Conditional probability: a week of ready-to-teach maths

Five lessons for Year 10 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 twelve probability cards to match, twelve which-probability cards to sort, 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. Cut the 12 probability cards before Day 2 and the 12 which-probability cards before Day 4, and keep each set in an envelope.
4. Calculators help but are not essential. A pair of dice and a bag of coloured counters make Days 1 and 3 real.
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 conditional probability is a chance worked out inside a smaller group. “Given a student plays sport”, “of the students who play sport”, “if a student plays sport, then” and “knowing that a student plays sport” all say the same thing: count only the students who play sport, so their total becomes the denominator. The week trains students to hear the condition in each wording, to find the right total in a Venn diagram, a table or a tree, and to notice that turning a condition around, P(A | B) against P(B | A), usually changes the answer.

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

P(A | B) is read “the probability of A given B”. Fractions are left unsimplified first, so the denominator still shows which group was counted (24/40), then simplified (3/5). Decimals and percentages appear where the context uses them.

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 shows what a condition does to the group; Day 2 reads conditions off a table; Day 3 writes them on the branches of a tree; Day 4 turns a condition around; and Day 5 asks when knowing one thing changes nothing at all. It builds on Year 9, where students listed sample spaces, drew two-way tables and trees and worked with “and” and “or”, and it prepares the next unit, AC9M10P02, where students simulate conditional probability.

Materials for the week (one trip)

• From the classroom: calculators, scissors, envelopes for the cards, this pack printed.
• Cut-out sheet 1, one per pair: 12 probability cards (Day 2).
• Cut-out sheet 2, one per pair: 12 which-probability cards (Day 4).
• If you have them: two dice per pair (Day 1) and a bag of 5 red and 3 blue counters per pair (Day 3). No other equipment to buy.

Words for the word wall

conditional probability, given, of, if ... then, knowing that, sample space, two-way table, Venn diagram, tree diagram, with replacement, without replacement, denominator, independent

Dear families

This week in maths, Year 10 is learning conditional probability: how a chance changes once we know something, and how to hear that in everyday words such as “given”, “of the”, “if ... then” and “knowing that”.

Try this at home

• Ask: two dice are rolled and the total is 8. What is the chance one of them is a 6? (2 of the 5 ways, 2/5.)
• Ask: a news story says most crashes happen close to home. Does that make driving near home more dangerous? (Not by itself: most driving is near home.)
• Ask: if it is raining, is the bus more likely to be late? Which would you need to know, P(late | rain) or P(rain | late)?
• Ask: after a coin lands heads five times, is tails “due”? (No: knowing the past tosses changes nothing.)

My given week

Fill one row a day. Write a sentence you heard or read with a condition in it, the group it counts in, and the chance it reports.

Given narrows the group

A condition keeps only part of the group, so that part’s total becomes the denominator. Students read conditional probabilities off a Venn diagram, name the group each question counts in, and narrow the 36 outcomes of two dice.

We are learning to

• hear the condition in a question worded with given, of, if ... then or knowing that,
• name the group a conditional probability counts in and use its total as the denominator,
• narrow a sample space to the outcomes a condition allows.

Success criteria

• I can find the chance a team sport player swims, 5/14, and say why it is out of 14.
• I can find the chance of a 6 given that two dice total 8.

You need

Two dice per pair if you have them (one red, one blue is best); the worksheet.

Lesson flow (about 60 minutes)

Two shorter sessions? Stop after Part A; start the next with Part B.

Watch for these ideas

• Dividing by the whole class: given a team sport, the denominator is 14, not 30.
• Reading given as and: P(swims and team sport) is 5/30, not 5/14.
• Counting (6, 2) and (2, 6) as one outcome: a red and a blue die make them different.

Answers

• Part A: the whole class, 30: 11/30; team sport players, 14: 5/14; swimmers, 11: 5/11; no team sport, 16: 6/16 = 3/8; non-swimmers, 19: 9/19.
• Part B: 5/30 = 1/6 against 5/14: both count the 5 who do both, but the second counts only in the 14 team players. Rewrites: given a student swims, the chance they play a team sport; if a student swims, then the chance; knowing a student swims.
• Part C: 5 outcomes total 8, 2 with a 6, so 2/5; knowing nothing, 11/36.
• Part D: 11 outcomes have a 6 and 1 has two, so 1/11: the condition leaves 11 outcomes, not 6.

Given narrows the group

A. Read the Venn diagram

For each question, write the group it counts in (with its size) and the probability. The first row is done for you.

B. Given is not and

C. Two dice

D. Stretch: at least one 6

Two-way tables

A two-way table holds every count at once; a condition picks one row or one column, and that total becomes the denominator. Students complete a table, read conditional probabilities off it, and match six questions in words to their probabilities.

We are learning to

• complete a two-way table from some of its counts,
• use the total of the given row or column as the denominator,
• match the four wordings of a condition to the same calculation.

Success criteria

• I can find the chance a music player studies a language and say which total I divided by.
• I can tell “of the music players” from “of the language students”.

You need

Calculators; the probability cards (cut-out sheet 1), one set per pair; the worksheet.

Lesson flow (about 60 minutes)

Two shorter sessions? Stop after Part A; start the next with Part B.

Watch for these ideas

• Dividing by the grand total: that gives P(A and B), not P(A | B).
• Using a row total for a column condition: of the language students means the language column.
• Simplifying too soon: 24/40 shows the group, 3/5 hides it.

Answers

• Part A: the missing counts are 16, 30, 80, 66 (reading across the rows).
• Part B: 54/120 = 9/20; 50/80 = 5/8; 66/120 = 11/20. Knowing a student plays no music raises the chance of no language from 66/120 to 50/80.
• Part C: the card key is with the mini-check answers.
• Part D: any table with the same language share in both rows, such as 18, 22 and 36, 44: 18/40 = 36/80.

Two-way tables

A. Complete the table

A survey of the 120 students in Year 10 asked whether they play a musical instrument and whether they study a language. Fill in the gaps.

B. Read the table

Write each as a fraction first, then simplify.

C. Match the cards

Match each question card to its probability card. Write the letters.

D. Stretch: when knowing changes nothing

If ... then: trees

In a two-stage tree the second branches are conditional: they say what happens if the first stage went a certain way. Students write the branches for two marbles taken without replacement, multiply along paths, and use a rain-and-bus tree to work back to the chance it rained.

We are learning to

• write the second branches of a tree as conditional probabilities,
• multiply along a path, and add the paths that give the same result,
• explain what changes between drawing with and without replacement.

Success criteria

• I can write “if the first marble is red, then the second is red” as 4/7.
• I can find the chance the bus is late from a tree.

You need

A bag of 5 red and 3 blue counters per pair if you have them; the worksheet.

Lesson flow (about 60 minutes)

Two shorter sessions? Stop after Part A; start the next with Part B.

Watch for these ideas

• Keeping the first fraction for the second draw: without replacement the bag has one fewer.
• Adding along a path: multiply along a path, add across paths.
• Reading P(late | rain) as P(rain | late): the tree gives the first; the second takes working back.

Answers

• Part A: after red, red 4/7 and blue 3/7; after blue, red 5/7 and blue 2/7.
• Part B: red, red 20/56 = 5/14; red, blue 15/56; blue, red 15/56; blue, blue 6/56 = 3/28. If the first is red, then the second is red with 4/7; one of each is 30/56 = 15/28.
• Part C: 0.4 × 0.3 + 0.6 × 0.1 = 0.18.
• Part D: P(rain | late) = 0.12 ÷ 0.18 = 12/18 = 2/3 (about 67%). Put back first: red, red is 25/64, against 5/14 without.

If ... then: trees

A. Two marbles, no putting back

A bag holds 5 red and 3 blue marbles. One is taken and kept out, then a second is taken. Fill in the second branches.

B. Multiply along the paths

The first row is done for you.

C. Rain and the bus

D. Stretch: working back

Turning a condition around

P(A | B) and P(B | A) share the same overlap but count it in different groups, so they are rarely equal. Students build a table for a spam filter that catches most spam, find that a flagged email is still often not spam, and sort twelve sentences by the probability they describe.

We are learning to

• build a two-way table from percentages,
• tell P(A | B) from P(B | A), in words and in numbers,
• explain why a large group of ordinary cases keeps P(spam | flagged) down.

Success criteria

• I can show the filter flags 90% of spam, yet only 90/135 of flagged emails are spam.
• I can sort a sentence into P(flagged | spam), P(spam | flagged) or P(spam and flagged).

You need

Calculators; the which-probability cards (cut-out sheet 2), one set per pair; the worksheet.

Lesson flow (about 60 minutes)

Two shorter sessions? Stop after Part A; start the next with Part B.

Watch for these ideas

• Swapping the condition: P(spam | flagged) is not P(flagged | spam).
• Forgetting the big group: 5% of 900 good emails is 45 false alarms, half as many as the spam caught.
• Reading “of the flagged emails” as a row: it is the Flagged column.

Answers

• Part A: spam 90, 10, 100; not spam 45, 855, 900; totals 135, 865, 1 000.
• Part B: P(flagged | spam) = 90/100 = 9/10; P(spam | flagged) = 90/135 = 2/3; P(spam and flagged) = 90/1000 = 9/100. The first two count the same 90 emails in different groups.
• Part C: the card key is with the mini-check answers.
• Part D: 90 spam flagged and 495 good emails flagged, so P(spam | flagged) = 90/585 = 2/13, about 15%: the rarer the spam, the more a flag is a false alarm.

Turning a condition around

A. Build the table

A school tests its email filter on 1 000 emails. 10% are spam. The filter flags 90% of the spam and 5% of the good emails. Complete the table. The first row is done for you.

B. Three probabilities from one table

C. Sort the cards

Write each card’s letter under the probability its sentence describes.

D. Stretch: rarer spam

Knowing that

Sometimes knowing one thing changes nothing: when P(A | B) equals P(A | not B), A and B are independent. Students test three tables, rewrite claims from the news in the language of conditions, and play the self-check quiz as a class game.

We are learning to

• decide whether knowing one thing changes the chance of another,
• rewrite a claim with given, of, if ... then or knowing that, and say which conditional it reports,
• say what more we would need to know before trusting a claim.

Success criteria

• I can show from a table that knowing a birth month makes no difference to left-handedness.
• I can explain why “most crashes are close to home” does not show that roads near home are more dangerous.

You need

Two coins per pair if you have them; the worksheet; the self-check quiz on the board.

Lesson flow (about 60 minutes)

Two shorter sessions? Stop after Part A; start the next with Part B.

Watch for these ideas

• Thinking a coin remembers: knowing past tosses does not change the next one.
• Comparing counts, not fractions: compare each count with its own row total.
• Reading “of the drivers in crashes” as the chance of a crash.

Answers

• Part A: the bus is late given it rains 0.3, otherwise 0.1: knowing changes it; a student is left-handed given they were born from January to June 0.12, otherwise 0.12: no change, independent; a student passes the quiz given they did the homework 0.75, otherwise 0.5: knowing changes it.
• Part B: 1. given a driver crashed, the chance they were near home is 0.7; we need how much driving is near home. 2. P(win | ticket) = 1/50 000. 3. P(pass | homework) = 45/60, not P(homework | pass) = 45/65 = 9/13.
• Part C: knowing the first is heads, 1/2 (HH, HT); knowing at least one is heads, 1/3 (HH of HH, HT, TH).
• Part D: P(born January to June | left-handed) = 12/24 = 1/2, the same as P(born January to June) = 100/200 = 1/2, so it works both ways.

Knowing that

A. Does knowing B change the chance of A?

Each line comes from a table of counts (in brackets: A, not A). The first row is done for you.

B. Claims in the news

Suppose a news story makes each claim. Rewrite it with “given”, and say what else you would need to know.

C. Two coins

D. Stretch: both ways

Questions and their probabilities

For Day 2, one set per pair, with the completed music and language table from Part A. Six cards ask a question in words; six give a probability in lowest terms. Match them.

Teacher note: the key is with the mini-check answers.

Which probability is it?

For Day 4, one set per pair. Every card is about the spam filter. Sort the cards into three piles: P(flagged | spam), P(spam | flagged) and P(spam and flagged).

Teacher note: the key is with the mini-check answers.

What we know: Conditional probability

Work on your own. A calculator is allowed. Show your working.

1. Of 40 students, 18 play guitar, 15 sing and 6 do both. Given a student plays guitar, what is the chance they sing? ______________________
2. Of the singers, what fraction play guitar? ______________________
3. Rewrite with “given”: “Of the students who walk to school, 40% live within 2 km.” ______________________
4. A bag holds 4 green and 2 yellow counters. Two are taken without putting the first back. If the first is green, then what is the chance the second is green? ______________________
5. What is the chance both counters are green? ______________________
6. Knowing a fair coin has just landed heads, the chance the next toss lands heads is, circle one:   one half  ·  one quarter  ·  one
7. A metal detector beeps over 95 of 100 buried coins and over 30 of 600 buried bottle tops. Given it beeps, what is the chance it has found a coin? ______________________
8. Of 30 tennis players, 6 are left-handed; of 70 other students, 14 are. Does knowing a student plays tennis change the chance they are left-handed? Show why. ______________________

Answers and marking guide

Answers with working

1. 6/18 = 1/3: count only the 18 guitar players.
2. 6/15 = 2/5: now the group is the 15 singers.
3. Given a student walks to school, the probability they live within 2 km is 0.4.
4. 3/5: one green is gone, so 3 green of 5 counters remain.
5. 4/6 × 3/5 = 12/30 = 2/5.
6. It is one half: the coin does not remember, so knowing the last toss changes nothing.
7. It beeps 125 times, 95 over coins: 95/125 = 19/25.
8. No: 6/30 = 1/5 and 14/70 = 1/5, the same 0.2 either way.

A quick three-level guide

Eight questions, four ideas. A student at standard names the group before dividing, in every wording.

Card keys

Probability cards (Day 2): 3/5: G and A; 4/9: D and J; 3/8: E and L; 8/33: B and H; 1/5: K and F; 2/5: I and C. Which-probability cards (Day 4): P(flagged | spam): M, P, R, U, X; P(spam | flagged): N, Q, T, W; P(spam and flagged): O, S, V.

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: given narrows the group, two-way tables, trees, turning a condition around, and knowing that.