Single-stage outcomes and probability

A bag has three red tokens and one blue token. Every individual token is equally likely to be selected once. Which statement is correct?

Name the experiment, trial, outcome and event

A chance experiment has a result that is not known in advance, such as one roll of a fair die

One performance of the experiment is a trial. In a single-stage experiment, one selection or action produces the recorded result

An outcome is one possible result. The sample space contains every possible recorded outcome under the stated model

An event is a specified collection of outcomes, such as rolling an even number. It can contain one result, several results, no results or all results

State what is recorded. Recording the colour of a token is different from recording the identity of the individual token

List all results in the stated coin model

A fair coin is tossed once. The model records only Heads or Tails, each equally likely

List every possible recorded result of the stated coin experiment, with each result written once

One trial produces one recorded label

Read every label that can result: Heads, Tails

List each possible recorded label once, even if several equally likely items carry it

The recorded sample space is \(\{\text{Heads},\,\text{Tails}\}\)

Label multiplicities: Heads: 1; Tails: 1

Here each label occurs equally often, so the label chances are equal

A list of distinct labels does not by itself make their probabilities equal; use the number of elementary items carrying each label

Select the complete die sample space

Repeated labels do not create new recorded results

A spinner can have several sectors with the same label. If the result records only the label, write that label once in the sample space

For six equal sectors labelled red, red, red, blue, blue and green, the colour sample space is {red, blue, green}

The six sectors are equally likely, but the three recorded colours are not: they occupy three, two and one sectors

When calculating probabilities, keep the counts of equally likely sectors or items instead of dividing only by the number of distinct labels

Separate sector identities from recorded colours

An ideal spinner has 6 equal-sized sectors labelled as shown. It is spun once; each sector is equally likely. Record its label

List every possible recorded colour for this spinner and explain why the colour probabilities need not be equal

One trial produces one recorded label

Read every label that can result: red, blue, green

List each possible recorded label once, even if several equally likely items carry it

The recorded sample space is \(\{\text{red},\,\text{blue},\,\text{green}\}\)

Label multiplicities: red: 3; blue: 2; green: 1

Here the multiplicities differ, so the recorded labels have unequal chances even though every elementary outcome is equally likely

A list of distinct labels does not by itself make their probabilities equal; use the number of elementary items carrying each label

Retain every possible label exactly once

Divide favourable outcomes by all equally likely outcomes

When every elementary outcome is equally likely, event probability equals favourable elementary outcomes divided by all elementary outcomes

For three red tokens and one blue token, \(P(\text{red})=\frac{3}{4}\) and \(P(\text{blue})=\frac{1}{4}\)

Reduce a fraction without changing its value. A probability of zero is impossible under the model; a probability of one is certain

No probability is negative or greater than one. These boundaries refer to the stated model and recorded result

Fairness must be supplied or justified. A list of possible labels alone does not show that the labels are equally likely

Assign both probabilities for a fair coin

A fair coin is tossed once. The model records only Heads or Tails, each equally likely

Assign an exact probability to every possible recorded label

Give probabilities as simplified fractions or whole numbers. Keep every requested calculation exact without rounding

There are 2 equally likely elementary outcomes, counted before combining repeated labels

Heads occurs on 1 of the 2 equally likely outcomes, so its probability is \(\frac{1}{2}\)

Tails occurs on 1 of the 2 equally likely outcomes, so its probability is \(\frac{1}{2}\)

The labels account for every elementary outcome exactly once

Their probabilities add to \(\frac{2}{2} = 1\) and each lies between zero and one

Assign unequal probabilities to two colours

A bag holds 4 tokens identical except for their labels. Mix them and select one without looking; each individual token is equally likely. Record its label

Assign an exact probability to every possible recorded label

Give probabilities as simplified fractions or whole numbers. Keep every requested calculation exact without rounding

There are 4 equally likely elementary outcomes, counted before combining repeated labels

red occurs on 3 of the 4 equally likely outcomes, so its probability is \(\frac{3}{4}\)

blue occurs on 1 of the 4 equally likely outcomes, so its probability is \(\frac{1}{4}\)

The labels account for every elementary outcome exactly once

Their probabilities add to \(\frac{4}{4} = 1\) and each lies between zero and one

Complete both colour probabilities

Account for every category in a bag

A bag holds 6 tokens identical except for their labels. Mix them and select one without looking; each individual token is equally likely. Record its label

Assign an exact probability to every possible recorded label

Give probabilities as simplified fractions or whole numbers. Keep every requested calculation exact without rounding

There are 6 equally likely elementary outcomes, counted before combining repeated labels

plain occurs on 2 of the 6 equally likely outcomes, so its probability is \(\frac{2}{6} = \frac{1}{3}\)

striped occurs on 3 of the 6 equally likely outcomes, so its probability is \(\frac{3}{6} = \frac{1}{2}\)

spotted occurs on 1 of the 6 equally likely outcomes, so its probability is \(\frac{1}{6}\)

The labels account for every elementary outcome exactly once

Their probabilities add to \(\frac{6}{6} = 1\) and each lies between zero and one

Use the total item count for every category

Check a complete probability assignment

The possible recorded labels are mutually separate results of one trial: exactly one recorded label occurs

Count all equally likely elementary outcomes once when forming the denominator

Add the probability assigned to every recorded label. The complete assignment must total one

This check does not make labels equally likely. Equal probabilities occur only when the corresponding favourable counts are equal

Combine repeated equal sectors

An ideal spinner has 6 equal-sized sectors labelled as shown. It is spun once; each sector is equally likely. Record its label

Assign an exact probability to every possible recorded label

Give probabilities as simplified fractions or whole numbers. Keep every requested calculation exact without rounding

There are 6 equally likely elementary outcomes, counted before combining repeated labels

red occurs on 3 of the 6 equally likely outcomes, so its probability is \(\frac{3}{6} = \frac{1}{2}\)

blue occurs on 2 of the 6 equally likely outcomes, so its probability is \(\frac{2}{6} = \frac{1}{3}\)

green occurs on 1 of the 6 equally likely outcomes, so its probability is \(\frac{1}{6}\)

The labels account for every elementary outcome exactly once

Their probabilities add to \(\frac{6}{6} = 1\) and each lies between zero and one

Assign probabilities from eight equal sectors

Check four probabilities against the whole spinner

An ideal spinner has 10 equal-sized sectors labelled as shown. It is spun once; each sector is equally likely. Record its label

Assign an exact probability to every possible recorded label

Give probabilities as simplified fractions or whole numbers. Keep every requested calculation exact without rounding

There are 10 equally likely elementary outcomes, counted before combining repeated labels

red occurs on 4 of the 10 equally likely outcomes, so its probability is \(\frac{4}{10} = \frac{2}{5}\)

blue occurs on 3 of the 10 equally likely outcomes, so its probability is \(\frac{3}{10}\)

green occurs on 2 of the 10 equally likely outcomes, so its probability is \(\frac{2}{10} = \frac{1}{5}\)

yellow occurs on 1 of the 10 equally likely outcomes, so its probability is \(\frac{1}{10}\)

The labels account for every elementary outcome exactly once

Their probabilities add to \(\frac{10}{10} = 1\) and each lies between zero and one

Assign probabilities to four recorded labels

A result named zero can still have positive probability

There are 6 same-size cards with the distinct numbers listed below. Select one card without looking after mixing; each card is equally likely

Assign an exact probability to every possible recorded label

Give probabilities as simplified fractions or whole numbers. Keep every requested calculation exact without rounding

There are 6 equally likely elementary outcomes, counted before combining repeated labels

0 occurs on 1 of the 6 equally likely outcomes, so its probability is \(\frac{1}{6}\)

1 occurs on 1 of the 6 equally likely outcomes, so its probability is \(\frac{1}{6}\)

2 occurs on 1 of the 6 equally likely outcomes, so its probability is \(\frac{1}{6}\)

3 occurs on 1 of the 6 equally likely outcomes, so its probability is \(\frac{1}{6}\)

4 occurs on 1 of the 6 equally likely outcomes, so its probability is \(\frac{1}{6}\)

5 occurs on 1 of the 6 equally likely outcomes, so its probability is \(\frac{1}{6}\)

The labels account for every elementary outcome exactly once

Their probabilities add to \(\frac{6}{6} = 1\) and each lies between zero and one

One of six equally likely cards is labelled 0 and the others are labelled 1, 2, 3, 4 and 5. What is the probability of selecting the card labelled 0?

Identify all outcomes satisfying the event

An event condition must be checked against the actual listed outcomes, not against an assumed standard device

For cards numbered 1 to 6, the event even number contains 2, 4 and 6

Use the count of matching elementary outcomes, not the sum of their number labels

For repeated category labels, add the counts of all items carrying the requested label

Count favourable tokens before reducing

A bag holds 6 tokens identical except for their labels. Mix them and select one without looking; each individual token is equally likely. Record its label

The event is: the recorded label is red

Count the outcomes and calculate the event probability

Give probabilities as simplified fractions or whole numbers. Keep every requested calculation exact without rounding

Count all 6 equally likely elementary outcomes

The labels satisfying the event are red

These labels account for 2 favourable elementary outcomes

Probability = favourable count divided by total count = \(\frac{2}{6} = \frac{1}{3}\)

Distinguish a label count from the number of categories

Use the earlier number definitions carefully

An even whole number is divisible by two. Zero is even

A prime number is an integer greater than one with exactly two positive factors. Zero and one are not prime

A multiple of three is obtained by three times a whole number; zero is a multiple of three

Less than and greater than are strict conditions: the boundary value itself does not qualify

List every even number in the given record

There are 6 same-size cards with the distinct numbers listed below. Select one card without looking after mixing; each card is equally likely

The event is: the card number is even

Count the outcomes and calculate the event probability

Give probabilities as simplified fractions or whole numbers. Keep every requested calculation exact without rounding

Count all 6 equally likely elementary outcomes

The labels satisfying the event are 2, 4, 6

These labels account for 3 favourable elementary outcomes

Probability = favourable count divided by total count = \(\frac{3}{6} = \frac{1}{2}\)

Use the actual card labels, including zero

Check prime labels rather than odd labels

There are 6 same-size cards with the distinct numbers listed below. Select one card without looking after mixing; each card is equally likely

The event is: the card number is prime

Count the outcomes and calculate the event probability

Give probabilities as simplified fractions or whole numbers. Keep every requested calculation exact without rounding

Count all 6 equally likely elementary outcomes

The labels satisfying the event are 2, 3, 5

These labels account for 3 favourable elementary outcomes

Probability = favourable count divided by total count = \(\frac{3}{6} = \frac{1}{2}\)

Include two and exclude one and composite odd values

Count multiples without adding their labels

There are 6 same-size cards with the distinct numbers listed below. Select one card without looking after mixing; each card is equally likely

The event is: the card number is a multiple of 3

Count the outcomes and calculate the event probability

Give probabilities as simplified fractions or whole numbers. Keep every requested calculation exact without rounding

Count all 6 equally likely elementary outcomes

The labels satisfying the event are 3, 6

These labels account for 2 favourable elementary outcomes

Probability = favourable count divided by total count = \(\frac{2}{6} = \frac{1}{3}\)

Retain zero when it satisfies the multiple condition

Exclude a strict boundary from the event

There are 6 same-size cards with the distinct numbers listed below. Select one card without looking after mixing; each card is equally likely

The event is: the card number is less than 5

Count the outcomes and calculate the event probability

Give probabilities as simplified fractions or whole numbers. Keep every requested calculation exact without rounding

Count all 6 equally likely elementary outcomes

The labels satisfying the event are 2, 3, 4

These labels account for 3 favourable elementary outcomes

Probability = favourable count divided by total count = \(\frac{3}{6} = \frac{1}{2}\)

Count only labels strictly above the boundary

Calculate impossible and certain cases directly

If no available outcome satisfies the stated event, its favourable count is zero and its probability is zero

If every available outcome satisfies the event, favourable and total counts are equal and the probability is one

These results follow directly from counting. No complement calculation is needed

Keep the stated model in view: an unlisted colour cannot appear when the record lists every possible token label

No favourable outcomes gives probability zero

A bag holds 4 tokens identical except for their labels. Mix them and select one without looking; each individual token is equally likely. Record its label

The event is: the selected item has a label absent from the list

Count the outcomes and calculate the event probability

Give probabilities as simplified fractions or whole numbers. Keep every requested calculation exact without rounding

Count all 4 equally likely elementary outcomes

The labels satisfying the event are none

These labels account for 0 favourable elementary outcomes

Probability = favourable count divided by total count = \(\frac{0}{4} = 0\)

A different total does not change impossibility

All outcomes favourable gives probability one

A bag holds 4 tokens identical except for their labels. Mix them and select one without looking; each individual token is equally likely. Record its label

The event is: the selected item has any of the listed labels

Count the outcomes and calculate the event probability

Give probabilities as simplified fractions or whole numbers. Keep every requested calculation exact without rounding

Count all 4 equally likely elementary outcomes

The labels satisfying the event are red, blue

These labels account for 4 favourable elementary outcomes

Probability = favourable count divided by total count = \(\frac{4}{4} = 1\)

Count the whole collection once

A bag has r red tokens and three blue tokens, each individual token equally likely. Move r from one to five. The decimal probability readout is a rounded display; the exact red probability is r divided by (r plus three)

Which statement is true for these settings?

Predict a proportion and an expected count

With a fixed chance model, the predicted long-run relative frequency of an event is its probability

If the probability is p and the experiment is repeated n times under unchanged conditions, the expected number is n × p

For probability 3/4 over twenty trials, the expected number is 20 × 3/4 = 15

This is a model prediction, not a promise that exactly fifteen successes will be observed

When drawing items repeatedly, replace and remix after every trial so that the same one-stage model applies again

Turn three quarters into an expected count

A bag holds 4 tokens identical except for their labels. Mix them and select one without looking; each individual token is equally likely. Record its label

The event is: the recorded label is red

Repeat the same trial 20 times under unchanged conditions, replacing and remixing the selected item after every trial

Predict the long-run relative frequency and the expected number, not a guaranteed observed result

Calculate the model predictions for the stated repeated experiment

Give probabilities as simplified fractions or whole numbers. Keep every requested calculation exact without rounding

Count all 4 equally likely elementary outcomes

The labels satisfying the event are red

These labels account for 3 favourable elementary outcomes

Probability = favourable count divided by total count = \(\frac{3}{4}\)

The predicted long-run relative frequency equals the model probability \(\frac{3}{4}\)

Multiply by one hundred to express it as 75%

Expected number = \(20 \times \frac{3}{4} = 15\)

An expectation can be fractional; it is an average prediction, not part of one observed event

A particular finite run need not match these predictions exactly

Use the probability rather than the favourable count alone

Predict a fair coin proportion and count

A fair coin is tossed once. The model records only Heads or Tails, each equally likely

The event is: the recorded label is Heads

Repeat the same trial 60 times under unchanged conditions

Predict the long-run relative frequency and the expected number, not a guaranteed observed result

Calculate the model predictions for the stated repeated experiment

Give probabilities as simplified fractions or whole numbers. Keep every requested calculation exact without rounding

Count all 2 equally likely elementary outcomes

The labels satisfying the event are Heads

These labels account for 1 favourable elementary outcomes

Probability = favourable count divided by total count = \(\frac{1}{2}\)

The predicted long-run relative frequency equals the model probability \(\frac{1}{2}\)

Multiply by one hundred to express it as 50%

Expected number = \(60 \times \frac{1}{2} = 30\)

An expectation can be fractional; it is an average prediction, not part of one observed event

A particular finite run need not match these predictions exactly

The same probability can predict a different count

Predict from repeated equal-sector labels

An ideal spinner has 8 equal-sized sectors labelled as shown. It is spun once; each sector is equally likely. Record its label

The event is: the recorded label is blue

Repeat the same trial 80 times under unchanged conditions

Predict the long-run relative frequency and the expected number, not a guaranteed observed result

Calculate the model predictions for the stated repeated experiment

Give probabilities as simplified fractions or whole numbers. Keep every requested calculation exact without rounding

Count all 8 equally likely elementary outcomes

The labels satisfying the event are blue

These labels account for 3 favourable elementary outcomes

Probability = favourable count divided by total count = \(\frac{3}{8}\)

The predicted long-run relative frequency equals the model probability \(\frac{3}{8}\)

Multiply by one hundred to express it as 75/2%

Expected number = \(80 \times \frac{3}{8} = 30\)

An expectation can be fractional; it is an average prediction, not part of one observed event

A particular finite run need not match these predictions exactly

Keep the same spinner model across trials

Keep an exact expectation when it is not a whole number

An observed number of successes is a whole-number count, but an expected number is an average prediction and can be fractional

For probability \(\frac{2}{3}\) in ten trials, the expectation is \(\frac{20}{3}\), not a guarantee of six or seven successes

A recurring percentage can also be written exactly as a fraction: \(\frac{2}{3}\) of one hundred is \(\frac{200}{3}\%\)

For the exact questions here, do not round a predicted percentage or expectation. Enter a simplified fraction when needed

An exact prediction need not be an observed count

There are 6 same-size cards with the distinct numbers listed below. Select one card without looking after mixing; each card is equally likely

The event is: the card number is prime

Repeat the same trial 10 times under unchanged conditions, replacing and remixing the selected item after every trial

Predict the long-run relative frequency and the expected number, not a guaranteed observed result

Calculate the model predictions for the stated repeated experiment

Give probabilities as simplified fractions or whole numbers. Keep every requested calculation exact without rounding

Count all 6 equally likely elementary outcomes

The labels satisfying the event are 2, 3, 5, 7

These labels account for 4 favourable elementary outcomes

Probability = favourable count divided by total count = \(\frac{4}{6} = \frac{2}{3}\)

The predicted long-run relative frequency equals the model probability \(\frac{2}{3}\)

Multiply by one hundred to express it as 200/3%

Expected number = \(10 \times \frac{2}{3} = \frac{20}{3}\)

An expectation can be fractional; it is an average prediction, not part of one observed event

A particular finite run need not match these predictions exactly

Retain the recurring percentage and expected value

A model gives event probability 1/4. Forty unchanged trials therefore have expected count ten. Which statement is justified before conducting the trials?

Describe a one-stage experiment and its predictions

Written planning task

Design a bag with six labelled tokens in at least two categories, including repeated labels. State that the tokens differ only by label and are selected after mixing

List the recorded-label sample space and every label probability. Check the assignment totals one

Choose an event, identify all its favourable tokens and calculate its probability

Predict the relative frequency and expected number for sixty trials, stating replacement and remixing after each draw

Explain why these predictions do not determine the actual record

The independent written design and explanation are not automatically checked; the numerical pages assess complete supplied records

Connect outcomes, probabilities and predictions

Define the result being recorded and list every possible recorded outcome

Distinguish equally likely elementary items or sectors from recorded labels that may have different multiplicities

For a specified event, count all favourable elementary outcomes and divide by the complete total

Assign every possible recorded label a probability from zero to one; the complete assignment totals one

Predict long-run relative frequency with p and expected count with n × p under an unchanged trial model

Retain exact answers and distinguish an expectation from a guaranteed finite-run result