NCEA Level 2Two-way tablesAS 91267

Two-way tables: 7 exam questions solved step by step

What this lesson covers

  • A conditional probability restricts attention to one row or column of the table.
  • P(A given B) and P(B given A) use different totals and are usually different.
Key idea

7 questions, one rule

  • Check the totals first.
  • No condition: ÷ the grand total.
  • "Given": ÷ that row or column total.
  • Percentages: turn into counts.
Your turn

Question 1 · Achieved, in context

  • A Cook Strait ferry survey of 200 passengers.
  • Find P(seasick), and
  • P(sat outside and was not seasick).
SeasickNot seasickTotal
Inside18102120
Outside87280
Total26174200
Worked example

Q1: out of everyone

  • No condition: ÷ 200.
  • "And" is one cell.
  1. 120 + 80 = 26 + 174 = 200totals agree
  2. P(seasick) = 26 ÷ 200column total
  3. = 0.13
  4. P(outside and not seasick)
  5. = 72 ÷ 200one cell
  6. = 0.36
Your turn

Question 2 · Achieved, in context

  • The same ferry survey.
  • A passenger who sat inside is chosen.
  • Find P(seasick | inside).
SeasickNot seasickTotal
Inside18102120
Outside87280
Total26174200
Worked example

Q2: given inside

  • "Given inside": that row.
  • The total becomes 120.
  1. inside row: 18 + 102 = 120reduced total
  2. P(seasick | inside) = 18 ÷ 120
  3. = 0.15
  4. 18 ÷ 200 = 0.09wrong total
  5. of those inside, 15% seasick
Your turn

Question 3 · Merit, in context

  • The same survey. A seasick passenger
  • is chosen. Find P(inside | seasick).
  • Is it the same as P(seasick | inside)?
SeasickNot seasickTotal
Inside18102120
Outside87280
Total26174200
Worked example

Q3: the other way round

  • "Given seasick": a column.
  • Order matters.
  1. seasick column: 18 + 8 = 26reduced total
  2. P(inside | seasick) = 18 ÷ 26
  3. = 0.69230…calculator
  4. = 0.69234 d.p.
  5. P(seasick | inside) = 0.15
  6. 0.6923 ≠ 0.15different totals
  7. of the seasick, 69% sat inside
Your turn

Question 4 · Merit, in context

  • A packhouse sorts 400 trays of avocados.
  • 60% come from Orchard A.
  • 5% of Orchard A's trays are rejected.
  • 30 trays are rejected altogether.
  • Complete the table. Pause and try it.
RejectedPassedTotal
Orchard A???
Orchard B???
Total30?400
Worked example

Q4: complete the table

  • Percentages to counts.
  • Then subtract to fill gaps.
  1. A: 60% of 400 = 240
  2. B: 400 − 240 = 160
  3. A rejected: 5% of 240 = 12
  4. 5% of 400 = 205% of A only
  5. A passed: 240 − 12 = 228
  6. B rejected: 30 − 12 = 18
  7. B passed: 160 − 18 = 142
  8. passed: 400 − 30 = 370
  9. check: 228 + 142 = 370
Your turn

Question 5 · Merit, in context

  • The completed packhouse table.
  • Find P(rejected | Orchard B).
  • Compare with Orchard A in a sentence.
RejectedPassedTotal
Orchard A12228240
Orchard B18142160
Total30370400
Worked example

Q5: compare the orchards

  • Each orchard, its own row.
  • Answer in a sentence.
  1. P(rej | B) = 18 ÷ 160B row: 160
  2. = 0.1125
  3. P(rej | A) = 12 ÷ 240A row: 240
  4. = 0.05
  5. 0.1125 ÷ 0.05 = 2.25
  6. B: 2.25 times as likely
  7. P(B | rej) = 18 ÷ 30 = 0.6a different question
Your turn

Question 6 · Excellence, in context

  • A survey of 600 workers.
  • 40% had the flu jab; 10% of those got flu.
  • 22% of all the workers got flu.
  • Find P(had the jab | got flu).
FluNo fluTotal
Jab???
No jab???
Total??600
Worked example

Q6: build the table

  • Every % to a count.
  • Subtract to fill the rest.
  1. jab: 40% of 600 = 240
  2. jab and flu: 10% of 240 = 24
  3. flu: 22% of 600 = 132
  4. no jab: 600 − 240 = 360
  5. no jab and flu: 132 − 24 = 108
  6. jab, no flu: 240 − 24 = 216
  7. no jab, no flu: 360 − 108 = 252
  8. check: 216 + 252 = 468 = 600 − 132
Worked example

Q6: given flu

  • "Given flu": the flu column.
  • Not the 10%.
  1. flu column total = 132reduced total
  2. P(jab | flu) = 24 ÷ 132
  3. = 0.181818…calculator
  4. = 0.18184 d.p.
  5. P(jab | flu) = 0.1that is P(flu | jab)
  6. P(flu | no jab) = 108 ÷ 360 = 0.3
  7. about 18% of flu cases had a jab
Your turn

Question 7 · Excellence, in context

  • A survey of 300 Year 12 students.
  • Claim: "80% of late students had a phone
  • in their room, so a phone in your room
  • makes you 80% likely to be late."
  • Evaluate the claim. Pause and try it.
LateNot lateTotal
Phone in room60140200
No phone1585100
Total75225300
Worked example

Q7: test the claim

  • Work out both directions.
  • Then compare the rows.
  1. P(phone | late) = 60 ÷ 75late column
  2. = 0.8the 80% is right
  3. P(late | phone) = 60 ÷ 200phone row
  4. = 0.3
  5. claim: P(late | phone) = 0.8swaps the order
  6. P(late | no phone) = 15 ÷ 100
  7. = 0.15
  8. 0.3 ÷ 0.15 = 2twice as likely
  9. linked, but not shown to cause
Summary

The method that solved all seven

  • "Given" picks the total; then say it in words.
StepHow
No conditiondivide by the grand total
"and"one cell ÷ grand total
"given"that row or column total
OrderP(A | B) ≠ P(B | A)
Percentagesturn them into counts first
In wordsname the group, then compare

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