Probability with Venn Diagrams

Lessons

Senior 3 · Mathematics

Probability with Venn Diagrams

3 lessons

Lesson 1 of 3

Meaning of the probability of an event

25 min

A sample space is a universal set

Choose one of the whole numbers from 1 to 20 without looking, so that each is equally likely to be chosen. How likely is it that you choose a prime number? In unit 1 you drew sets inside a rectangle called the universal set. Probability uses exactly the same picture: the rectangle holds everything that can happen, and a circle inside it holds the outcomes you are interested in. The probability is then a comparison of two counts, the number of outcomes in the circle against the number in the rectangle.

In S1 and S2 you met probability as "favourable outcomes over possible outcomes". This unit writes the same idea in the language of sets, because once probability is a set problem, every tool from unit 1 (Venn diagrams, unions, intersections, complements and the formula for $n(A \cup B)$ ) becomes a probability tool.

Definition

An experiment is an action with an uncertain result, such as choosing a number at random.
An outcome is one possible result of the experiment.
The sample space $S$ is the set of all possible outcomes. It plays the part of the universal set.
An event is a subset of the sample space, written with a capital letter: $E \subseteq S$ . The event happens when the outcome is one of its elements.

Example: for "choose a number from 1 to 20", $S = \{1, 2, 3, \ldots, 20\}$ and $n (S) = 20$ . The event "the number is prime" is $E = \{2, 3, 5, 7, 11, 13, 17, 19\}$ , with $n (E) = 8$ .

Important

If all the outcomes in $S$ are equally likely, the probability of an event $E$ is

P(E) = \frac{n(E)}{n(S)}.

Why it lies between 0 and 1: every event satisfies $\varnothing \subseteq E \subseteq S$ , so $0 \le n(E) \le n(S)$ . Dividing by $n (S)$ gives $0 \le P(E) \le 1$ . The impossible event is the empty set, with $P(\varnothing) = 0$ ; the certain event is $S$ itself, with $P (S) = 1$ .

Worked example 1: the numbers 1 to 20

A number is chosen at random from 1 to 20. Find the probability that it is (a) prime, (b) a multiple of 6, (c) a multiple of 25, (d) less than 21.

Solution $n (S) = 20$ and every number is equally likely.

(a) $P(E) = \frac{8}{20} = \frac{2}{5}$ , using the eight primes in the diagram.

(b) The multiples of 6 are $\{6, 12, 18\}$ , so the probability is $\frac{3}{20}$ .

(c) No number from 1 to 20 is a multiple of 25. The event is $\varnothing$ , so the probability is 0: the event is impossible.

(d) Every number in $S$ is less than 21. The event is $S$ itself, so the probability is 1: the event is certain.

Definition

The complement of an event $E$ , written $E^{'}$ ("not $E$ "), is the set of outcomes in $S$ that are not in $E$ . On a Venn diagram it is everything inside the rectangle and outside the circle.

Example: in the diagram above, $E^{'}$ is the twelve numbers from 1 to 20 that are not prime, including 1, which is neither prime nor composite.

Important

The complement rule. Each outcome is either in $E$ or in $E^{'}$ , never both, so $n (E) + n (E^{'}) = n (S)$ . Divide by $n (S)$ :

P(E) + P(E') = 1 \qquad \text{so} \qquad P(E') = 1 - P(E).

For the primes, $P(E') = 1 - \frac{2}{5} = \frac{3}{5}$ , which matches counting directly: $\frac{12}{20} = \frac{3}{5}$ .

Worked example 2: choosing a month

One month of the year is chosen at random, each month equally likely. Let $A$ be "the month has 31 days" and $B$ be "the month's name begins with J". List $A$ , $A^{'}$ and $B$ , and find $P (A)$ , $P (A^{'})$ and $P (B)$ .

Solution $n (S) = 12$ .

$A = \{\text{Jan, Mar, May, Jul, Aug, Oct, Dec}\}$ , so $P(A) = \frac{7}{12}$ .

$A' = \{\text{Feb, Apr, Jun, Sep, Nov}\}$ , so $P(A') = \frac{5}{12}$ . Check: $\frac{7}{12} + \frac{5}{12} = 1$ .

$B = \{\text{Jan, Jun, Jul}\}$ , so $P(B) = \frac{3}{12} = \frac{1}{4}$ .

Notice that January and July belong to both $A$ and $B$ . Lesson 2 draws $A$ and $B$ as two overlapping circles and uses that overlap.

Definition

When outcomes are not equally likely, probability is estimated from data. The relative frequency of an event is

\text{relative frequency} = \frac{\text{number of times the event happened}}{\text{number of trials}}.

With many trials it is a good estimate of the probability, and the expected number of times the event happens in $N$ further trials is $N \times P(E)$ . When one person is chosen at random from a whole population, the probability of a property is simply the proportion of the population that has it.

Worked example 3: a person chosen from the 2022 census

The 2022 census counted 13,246,394 residents of Rwanda, of whom 6,817,068 were female and 1,745,555 lived in the City of Kigali (NISR). One resident is chosen at random. Find, to 3 decimal places, the probability that the person (a) is female, (b) lives in Kigali, (c) lives outside Kigali.

Solution Here $S$ is the set of all 13,246,394 residents, each equally likely to be chosen.

(a) $P(\text{female}) = \frac{6\,817\,068}{13\,246\,394} \approx 0.515$ .

(b) $P(\text{Kigali}) = \frac{1\,745\,555}{13\,246\,394} \approx 0.132$ .

(c) "Outside Kigali" is the complement of "in Kigali": $1 - 0.132 = 0.868$ . Using the unrounded value, $1 - 0.13178 \ldots \approx 0.868$ , the same to 3 decimal places.

Worked example 4: a germination test

Before planting, a farmer tests 200 bean seeds from a new bag, and 172 of them germinate (illustrative data). (a) Estimate the probability that a seed from the bag germinates. (b) The farmer plants 5,000 seeds from the same bag. About how many plants should the farmer expect?

Solution (a) The relative frequency is $\frac{172}{200} = 0.86$ . (b) Expected number $= 5000 \times 0.86 = 4300$ plants. This is an estimate: the real number will vary a little, and a larger test would make the estimate more reliable.

Warning

Equally likely is a condition, not a habit. "A football match can be won, drawn or lost, so the probability of a win is $\frac{1}{3}$ " is wrong: the three outcomes are not equally likely for two teams of different strength. Use $P(E) = \frac{n(E)}{n(S)}$ only when every outcome in $S$ has the same chance; otherwise use data.

Tip

Before calculating any probability, write two lines: $S = \ldots$ with $n (S)$ , and $E = \ldots$ with $n (E)$ . Most errors in probability come from a wrong count of $S$ , and writing it out catches them.

Khan Academy: Probability explained. Watch how the probability of an event is found by counting the outcomes in the event and in the whole sample space, the counting rule in the box above.

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Practice

A number is chosen at random from 1 to 30. Find the probability that it is (a) a multiple of 4, (b) not a multiple of 4.
A letter is chosen at random from the 26 letters of the English alphabet. Find the probability that it is (a) a vowel (a, e, i, o, u), (b) a consonant.
Which of these numbers cannot be the probability of an event? $0.9$ , $- 0.2$ , $\frac{7}{5}$ , $0$ , $1$ , $35 %$ . Give a reason.
The probability that a bulb from a box is faulty is 0.03. Find the probability that a bulb chosen from the box is not faulty.
A spinner has 8 equal sectors numbered 1 to 8. Let $A$ be "the number is greater than 5". List $A$ and $A^{'}$ , and find $P (A)$ and $P (A^{'})$ .
The 2022 census counted 3,563,145 people in the Eastern Province, out of 13,246,394 in Rwanda (NISR). A resident is chosen at random. Find, to 3 decimal places, the probability that the person lives (a) in the Eastern Province, (b) outside it.
In a germination test, 45 of 60 maize seeds germinated. How many plants should a farmer expect from 2,000 of these seeds?
Gatete says: "Tomorrow it will either rain or not rain, so the probability of rain is $\frac{1}{2}$ ." Explain his mistake.

Show answersHide answers

(a) Multiples of 4: $\{4, 8, 12, 16, 20, 24, 28\}$ , so $\frac{7}{30}$ . (b) $1 - \frac{7}{30} = \frac{23}{30}$ .
(a) $\frac{5}{26}$ . (b) $1 - \frac{5}{26} = \frac{21}{26}$ .
$- 0.2$ is less than 0, and $\frac{7}{5} = 1.4$ is greater than 1. All the others lie between 0 and 1 inclusive (35% = 0.35).
$1 - 0.03 = 0.97$ .
$A = \{6, 7, 8\}$ and $A' = \{1, 2, 3, 4, 5\}$ ; $P(A) = \frac{3}{8}$ , $P(A') = \frac{5}{8}$ .
(a) $\frac{3\,563\,145}{13\,246\,394} \approx 0.269$ . (b) $1 - 0.26899 \ldots \approx 0.731$ .
$P(\text{germinates}) \approx \frac{45}{60} = 0.75$ ; expected number $= 2000 \times 0.75 = 1500$ .
"Rain" and "no rain" are the only two outcomes, but they are not equally likely: on most days in the dry season rain is far less likely than no rain. The probability must come from weather data, not from counting outcomes.

Sources and credits

Census figures (total, females, City of Kigali and Eastern Province): Fifth Rwanda Population and Housing Census 2022, Main Indicators Report, National Institute of Statistics of Rwanda (NISR). Official statistics, cited as facts.
Video: Probability explained, Khan Academy, embedded from YouTube.
Written by Ganzaa. The seed and bulb data are illustrative, and the diagrams, worked examples and practice items were written by Ganzaa.

1A sample space is a universal set

A sample space is a universal set