NCEA Level 2Probability treesAS 91267

Probability trees from written information, forwards and backwards

What this lesson covers

  • Each branch of a tree carries a probability conditional on the branches before it.
  • Working backwards through a tree finds the probability of an earlier event given a later one.
Key idea

From words to a tree

  • The exam gives a paragraph, not a tree.
  • Drawing the tree is part of the question.
  • First event first, then what depends on it.
InformationMeansOn the tree
60% of customerspay by cardfirst stage
15% of card payersbuy a warrantysecond stage, card branch
5% of cash payersbuy a warrantysecond stage, cash branch
Worked example

Build the tree

  • First stage: card 0.6, cash 0.4.
  • After card: warranty 0.15, none 0.85.
  • After cash: warranty 0.05, none 0.95.
Build the tree
Worked example

Multiply along each path

  • Card and warranty: 0.6 × 0.15 = 0.09.
  • Every path gets its product.
  • The four products add to 1.
Multiply along each path
Worked example

The four paths

  • Along a path: multiply.
  • Check: the four total 1.
  1. card, W: 0.6 × 0.15 = 0.09
  2. card, none: 0.6 × 0.85 = 0.51
  3. cash, W: 0.4 × 0.05 = 0.02
  4. cash, none: 0.4 × 0.95 = 0.38
  5. check: 0.09 + 0.51 + 0.02 + 0.38
  6. = 1
Worked example

Add across paths

  • Two paths end in W.
  • Across paths: add.
  1. P(W) = 0.6 × 0.15 + 0.4 × 0.05
  2. = 0.09 + 0.02multiply along
  3. = 0.11add across
Worked example

Working backwards

  • Given W: only the W paths.
  • The path ÷ the total.
  1. P(card | W)
  2. = P(card and W) ÷ P(W)
  3. = 0.09 ÷ 0.11
  4. = 0.81818…calculator
  5. = 0.81824 d.p.
Key idea

Why the answer is so high

  • Only 60% pay by card.
  • But 82% of W buyers did.
  1. P(card) = 0.6forwards
  2. 0.15 ÷ 0.05 = 3card rate ÷ cash rate
  3. P(card | W) = 0.8182backwards: 82%
Worked example

An unknown on the tree

  • Call P(card) p.
  • Multiply along, add across.
  1. P(card) = pthe unknown
  2. P(cash) = 1 − pbranches add to 1
  3. P(card and W) = 0.15pmultiply along
  4. P(cash and W) = 0.05(1 − p)multiply along
  5. 0.15p + 0.05(1 − p) = 0.12add: P(W) = 0.12
Worked example

Solve for p

  • Expand, collect, solve.
  • Check in the equation.
  1. 0.15p + 0.05(1 − p) = 0.12
  2. 0.15p + 0.05 − 0.05p = 0.12expand
  3. 0.1p + 0.05 = 0.12collect p terms
  4. 0.1p = 0.07subtract 0.05
  5. p = 0.07 ÷ 0.1 = 0.7divide by 0.1
  6. P(card) = 0.7: 70% by card
  7. check: 0.15 × 0.7 + 0.05 × 0.3
  8. = 0.105 + 0.015 = 0.12
Key idea

Where the marks are

  • Achieved: a path probability.
  • Merit: adding paths, and working backwards.
  • Excellence: forming and solving an equation.
Summary

What to take away

  • Along, multiply. Across, add. Backwards, divide.
Buildfirst event, then conditional branches
Each set of branchesadds to 1
Along a pathmultiply
Across pathsadd
Backwardsthe path ÷ the total
Unknownan equation from the tree
Your turn

Your turn: four questions

  • Draw the tree first.
  • Pause here and answer each one.
  • Four decimal places.
Question
Info4% have it. Positive: 90% if they do, 8% if not.
1P(a random person tests positive)?
2P(has the condition | tests positive)?
3P(does not have it and tests negative)?
4Of 1000 people tested, how many positives expected?
Show answer

1. 0.04 × 0.9 + 0.96 × 0.08 = 0.036 + 0.0768 = 0.1128. 2. 0.036 ÷ 0.1128 = 0.3191. 3. 0.96 × 0.92 = 0.8832. 4. 1000 × 0.1128 = 112.8, about 113.

Worked example

The answers: questions 1 and 2

  1. 1. P(not C) = 1 − 0.04 = 0.96C: has the condition
  2. P(pos) = 0.04 × 0.9 + 0.96 × 0.08two positive paths
  3. = 0.036 + 0.0768multiply along
  4. = 0.1128add across
  5. 2. P(C | pos) = 0.036 ÷ 0.1128path ÷ total
  6. = 0.31914…
  7. = 0.31914 d.p.
Worked example

The answers: questions 3 and 4

  1. 3. P(neg | not C) = 1 − 0.08branches add to 1
  2. = 0.92
  3. P(not C and neg) = 0.96 × 0.92multiply along
  4. = 0.8832
  5. 4. expected = 1000 × P(pos)
  6. = 1000 × 0.1128
  7. = 112.8
  8. about 113 positiveswhole people

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