Compound events: a week of ready-to-teach maths

Five lessons for Year 9 Probability. Print this pack and the week is prepared: each day has a one-page plan and a student worksheet with practice and a stretch section, plus two spinners and draw cards, event sort cards, a mini-check and every answer with its working.

Start here: five minutes to Monday

1. Skim the week at a glance on the next page.
2. Print the five days. Each day is two A4 sheets: a plan and a worksheet.
3. Print the two cut-out sheets, one copy per pair: sheet 1 has two spinners and the draw cards (Days 2 to 4), sheet 2 the event sort cards (Day 5). Cut them before Day 2.
4. Each pair needs two coins, two dice, a paper clip and a pencil for the spinners, and an envelope or cup to draw cards from.
5. Open the free interactive unit on the board. Every plan says which picture to show and which buttons to press.

The idea in plain words

A compound event joins two or more simple events, such as tossing two coins or drawing two cards. To find its probability, first list every outcome in a system: a list, a table (for two stages) or a tree (for any number of stages). When the outcomes are equally likely, the probability of an event is the number of favourable outcomes over the total. On a tree, multiply the probabilities along each path. With replacement every stage has the same options; without replacement the later stages have fewer, and repeats can disappear.

One day, one lesson

The five lessons fill a week of maths, one lesson of about 60 minutes a day. Run them in order: each day stands on the one before. Every plan marks a split point for two shorter sessions, and every worksheet ends with a stretch section for students who finish early.

How we write these answers

Outcomes are written in stage order, so HT means a head and then a tail, and HT and TH are different outcomes. Probabilities are fractions, the count over the total first and then in lowest terms, as in 6/36 = 1/6; a tree’s branch probabilities are written on the branches. Two cards of the same colour are still two different cards.

The week at a glance

One lesson a day for a week. Each day stands on the day before, so run them in order.

How the week builds

Day 1 lists outcomes and counts them by multiplying; Day 2 uses tables for two-stage events; Day 3 draws trees and multiplies along their paths; Day 4 compares drawing with and without replacement; and Day 5 assigns probabilities to events and chooses a method. It builds on Year 8, where students used complementary events and listed the combinations of two events in tables and trees. It prepares estimating probabilities of “and” and “or” events from data, and running simulations, later in Year 9.

Materials for the week (one trip)

• From the classroom: two coins and two dice per pair, paper clips and pencils, scissors, envelopes or cups, this pack printed.
• Cut-out sheet 1, one per pair: a colour spinner, a number spinner and nine draw cards (Days 2 to 4).
• Cut-out sheet 2, one per pair: 12 event sort cards (Day 5).
• If you have them: coloured counters in a bag instead of the paper cards. No other equipment to buy.

Words for the word wall

outcome, compound event, sample space, systematic list, table, array, tree diagram, branch, equally likely, favourable, probability, with replacement, without replacement, independent, complement, at least one, fair game

Dear families

This week in maths, Year 9 is working out chances of events made of two or more steps, like tossing two coins or drawing two cards. The key is to list every possible outcome in a system, with a list, a table or a tree, before counting.

Try this at home

• Toss two coins 20 times and tally how often you get one head and one tail. Is it close to half the time?
• Ask: how many different ice-cream cones can you make with 3 flavours and 2 kinds of cone? (6.)
• Ask: what is the chance of rolling a double with two dice? (6 out of 36, which is 1/6.)
• Ask: from a bag of 2 red and 1 blue, if two are taken out without putting the first back, can you get two blues? (No.)

My chance week

Fill one row a day. Write an experiment from class, how many outcomes it has, and one probability.

Listing outcomes

A compound event joins simple events. Before any probability, list every outcome in a system so that none is missed or counted twice: hold the first stage still and run through the second. The number of outcomes is the product of the options at each stage.

We are learning to

• list every outcome of a two- or three-stage event in a system,
• count outcomes without listing them, by multiplying,
• find a probability from a list of equally likely outcomes.

Success criteria

• I can list every outcome of two stages without missing any, and check the count by multiplying.
• I can find the probability of an event from a list of equally likely outcomes.

You need

The worksheet; two coins per pair; the number spinner from cut-out sheet 1 if it is cut.

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after Part A; start the second with the board picture again.

Watch for these ideas

• Listing in no order, so an outcome is missed or written twice.
• Counting HT and TH as one outcome: they are different paths.
• Adding the options at each stage instead of multiplying: 4 shirts and 3 pants make 12 outfits, not 7.

Answers

• Part A: H and blue; H and green; H and gold; T and blue; T and green; T and gold (6). juice and apple; juice and muesli bar; juice and yoghurt; milk and apple; milk and muesli bar; milk and yoghurt (6). HHH, HHT, HTH, HTT, THH, THT, TTH, TTT (8).
• Part B: 4 × 3 = 12; 10 × 10 × 10 × 10 = 10000; 2 × 6 × 3 = 36.
• Part C: exactly one head 3/8; HTH 1/8; at least two heads 4/8 = 1/2.
• Part D: 2 × 2 × 2 × 2 = 16 for four coins; each coin doubles the count, so n coins give 2 multiplied by itself n times.

Listing outcomes

A. List every outcome

List the outcomes in a system, holding the first stage still. Then count them.

B. Count without listing

Multiply the number of options at each stage.

C. Chances from your list

Use your three-coin list. Every outcome is equally likely.

D. Stretch: more coins

How many outcomes are there when four coins are tossed? When n coins are tossed? Explain the pattern.

Tables and arrays

For two stages, a table lists every outcome at once: the first stage down the side, the second across the top, one cell for each outcome. Writing a sum, a product or a difference in each cell turns the table into a count of the ways each value can happen.

We are learning to

• set out a two-stage experiment as a table of outcomes,
• fill the cells with a sum, product or difference and count the ways,
• find and compare probabilities from a table.

Success criteria

• I can complete a table for two dice and find a probability from it.
• I can explain why two dice have 36 equally likely outcomes but only 11 possible sums.

You need

The worksheet; two dice per pair; the two spinners from cut-out sheet 1.

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after Part A; start the second with the board picture again.

Watch for these ideas

• Thinking two dice make 6 + 6 = 12 outcomes: the table has 6 × 6 cells.
• Counting 3 then 4 and 4 then 3 as one way: they are two cells.
• Treating the 11 sums as equally likely: a sum of 7 has six cells, a sum of 2 has one.

Answers

• Part A: product 12 in 4/36 = 1/9; odd product 9/36 = 1/4; more than 20 6/36 = 1/6.
• Part B: 3 × 4 = 12 cells; blue and even 2/12 = 1/6; gold 4/12 = 1/3.
• Part C: differences 0 to 5 appear 6, 10, 8, 6, 4, 2 times; the most likely is 1 (10/36 = 5/18); P(difference 0) = 6/36 = 1/6.
• Part D: a 1 to 4 die and a 1 to 6 die give 24 outcomes, and sums of 5, 6, 7 each come up 4 times: three most likely sums.

Tables and arrays

A. The product table

B. Two spinners

C. The difference table

D. Stretch: two different dice

A 1 to 4 die and a 1 to 6 die are rolled. Which sum is most likely? Is there just one?

Tree diagrams

A tree draws each stage as branches, with a probability on every branch. Multiply along a path for that outcome’s probability, then add the paths that make up an event. The ends of every tree add to 1, which makes a good check.

We are learning to

• draw a tree for two or three stages and write a probability on every branch,
• multiply along a path, and add the paths that make an event,
• check a tree by adding its ends to 1.

Success criteria

• I can label every branch of a tree with its probability and multiply along a path.
• I can find the chance of one of each colour by adding two paths.

You need

The worksheet; three coins per pair; the five R and B draw cards (cut-out sheet 1) in an envelope.

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after Part A; start the second with the board picture again.

Watch for these ideas

• Adding along a path instead of multiplying.
• Finding one of each from one path only: red then blue, and blue then red, are two paths.
• Branches from one point that do not add to 1.

Answers

• Part A: three heads 1/8; exactly two heads 3/8.
• Part B: branches 3/5 and 2/5 each time. RR 9/25, BB 4/25, one of each 12/25; together 25/25 = 1.
• Part C: both in 0.49, neither 0.09, exactly one 0.21 + 0.21 = 0.42; together 1.
• Part D: every path is a different outcome and the paths are all of them, so their chances make the whole: 1.

Tree diagrams

A. Three coins

Draw a tree for three coins, with 1/2 on every branch. Find P(three heads) and P(exactly two heads).

B. A bag of cards

An envelope holds 3 red cards and 2 blue cards. A card is drawn, its colour noted, and it is put back. Then a second card is drawn. Draw the tree and find:

C. Two free throws

A player gets a free throw in 7 times out of 10. Taking each throw to be like the last, draw a tree for two throws and find the chance of both going in, neither, and exactly one.

D. Stretch: why 1?

Explain why the probabilities at the ends of any tree add to 1.

With and without replacement

With replacement every stage is the same. Without replacement, what is taken out stays out: later stages have fewer options, the fractions on the later branches change, and a single item can no longer turn up twice.

We are learning to

• list the outcomes of two draws with and without replacement and compare them,
• draw a tree without replacement, where the second-stage fractions change,
• test a probability with repeated draws.

Success criteria

• I can explain why AA can happen with replacement but not without.
• I can draw a tree without replacement and multiply its changing fractions.

You need

The worksheet; the draw cards from cut-out sheet 1 (letters A to D, and three R and two B cards), an envelope or cup.

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after Part A; start the second with the demo draw.

Watch for these ideas

• Using the first-stage fractions again on the second stage of a draw without replacement.
• Forgetting that the total drops by one as well as the colour taken.
• Listing AA, BB and so on when nothing is put back.

Answers

• Part A: with replacement 16 outcomes, without 12. P(AA) 1/16 with, 0 without; both from A and B 4/16 = 1/4 with, 2/12 = 1/6 without.
• Part B: RR 3/5 × 2/4 = 6/20 = 3/10; BB 2/5 × 1/4 = 2/20 = 1/10; one of each 12/20 = 3/5; together 1.
• Part C: results vary; about 3 in every 10 pairs should be RR, and the class total should be closer than one pair’s 20 draws.
• Part D: 3/25 × 2/24 = 6/600 = 1/100.

With and without replacement

A. Two letters from A, B, C, D

List the outcomes of drawing two letter cards, first putting the first card back, then not putting it back.

B. The changing tree

The envelope holds 3 red and 2 blue cards. Two are drawn without replacement. Draw the tree, with the second-stage fractions changed, and find P(two reds), P(two blues) and P(one of each). Check that they add to 1.

C. Test it

Draw two cards without replacement 20 times. Tally the results.

D. Stretch: bruised apples

A box of 25 apples has 3 bruised ones. Two apples are taken without looking. What is the chance that both are bruised?

Assigning probabilities

Every compound probability rests on a full list of outcomes, so choose the tool that keeps the list honest: a list for small cases, a table for two stages, a tree for more stages or unequal chances. For “at least one”, the complement is quicker: 1 minus the chance of none.

We are learning to

• assign probabilities to compound events and compare them with 1/2,
• use the complement for “at least one”,
• decide whether a game is fair by counting each player’s outcomes.

Success criteria

• I can find the chance of at least one of something using the complement.
• I can decide whether a dice game is fair and explain why.

You need

The worksheet; the event sort cards (cut-out sheet 2); two dice per pair; the unit’s self-check quiz.

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after the sort; start the second with Part B and the quiz.

Watch for these ideas

• Adding chances for “at least one” (1/2 + 1/2 + 1/2 is more than 1).
• Calling a game fair because each player has the same number of totals: count the outcomes, not the totals.
• Choosing a table for three stages: a table holds two; use a tree or a list.

Answers

• Part A: the card key is with the mini-check answers.
• Part B: 3/12 = 1/4; 1 − 16/36 = 4/9 = 20/36 = 5/9; 1/27; 1 − 8/27 = 19/27.
• Part C: A wins on 16/36 = 4/9, B on 20/36 = 5/9: not fair, B is favoured although A’s totals are the likeliest ones.
• Part D: many answers. One: A wins if the two coins match (HH or TT, 2/4 = 1/2) and B wins if they do not; the die is not needed.

Assigning probabilities

A. Event sort

Work out the chance on each card, then write its letter in its pile.

B. Choose a method

Name your method (list, table, tree or complement), then find the chance.

C. Is it fair?

Two dice are rolled. Player A wins if the sum is 6, 7 or 8. Player B wins on any other sum. Is the game fair? Use a table to decide, and explain.

D. Stretch: make it fair

Design a fair game for two players using two coins and a die. Show that it is fair.

Two spinners and nine draw cards

Spinners (Days 1 to 3), one set per pair

Hold a pencil point on the centre dot, through the end of a paper clip, and flick the clip.

Draw cards (Days 3 and 4)

Cut out the nine cards. Keep the letters and the colours in two separate envelopes.

R stands for a red card and B for a blue one; coloured counters in a bag work just as well.

Event sort cards

For Day 5, one set per pair: 12 cards

Work out the chance of each event. Sort the cards: less than 1/2, exactly 1/2, or more than 1/2.

Teacher note: the key is with the mini-check answers, so this sheet can go to pairs whole.

What we know: Compound events

Work on your own. Give each probability as a fraction in lowest terms. Show your list, table or tree.

1. A coin is tossed and a 1 to 5 spinner is spun. List every outcome. How many are there? ______________________
2. A uniform comes in 3 shirts, 4 hats and 2 scarves. How many different uniforms? ______________________
3. Two 1 to 4 spinners are spun. What is P(the sum is 6)? ______________________
4. Two dice are rolled. What is P(both numbers are even)? ______________________
5. A spinner lands on red 1 time in 4. It is spun twice. Find P(both red) and P(exactly one red). ______________________
6. Two of the letters A, B, C, D and E are drawn without replacement. How many outcomes are there? ______________________
7. A bag holds 4 green counters and 1 white. Two are taken without replacement. What is P(both green)? ______________________
8. Two dice are rolled. Use the complement to find P(at least one even number). ______________________

Answers and marking guide

Answers with working

1. H1, H2, H3, H4, H5, T1, T2, T3, T4, T5: 2 × 5 = 10.
2. 3 × 4 × 2 = 24.
3. The 4 by 4 table has 3 cells with sum 6: 3/16.
4. 3 × 3 = 9 of 36 cells: 9/36 = 1/4.
5. Both red 1/4 × 1/4 = 1/16; exactly one 1/4 × 3/4 + 3/4 × 1/4 = 6/16 = 3/8.
6. 5 × 4 = 20 (no letter twice).
7. 4/5 × 3/4 = 12/20 = 3/5.
8. No even number means both odd, 9/36 = 1/4; so 1 − 1/4 = 3/4.

A quick three-level guide

Eight questions, four ideas. A student at standard answers most questions with a list, table or tree that shows every outcome.

Card key

Event sort (Day 5): less than 1/2: B (1/4), D (1/6), E (1/12), I (11/36), K (1/8); exactly 1/2: A (1/2), F (1/2), G (1/2), L (1/2); more than 1/2: C (3/4), H (11/12), J (7/8).

Weekly class record

Jot a tick as you move around the room; the mini-check fills any gaps. A tick a day is plenty.

The five columns are the five days: lists, tables, trees, with and without replacement, and assigning probabilities.