Probability as a Measure of Chance

Probability measures chance by a number from 0 to 1: equally likely outcomes, the probability formula, impossible and certain events, and the probability scale.

What You Should Already Know

From lesson 13.1: the sample space is the list of all possible outcomes of a probability experiment. The sample space of rolling a die is $\{1, 2, 3, 4, 5, 6\}$ , with 6 outcomes.
Simplifying a fraction, and changing a fraction to a decimal and a percentage: $\frac{6}{8} = \frac{3}{4} = 0.75 = 75\%\text{.}$

Measuring Chance

In Primary school you described chance in words: an event is likely or unlikely. Words are not exact. "Unlikely" could mean a 1 in 3 chance or a 1 in 1000 chance.

Probability measures chance by a number from 0 to 1. An event that cannot happen has probability 0; an event that must happen has probability 1. Every other event has a probability between them, and the larger the probability, the more likely the event.

Word	Probability
Impossible	exactly $0$
Unlikely	more than $0$ and less than $\frac{1}{2}$
Even chance	exactly $\frac{1}{2}$
Likely	more than $\frac{1}{2}$ and less than $1$
Certain	exactly $1$

"Unlikely" and "likely" are ranges, not points: a probability of $\frac{1}{10}$ and a probability of $\frac{2}{5}$ are both unlikely, though the first is far less likely than the second.

Equally Likely Outcomes and the Probability Formula

When a coin is tossed, a head and a tail have the same chance if the coin is evenly made. We say the two outcomes are equally likely, and that the coin is fair (or unbiased).

A coin that has been bent, or a die with a weight hidden in one side, may land one way more often than another. Its outcomes are not equally likely, and it is biased. In this chapter a coin or a die is taken to be fair only when the question says so.

Event and favourable outcomes

An event is a group of outcomes of an experiment, described in words: "an even number" when a die is rolled is the event made of the outcomes 2, 4 and 6. An event may have no outcomes at all, such as rolling a 7 with a die.

The outcomes that make up an event are its favourable outcomes. The probability of an event $E$ is written $P(E)$ .

The probability of an event

When all the possible outcomes are equally likely,

$P(E) = \dfrac{\text{number of favourable outcomes}}{\begin{matrix}\text{total number of} \\ \text{possible outcomes}\end{matrix}}$

This holds only for equally likely outcomes. Lesson 13.5 deals with outcomes that are not.

Example 1: A fair coin and a fair die

Find the probability of (a) getting a head when a fair coin is tossed; (b) getting a 4 when a fair die is rolled; (c) getting an even number when a fair die is rolled.

Solution

(a) Sample space $= \{H, T\}$ : 2 equally likely outcomes, since the coin is fair. One of them, H, is favourable.

$P(\text{head}) = \frac{1}{2}$ .

(b) Sample space $= \{1, 2, 3, 4, 5, 6\}$ : 6 equally likely outcomes. One of them, 4, is favourable.

$P(4) = \frac{1}{6}$ .

$P(\text{even number}) = \frac{3}{6} = \frac{1}{2}\text{.}$

(a) $\frac{1}{2}$ ; (b) $\frac{1}{6}$ ; (c) $\frac{1}{2}$ .

Example 2: Why every ball is counted

A bag contains 2 identical black balls and 3 identical white balls. One ball is drawn at random. Find the probability that it is (a) black; (b) white.

Solution

Sample space $= \{B_1, B_2, W_1, W_2, W_3\}$ (lesson 13.1). The ball is drawn at random, so the 5 outcomes are equally likely.

(a) Favourable outcomes: $B_1$ , $B_2$ . $P(\text{black}) = \frac{2}{5}$ .

(b) Favourable outcomes: $W_1$ , $W_2$ , $W_3$ . $P(\text{white}) = \frac{3}{5}$ .

Listing the sample space as {Black, White} and answering $\frac{1}{2}$ would be wrong: black and white are not equally likely, but the 5 balls are.

(a) $P(\text{black}) = \frac{2}{5}$ ; (b) $P(\text{white}) = \frac{3}{5}$ .

Impossible and Certain Events

From 0 to 1

An impossible event has no favourable outcomes: $P(E) = \frac{0}{\text{total}} = 0\text{.}$ Example: rolling a 7 with a fair die.

A certain event has every outcome favourable: $P(E) = \frac{\text{total}}{\text{total}} = 1\text{.}$ Example: rolling a number less than 7 with a fair die.

The number of favourable outcomes is never less than 0 and never more than the total, so for every event $0 \le P(E) \le 1$ .

Example 3: Impossible, certain or neither

A bag contains 5 red balls and 3 blue balls. One ball is drawn at random. Say whether each event is impossible, certain or neither, and find its probability. (a) A red or blue ball. (b) A green ball. (c) A red ball. (d) A ball that is not purple.

Solution

Total number of possible outcomes $= 5 + 3 = 8$ , all equally likely.

(a) Every ball is red or blue: 8 favourable outcomes. $P = \frac{8}{8} = 1$ . Certain.

(b) There is no green ball: 0 favourable outcomes. $P = \frac{0}{8} = 0$ . Impossible.

(d) No ball is purple, so every ball is "not purple": $P = \frac{8}{8} = 1$ . Certain.

(a) certain, 1; (b) impossible, 0; (c) neither, $\frac{5}{8}$ ; (d) certain, 1.

Example 4: Which numbers can be probabilities?

Which of these could be the probability of an event? Give a reason for each. (a) $0.75$ (b) $1.5$ (c) $-0.2$ (d) $0$ (e) $1$ (f) $\frac{7}{5}$ (g) $105\%$

Solution

(a) $0.75$ is between 0 and 1: yes.

(b) $1.5$ is more than 1: no. An event cannot have more favourable outcomes than there are outcomes.

(d) $0$ : yes, an impossible event.

(e) $1$ : yes, a certain event.

(f) $\frac{7}{5} = 1.4$ , more than 1: no.

(g) $105\% = 1.05$ , more than 1: no.

(a), (d) and (e) could be probabilities; (b), (c), (f) and (g) could not.

Fraction, Decimal or Percentage

Writing a probability

A probability is usually given as a fraction in its simplest form: $\frac{6}{8}$ is written $\frac{3}{4}$ .

It may also be written as a decimal or a percentage when a question asks: $\frac{3}{4} = 0.75 = 75\%$ . All three say the same chance.

Example 5: Three ways to write one probability

A box contains 25 counters, of which 9 are green. One counter is taken at random. Find the probability that it is green, as a fraction, a decimal and a percentage.

Solution

Favourable outcomes: 9 green counters. Total: 25 counters, equally likely.

$P(\text{green}) = \frac{9}{25}\text{,}$ already in simplest form (9 and 25 have no common factor other than 1).

As a decimal: $\frac{9}{25} = \frac{36}{100} = 0.36\text{.}$ As a percentage: $0.36 = 36\%$ .

$P(\text{green}) = \frac{9}{25} = 0.36 = 36\%\text{.}$

Placing and Comparing Probabilities

Example 6: Events on the probability scale

A fair die is rolled once. Find the probability of each event, describe it in words, and mark it on the probability scale. A: a 7. B: a number less than 7. C: an even number. D: a 6. E: a number greater than 2.

Solution

Total number of possible outcomes $= 6$ , equally likely as the die is fair.

A: no face shows 7. $P(A) = 0$ : impossible.

B: all 6 faces. $P(B) = \frac{6}{6} = 1$ : certain.

C: 2, 4, 6. $P(C) = \frac{3}{6} = \frac{1}{2}$ : even chance.

D: one face. $P(D) = \frac{1}{6}$ : unlikely.

E: 3, 4, 5, 6. $P(E) = \frac{4}{6} = \frac{2}{3}$ : likely.

$P(A) = 0$ , $P(B) = 1$ , $P(C) = \frac{1}{2}$ , $P(D) = \frac{1}{6}$ , $P(E) = \frac{2}{3}$ , placed as shown.

Example 7: Which is most likely?

Arrange these events in order, from the least likely to the most likely. A: getting a head when a fair coin is tossed. B: getting a number greater than 4 when a fair die is rolled. C: drawing a red ball at random from a bag of 7 red and 3 yellow balls.

Solution

$P(A) = \frac{1}{2}$ .

B: 5 and 6 are favourable. $P(B) = \frac{2}{6} = \frac{1}{3}$ .

C: $P(C) = \frac{7}{10}$ .

Compare as decimals: $\frac{1}{3} = 0.333\ldots$ , $\frac{1}{2} = 0.5$ , $\frac{7}{10} = 0.7$ . (Or with a common denominator of 30: $\frac{10}{30}$ , $\frac{15}{30}$ , $\frac{21}{30}$ .)

From least to most likely: B, A, C.

Common Mistakes

Using the formula when the outcomes are not equally likely. "It rains tomorrow or it does not, so the probability of rain is $\frac{1}{2}$ " is wrong: the two outcomes need not have the same chance.
Counting colours instead of objects: for 2 black and 3 white balls, $P(\text{black}) = \frac{2}{5}$ , not $\frac{1}{2}$ .
Giving a probability above 1 or below 0. Such an answer shows a mistake in the counting.
Leaving a fraction unsimplified: $\frac{3}{6}$ should be written $\frac{1}{2}$ .
Treating "unlikely" as one value such as $\frac{1}{4}$ . Every probability between 0 and $\frac{1}{2}$ is unlikely.

Practice

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

Question 1

Describe each event as impossible, unlikely, even chance, likely or certain. (a) A fair die shows a number less than 10. (b) A fair coin shows a tail. (c) A fair die shows a 6. (d) A ball drawn at random from a bag of 9 blue balls and 1 red ball is blue. (e) A spinner coloured only red and blue lands on yellow.

ShowHide solution

(a) Every face is less than 10: $P = 1$ , certain.

(b) $P = \frac{1}{2}$ : even chance.

(d) $P = \frac{9}{10}$ , between $\frac{1}{2}$ and 1: likely.

(e) There is no yellow sector: $P = 0$ , impossible.

(a) certain; (b) even chance; (c) unlikely; (d) likely; (e) impossible.

Question 2

Are the outcomes equally likely? Explain. (a) Head and tail when a fair coin is tossed. (b) Head and tail when a bent coin is tossed. (c) The four colours of a spinner with four equal sectors coloured red, blue, green and yellow. (d) "Rain" and "no rain" tomorrow.

ShowHide solution

(a) Yes: "fair" means the outcomes are equally likely.

(b) Not necessarily: bending may make it land one way more often. It may be biased.

(d) Not necessarily: there are two outcomes, but nothing makes them equally likely. Weather depends on the season and the clouds, not on a count of outcomes.

(a) yes; (b) not necessarily; (c) yes; (d) not necessarily (we cannot assume so).

Question 3

Ten cards are numbered 1 to 10, and one card is drawn at random. For each event, say whether it is impossible, certain or neither, and find its probability. (a) A number less than 11. (b) A number greater than 10. (c) A factor of 10. (d) A whole number.

ShowHide solution

Total number of possible outcomes $= 10$ , equally likely as the card is drawn at random.

(a) 1 to 10 are all less than 11: $\frac{10}{10} = 1$ , certain.

(b) No card is greater than 10: $\frac{0}{10} = 0$ , impossible.

(d) Every card shows a whole number: $\frac{10}{10} = 1$ , certain.

(a) certain, 1; (b) impossible, 0; (c) neither, $\frac{2}{5}$ ; (d) certain, 1.

Question 4

Which of these could be probabilities? (a) $0.3$ (b) $\frac{4}{3}$ (c) $-\frac{1}{2}$ (d) $100\%$ (e) $0.05$ (f) $1.01$

ShowHide solution

A probability must be from 0 to 1, both included.

(a) yes. (b) $\frac{4}{3} > 1$ : no. (c) negative: no. (d) $100\% = 1$ : yes, a certain event. (e) yes. (f) more than 1: no.

(a), (d) and (e) could be probabilities; (b), (c) and (f) could not.

Question 5

A bag contains 4 red counters and 6 white counters. One counter is taken at random. Find the probability that it is (a) red; (b) white; (c) red or white; (d) black.

ShowHide solution

Total $= 4 + 6 = 10$ counters, equally likely.

(a) $\frac{4}{10} = \frac{2}{5}$ . (b) $\frac{6}{10} = \frac{3}{5}$ . (c) every counter: $\frac{10}{10} = 1$ . (d) none: $0$ .

(a) $\frac{2}{5}$ ; (b) $\frac{3}{5}$ ; (c) $1$ ; (d) $0$ .

Question 6

In a survey of how pupils travel to school, 14 of the 40 pupils in a class said they walk. One pupil of the class is chosen at random. Find the probability that the pupil walks to school, as a fraction in simplest form, a decimal and a percentage.

ShowHide solution

$P(\text{walks}) = \frac{14}{40} = \frac{7}{20}$ (dividing both by 2).

$\frac{7}{20} = \frac{35}{100} = 0.35 = 35\%\text{.}$

$\frac{7}{20} = 0.35 = 35\%\text{.}$

Question 7

The spinner shown has 8 equal sectors numbered 1 to 8. It is spun once. Find the probability of each event and mark the events on a probability scale. A: a 9. B: a number less than 4. C: an odd number. D: a number greater than 1.

ShowHide solution

8 equally likely outcomes, as the sectors are equal.

A: no sector shows 9. $P(A) = 0$ .

B: 1, 2, 3. $P(B) = \frac{3}{8}$ .

C: 1, 3, 5, 7. $P(C) = \frac{4}{8} = \frac{1}{2}$ .

D: 2 to 8. $P(D) = \frac{7}{8}$ .

$P(A) = 0$ , $P(B) = \frac{3}{8}$ , $P(C) = \frac{1}{2}$ , $P(D) = \frac{7}{8}$ , placed as shown.

Question 8

Arrange these events from the least likely to the most likely. P: drawing a red marble at random from a bag of 3 red and 9 blue marbles. Q: drawing a multiple of 3 at random from cards numbered 1 to 10. R: a spinner with 5 equal sectors, one of them green, landing on green.

ShowHide solution

P: $\frac{3}{12} = \frac{1}{4} = 0.25\text{.}$

Q: the multiples of 3 are 3, 6, 9: $\frac{3}{10} = 0.3$ .

R: $\frac{1}{5} = 0.2$ .

$0.2 < 0.25 < 0.3$ .

R, P, Q.

Question 9

For Teachers' Day, the names of all 200 pupils in Secondary 2 are put in a box, and one name is drawn at random to present the flowers. 5 of the names are from Hui Min's class. Find the probability that the name drawn is from her class, as a fraction, a decimal and a percentage. Is it likely or unlikely?

ShowHide solution

The 200 names are equally likely to be drawn, and 5 of them are from Hui Min's class.

$P(\text{name from her class}) = \frac{5}{200} = \frac{1}{40}\text{.}$

$\frac{1}{40} = 0.025 = 2.5\%\text{.}$

It is between 0 and $\frac{1}{2}$ (and very close to 0).

$\frac{1}{40} = 0.025 = 2.5\%\text{;}$ unlikely.

Question 10

A bag has 1 red ball and 9 blue balls. Daniel says: "When I draw a ball at random, either I get red or I do not. That is 2 outcomes, so $P(\text{red}) = \frac{1}{2}$ ." Explain his mistake and find $P(\text{red})$ .

ShowHide solution

"Red" and "not red" are not equally likely: "not red" stands for 9 balls and "red" for 1. The formula needs equally likely outcomes.

The 10 balls are equally likely, and 1 of them is red: $P(\text{red}) = \frac{1}{10}$ .

His two outcomes are not equally likely; $P(\text{red}) = \frac{1}{10}$ .

Key Points

Probability measures chance by a number from 0 to 1: $0 \le P(E) \le 1$ .
Impossible: $P = 0$ . Certain: $P = 1$ . Even chance: $P = \frac{1}{2}$ . Unlikely and likely are the ranges between.
A fair coin or die has equally likely outcomes; a biased one need not.
For equally likely outcomes, $P(E) =$ number of favourable outcomes $\div$ total number of possible outcomes.
Count objects, not colours: every ball, card or sector is its own outcome.
Give a probability as a fraction in simplest form, or as a decimal or percentage when asked.
The larger the probability, the more likely the event: compare probabilities as decimals or with a common denominator.