C8.1; E8.1 - Single-event probability

C8.1; E8.1 - Single-event probability

Probability measures chance on a scale from impossible to certain. For Core and Extended, C8.1/E8.1 connects that scale to equally likely outcomes and information in tables, graphs and two-set Venn diagrams. You will calculate single-event probabilities and use complements to find the chance that an event does not happen.

Probability scale and equally likely outcomes

A bag contains three red counters and one blue counter. Is blue just as likely as red because there are two colours? Imagine choosing one counter with each individual counter equally likely. Three of the four possible counters are red: counting colour names would miss the different numbers of counters.

Outcome, event and probability

An outcome is one possible result of an experiment. An event is a specified collection of outcomes, such as “the counter is red”. Its probability measures how likely the event is, from 00 for impossible to 11 for certain.

For equally likely outcomes in a finite complete list,

probability of an event=number of favourable outcomestotal number of outcomes.\text{probability of an event}=\frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}.

“Favourable” means it satisfies the event, whether or not you personally want it to happen.

For the bag, red has probability 34\frac34 and blue 14\frac14. A probability may also be a decimal or percentage: 34=0.75=75%\frac34=0.75=75\%. These describe the same chance, not three different predictions.

[DIAGRAM: asset_name: The probability scale; asset_slug: probability_scale; caption: Probabilities run from zero to one, with an even chance at one half.; description: A horizontal evenly spaced scale from 0 to 1 has marked positions 0, one quarter, one half, three quarters and 1. The endpoints are labelled impossible and certain, the centre even chance, and the quarter and three-quarter positions less likely and more likely than not.]
Probabilities run from zero to one, with an even chance at one half.

An event with probability 12\frac12 has an even chance. A probability below 12\frac12 is less likely to happen than not; one above 12\frac12 is more likely to happen than not. Values such as −0.2-0.2 and 1.31.3 cannot be probabilities.

Worked example 1: equally likely die faces

Question. A fair six-sided die is numbered 11 to 66. Find the probability of rolling 55.

Plan. Count the favourable faces among the six equally likely faces.

favourable faces=1only the face labelled 5probability=16one of six equal chances\begin{aligned} \text{favourable faces}&=1 &&\text{only the face labelled 5}\\ \text{probability}&=\frac16 &&\text{one of six equal chances} \end{aligned}

Answer. 16\frac16.

Check. Each of the six face events has probability 16\frac16, and 6×16=16\times\frac16=1, accounting for every possible roll.

Keep the die, but change only the event. One event can contain several outcomes.

Worked example 2: an event with several outcomes

Question. Find the probability of rolling a number greater than 44 on the same die.

Plan. List the favourable faces before forming the fraction.

favourable faces={5,6}both exceed fourprobability=26=13two of six equal chances\begin{aligned} \text{favourable faces}&=\{5,6\} &&\text{both exceed four}\\ \text{probability}&=\frac26=\frac13 &&\text{two of six equal chances} \end{aligned}

Answer. 13\frac13.

Check. The other four faces give probability 46=23\frac46=\frac23; together 13+23=1\frac13+\frac23=1.

The equal-chance condition matters. On a biased die, the six face labels still exist, but dividing by six need not give their probabilities. You would need information about their actual chances. Random means the particular result is uncertain; it does not by itself make a chosen list of categories equally likely.

Watch out

“Red or blue, so the probability of blue is 12\frac12” is wrong for the opening bag. The four counters are equally likely, while the two colours are not. The correct probability is 14\frac14.

For the die below, the equal-chance condition is supplied; identify the favourable faces.

Counting a single event

When choosing a person at random from a group, make the intended model explicit: every person is equally likely to be chosen. Then a table, bar chart or Venn diagram can supply the favourable count and the whole-group count. Read what the numbers represent before calculating.

The chart shows the travel methods of 2020 students. Each student is in exactly one category, so the bar heights add to the total.

[DIAGRAM: asset_name: Travel-method counts; asset_slug: probability_travel_counts; caption: Use the frequency scale to count students, rather than counting the three bars.; description: A bar chart has categories Walk, Bus and Cycle with frequencies 8, 7 and 5. The vertical frequency axis starts at zero, is marked in steps of one and reaches ten. Each bar has its frequency printed above it.]
Use the frequency scale to count students, rather than counting the three bars.

Worked example 3: reading a bar chart

Question. One of these 2020 students is chosen, each equally likely. Find the probability that the student cycles.

Plan. Read the cycle count and divide by the total number of students.

total=8+7+5=20all three separate categoriesprobability of cycling=520=14five favourable students\begin{aligned} \text{total}&=8+7+5=20 &&\text{all three separate categories}\\ \text{probability of cycling}&=\frac5{20}=\frac14 &&\text{five favourable students} \end{aligned}

Answer. 14\frac14.

Check. Five is one quarter of twenty, and the other 1515 students form the remaining three quarters.

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