NCEA Level 3Conditional probabilityAS 91585

Conditional probability from tables, trees and Venn diagrams

What this lesson covers

  • A conditional probability restricts attention to one group, so its denominator is that group's total.
  • The notation P(A | B) is read 'A given B', and the order matters.
Key idea

A probability inside one group

  • Look only at outcomes that meet a condition.
  • The denominator changes.
  • It becomes that group's total.
Key idea

Reading P(A | B)

  • Say it: "the probability of A given B".
  • After the bar: the group you are in.
  • Denominator: the total of B.
Means
P(A | B)"A given B"
After the barthe group you are in
Before the barwhat you want in it
Denominatorthe group's total
Worked example

Cycling to work: 500 residents

  • 200 under 40: 60 of them cycle.
  • 300 aged 40 and over: 45 cycle.
  • 105 cyclists in all.
CycleDo notTotal
Under 4060140200
40 and over45255300
Total105395500
Worked example

The gaps, by subtraction

  • Each gap: total − known.
  • Check the grand total.
  1. under 40, do not: 200 − 60 = 140
  2. 40+, do not: 300 − 45 = 255
  3. cycle: 60 + 45 = 105
  4. do not: 140 + 255 = 395
  5. 105 + 395 = 500
Key idea

Words into notation

  • "Of cyclists" means given C.
  • "Of under-40s" means given U.
  • "Under 40 and cycle": no condition.
Question asksWrite
% of cyclists under 40P(U | C)
% of under-40s who cycleP(C | U)
% under 40 and cycleP(U and C)
Worked example

One numerator, three denominators

  • And: divide by everyone.
  • Given: divide by the group.
  1. P(C and U) = 60 ÷ 500everyone
  2. = 0.12
  3. P(C | U) = 60 ÷ 200under-40s only
  4. = 0.3
  5. P(U | C) = 60 ÷ 105cyclists only
  6. = 0.57144 decimal places
Key idea

The order matters

  • P(C | U) = 0.3: 30% of under-40s cycle.
  • P(U | C) = 0.5714: 57% of cyclists are under 40.
  • Say which group you mean.
Worked example

With probabilities, not counts

  • Divide by P(B).
  • Same as the counts.
  1. P(A | B) = P(A and B) ÷ P(B)
  2. P(U) = 200 ÷ 500 = 0.4from the table
  3. P(C | U) = 0.12 ÷ 0.4substitute
  4. = 0.3same as 60 ÷ 200
Worked example

In a Venn, the condition is a circle

  • Given coffee: only the coffee circle.
  • Its total: 0.35 + 0.15 = 0.5.
  • P(M | C) = 0.15 ÷ 0.5 = 0.3.
In a Venn, the condition is a circle
Worked example

Other conditions from the same Venn

  • Only the given regions.
  • Given no muffin: outside.
  1. P(M) = 0.15 + 0.1 = 0.25the muffin circle
  2. P(C | M) = 0.15 ÷ 0.25both ÷ P(M)
  3. = 0.6
  4. P(not M) = 0.35 + 0.4 = 0.75outside the circle
  5. P(C | not M) = 0.35 ÷ 0.75coffee only ÷ P(not M)
  6. = 0.46674 decimal places
Worked example

Second branches are conditional

  • Polluted 0.2, clean 0.8.
  • P(pos | polluted) = 0.85.
  • P(pos | clean) = 0.1.
Second branches are conditional
Worked example

Multiply along each path

  • Each path: multiply.
  • The four paths total 1.
  1. polluted, pos: 0.2 × 0.85 = 0.17
  2. polluted, neg: 0.2 × 0.15 = 0.03
  3. clean, pos: 0.8 × 0.1 = 0.08
  4. clean, neg: 0.8 × 0.9 = 0.72
  5. 0.17 + 0.03 + 0.08 + 0.72 = 1
Worked example

Reversing the conditional

  • Add the positive paths.
  • Path ÷ matching paths.
  1. P(pos) = 0.17 + 0.08both positive paths
  2. = 0.25
  3. P(polluted | pos)
  4. = P(polluted and pos) ÷ P(pos)
  5. = 0.17 ÷ 0.25substitute
  6. = 0.68
Key idea

Say what it means

  • The test catches 85% of polluted streams.
  • Of positives, 68% are polluted.
  • Clean streams are far more common.
Key idea

Where the marks are

  • Achieved: a conditional from a table or tree.
  • Merit: accurate, with correct notation.
  • Excellence: support or reject a claim.
Summary

What to take away

  • Given picks the group, and the group picks the denominator.
P(A | B)A given B
Denominatorthe given group
FormulaP(A and B) ÷ P(B)
OrderP(A | B) ≠ P(B | A)
Reversepath ÷ matching paths
Answerin context, 4 dp
Your turn

Your turn: four questions

  • Pause here and answer each one.
  • Name the condition before you divide.
  • Four decimal places, then context.
GivenFind
1the cycling tableP(C | 40+)
2the cycling tableP(40+ | C)
30.2, 0.1, 0.3, 0.4P(A | B)
4U 0.1; flag 0.9, 0.05P(U | flag)
Show answer

1. P(cycle | 40 and over) = 45 ÷ 300 = 0.15. 2. P(40 and over | cycle) = 45 ÷ 105 = 0.4286. 3. P(B) = 0.1 + 0.3 = 0.4; P(A | B) = 0.1 ÷ 0.4 = 0.25. 4. P(flag) = 0.09 + 0.045 = 0.135; P(U | flag) = 0.09 ÷ 0.135 = 0.6667.

Worked example

The answers: questions 1 to 3

  1. 1. P(C | 40+) = 45 ÷ 30040 and over only
  2. = 0.15
  3. 2. P(40+ | C) = 45 ÷ 105cyclists only
  4. = 0.42864 decimal places
  5. 3. P(B) = 0.1 + 0.3 = 0.4both + only B
  6. P(A | B) = 0.1 ÷ 0.4both ÷ P(B)
  7. = 0.25
Worked example

The answers: question 4

  1. 4. U and flag: 0.1 × 0.9 = 0.09
  2. not U, flag: 0.9 × 0.05 = 0.045
  3. P(flag) = 0.09 + 0.045 = 0.135both flagged paths
  4. P(U | flag) = 0.09 ÷ 0.135U path ÷ P(flag)
  5. = 0.66674 decimal places

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