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Probability · Lesson 1 of 2

Probability and sample spaces

Probability as favourable outcomes over possible outcomes: listing sample spaces, using tables of results, the complement, and 'or' with and without overlap.

16 minYou should already know: Number foundations & fractions Sets & Venn diagrams
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Probability measures how likely something is, on a scale from 0 (impossible) to 1 (certain). When all the outcomes are equally likely:

P(event)=outcomes in the eventpossible outcomesP(\text{event}) = \frac{\text{outcomes in the event}}{\text{possible outcomes}}

For example, a bag holds 3 red, 5 green and 4 blue beads, all the same size. There are 12 equally likely beads to pick, and 5 of them are green. So P(green)=512P(\text{green}) = \frac{5}{12}.

P(green) = 5⁄12
Favourable over possible5 green beads out of 12 equally likely beads

The complement: “not”

Something either happens or it doesn’t, so

P(not A)=1−P(A)P(\text{not } A) = 1 - P(A)

This is often the quickest way in, especially for “at least one” questions.

P(A)P(not A)together they make 1
The complementP(A) + P(not A) = 1

More: one event

Listing the sample space

The sample space is the list of all possible outcomes. For two dice there are 6×6=366 \times 6 = 36, and a table is the clearest way to list them.

Two dice: the sample spacePick an event
second diefirst die112233445566234567345678456789567891067891011789101112
6outcomes where the sum is 736outcomes in all1/6P(sum is 7)
6 of the 36 equally likely outcomes, so P(sum is 7) = 6/36 = 1/6. The sum 7 has the most ways of any total: it runs corner to corner.

Pick an event to see its outcomes shaded. Then press Your turn and tap them yourself before checking. Counting correctly is the whole skill.

More: listing the sample space

Probability from a table of results

When a die has been thrown many times, the probability of a score is estimated by its relative frequency:

P(score)=frequency of that scoretotal frequencyP(\text{score}) = \frac{\text{frequency of that score}}{\text{total frequency}}
182123104951467P(5) ≈ 14⁄60
Relative frequencyFrequency of the score ÷ total number of throws

Worked example · WAEC 2016

WAEC 2016 · Paper 2 · Q4

Score 1 2 3 4 5 6
Frequency 25 30 xx 28 40 32

The table shows the outcome when a die is thrown a number of times. If the probability of obtaining a 3 is 0.225:

How many times was the die thrown?

Calculate the probability that a trial chosen at random gives a score of an even number or a prime number.

  1. Form an equation

    x155+x=0.225\frac{x}{155 + x} = 0.225

    Think first. The total frequency is 155+x155 + x. Write P(3)P(3) as a fraction.

  2. (a) Solve

    • Multiply both sides by 155+x{155 + x}: x=0.225(155+x){x = 0.225(155 + x)}.
    • Multiply out: x=34.875+0.225x{x = 34.875 + 0.225x}.
    • 0.775x=34.875{0.775x = 34.875}, so x=45{x = 45}.
    • The die was thrown 155+45=200{155 + 45 = 200} times.

  3. (b) Even or prime

    • Even: 2, 4, 6. Prime: 2, 3, 5.
    • Together: 2, 3, 4, 5, 6. 2 is counted once, and 1 is neither.
    • Their frequencies: 30+45+28+40+32=175{30 + 45 + 28 + 40 + 32 = 175}.
    • P=175200=78{P = \frac{175}{200} = \frac78}.

    Think first. Which scores are even or prime? Is 1 one of them?

More: probability from a table or chart

“Or”: adding probabilities

If two events can’t happen together, they’re called mutually exclusive. Add their probabilities:

P(A or B)=P(A)+P(B)P(A \text{ or } B) = P(A) + P(B)

If they can happen together, the outcomes in both would be counted twice. So take them away once:

P(A or B)=P(A)+P(B)−P(A and B)\begin{aligned} &P(A \text{ or } B) \\ &= P(A) + P(B) - P(A \text{ and } B) \end{aligned}
UAB
Mutually exclusiveNo overlap: P(A or B) = P(A) + P(B)
UAB
OverlappingAdding P(A) and P(B) counts the middle twice, so subtract P(A and B) once

This is the same idea as the overlap in a Venn diagram.

Worked example · WAEC 2020

WAEC 2020 · Paper 2 · Q5

A number is selected at random from the set S={1,2,3,…,24,25}S = \{1, 2, 3, \ldots, 24, 25\}. Find the probability that the number selected is:

even;

prime;

either even or prime;

both even and prime.

  1. (a) Even

    2, 4, …, 24: 12 numbers. P(even)=1225{P(\text{even}) = \frac{12}{25}}.

    Think first. How many even numbers from 1 to 25?

  2. (b) Prime

    2, 3, 5, 7, 11, 13, 17, 19, 23: 9 numbers. P(prime)=925{P(\text{prime}) = \frac{9}{25}}.

    Think first. List the primes up to 25.

  3. (d) Both

    Only 2. P(both)=125{P(\text{both}) = \frac{1}{25}}.

    Think first. Which numbers are even and prime?

  4. (c) Either

    P(even or prime)=1225+925−125=2025=45\begin{aligned} &P(\text{even or prime}) \\ &= \frac{12}{25} + \frac{9}{25} - \frac{1}{25} \\ &= \frac{20}{25} = \frac45 \end{aligned}

    Taking away 125\frac{1}{25} stops the number 2 being counted twice.

More: or, and Venn diagrams

Finding a missing number

Some questions give the probability and ask how many there are. Write the probability as a fraction with the unknown in it and solve.

Your turn

WAEC 2012 · Paper 2 · Q4 (a)

  1. (a)

    A box contains 40 identical discs which are either red or white. If the probability of picking a red disc is 14\frac14, calculate the number of: (i) white discs; (ii) red discs that should be added such that the probability of picking a red disc will be 13\frac13.

    Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

  1. P(red)=red discs40P(\text{red}) = \frac{\text{red discs}}{40}
    =14= \frac14, so there are 14×40=10\frac14 \times 40 = 10 red discs and 40−10=3040 - 10 = 30 white discs.

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