$ before dividing.
Dividing by the number of sectors instead of by $360^\circ$ .
Using $x$ as the angle after solving: when the sectors are $x^\circ$ , $2x^\circ$ , $3x^\circ$ , sector B is $2x = 120^\circ$ , not $60^\circ$ .
Leaving $\frac{150}{360}$ unsimplified: it is $\frac{5}{12}$ .

Practice

Try each question first, then open the solution to check your working. Give each probability as a fraction in its simplest form.

Question 1

A spinner has a red sector of $72^\circ$ , a blue sector of $108^\circ$ and a green sector of $180^\circ$ . It is spun once. Find the probability of landing on each colour, and of not landing on blue. Check that the three probabilities add up to 1.

ShowHide solution

$P(\text{red}) = \frac{72}{360} = \frac{1}{5}\text{,}$ $P(\text{blue}) = \frac{108}{360} = \frac{3}{10}\text{,}$ $P(\text{green}) = \frac{180}{360} = \frac{1}{2}\text{.}$

$P(\text{not blue}) = 1 - \frac{3}{10} = \frac{7}{10}\text{.}$

Check: $\frac{2}{10} + \frac{3}{10} + \frac{5}{10} = 1\text{.}$

Red $\frac{1}{5}$ , blue $\frac{3}{10}$ , green $\frac{1}{2}$ ; not blue $\frac{7}{10}$ .

Question 2

A spinner has a yellow sector of $100^\circ$ , a purple sector of $140^\circ$ and an orange sector. Find the probability of landing on (a) orange; (b) yellow; (c) purple.

ShowHide solution

Orange angle $= 360^\circ - 100^\circ - 140^\circ = 120^\circ\text{.}$

(a) $\frac{120}{360} = \frac{1}{3}$ . (b) $\frac{100}{360} = \frac{5}{18}$ . (c) $\frac{140}{360} = \frac{7}{18}$ .

(a) $\frac{1}{3}$ ; (b) $\frac{5}{18}$ ; (c) $\frac{7}{18}$ .

Question 3

On a spinner, the blue sector takes up $\frac{1}{6}$ of the circle, the red sector $\frac{1}{2}$ , and the rest is white. Find the probability of landing on white, and the angle of the white sector.

ShowHide solution

$P(\text{white}) = 1 - \frac{1}{6} - \frac{1}{2} = \frac{6}{6} - \frac{1}{6} - \frac{3}{6} = \frac{2}{6} = \frac{1}{3}\text{.}$

Angle $= \frac{1}{3} \times 360^\circ = 120^\circ\text{.}$

$\frac{1}{3}$ ; $120^\circ$ .

Question 4

A spinner has four sectors A, B, C and D with angles $x^\circ$ , $(x + 30)^\circ$ , $2x^\circ$ and $90^\circ$ . Find $x$ and the probability of landing on each sector.

ShowHide solution

$x + (x + 30) + 2x + 90 = 360\text{,}$ so $4x + 120 = 360$ , $4x = 240$ , $x = 60$ .

Angles: A $60^\circ$ , B $90^\circ$ , C $120^\circ$ , D $90^\circ$ (total $360^\circ$ ).

$P(A) = \frac{60}{360} = \frac{1}{6}\text{,}$ $P(B) = \frac{90}{360} = \frac{1}{4}\text{,}$ $P(C) = \frac{120}{360} = \frac{1}{3}\text{,}$ $P(D) = \frac{1}{4}$ .

$x = 60$ ; $P(A) = \frac{1}{6}$ , $P(B) = \frac{1}{4}$ , $P(C) = \frac{1}{3}$ , $P(D) = \frac{1}{4}$ .

Question 5

A spinner has sectors numbered 1 to 5 with angles $40^\circ$ , $80^\circ$ , $100^\circ$ , $60^\circ$ and $80^\circ$ . It is spun once. Find the probability of getting (a) an even number; (b) a prime number; (c) a number less than 3.

ShowHide solution

(a) 2 and 4: $80 + 60 = 140$ . $\frac{140}{360} = \frac{7}{18}$ .

(b) 2, 3 and 5: $80 + 100 + 80 = 260$ . $\frac{260}{360} = \frac{13}{18}$ .

(a) $\frac{7}{18}$ ; (b) $\frac{13}{18}$ ; (c) $\frac{1}{3}$ .

Question 6

A board is made of 20 equal squares: 6 are red, 4 are blue and the rest are white. A point on the board is chosen at random. Find the probability that it lies in (a) a red square; (b) a blue square; (c) a square that is neither red nor blue.

ShowHide solution

(a) $\frac{6}{20} = \frac{3}{10}$ . (b) $\frac{4}{20} = \frac{1}{5}$ .

(a) $\frac{3}{10}$ ; (b) $\frac{1}{5}$ ; (c) $\frac{1}{2}$ .

Question 7

At a Children's Day class party, pupils take turns to spin a spinner, and a pupil gets a sticker when the pointer stops on the yellow sector, whose angle is $40^\circ$ . (a) Find the probability of getting a sticker on one spin. (b) During the party, the spinner is spun 180 times. About how many stickers are given out?

ShowHide solution

(a) $\frac{40}{360} = \frac{1}{9}$ .

(b) $\frac{1}{9} \times 180 = 20$ (lesson 13.4).

(a) $\frac{1}{9}$ ; (b) about 20 stickers.

Question 8

A spinner has a red sector of $180^\circ$ , a blue sector of $120^\circ$ and a green sector of $60^\circ$ . Jia Hui says $P(\text{red}) = \frac{1}{3}$ , because there are 3 colours. Explain her mistake and find $P(\text{red})$ .

ShowHide solution

The three sectors are not equal, so the three colours are not equally likely, and the probability formula of lesson 13.2 does not apply.

$P(\text{red}) = \frac{180}{360} = \frac{1}{2}\text{.}$

The colours are not equally likely; $P(\text{red}) = \frac{1}{2}$ .

Question 9

A spinner has a green sector of $180^\circ$ , and a red and a blue sector. The angle of the red sector is twice the angle of the blue sector. Find the probability of landing on each colour.

ShowHide solution

Let the blue angle be $x^\circ$ ; the red angle is $2x^\circ$ .

$x + 2x + 180 = 360$ , so $3x = 180$ and $x = 60$ . Blue $60^\circ$ , red $120^\circ$ .

$P(\text{blue}) = \frac{60}{360} = \frac{1}{6}\text{,}$ $P(\text{red}) = \frac{120}{360} = \frac{1}{3}\text{,}$ $P(\text{green}) = \frac{180}{360} = \frac{1}{2}\text{.}$

Blue $\frac{1}{6}$ , red $\frac{1}{3}$ , green $\frac{1}{2}$ .

Key Points

The probability formula of lesson 13.2 needs equally likely outcomes; the colours of a spinner with unequal sectors are not.
$P(\text{sector}) = \text{angle of the sector} \div 360^\circ\text{:}$ the fraction of the circle it takes up.
The angles add up to $360^\circ$ and the probabilities to 1: use this to find a missing angle or check your answers.
For an event made of several sectors, add their angles, then divide by $360^\circ$ .
With angles in algebra, solve for $x$ from the total $360^\circ$ , then work out each angle.
On a board of equal squares, $P(\text{shaded}) =$ shaded squares $\div$ all the squares.