Probability estimates: 7 exam questions solved step by step

← Previous10 of 10Next →

What this lesson covers

  • The true probability is usually unknown; models and experiments both estimate it.
  • Which estimate to trust depends on the assumptions behind the model and the size of the experiment.
Key idea

7 questions, three probabilities

  • 1. True: the long run, unknown.
  • 2. Model: from assumptions.
  • 3. Experimental: from real trials.
  • 4. Trust: assumptions and size.
Your turn

Question 1 · Achieved, in context

  • A school fair game: draw a ball from a
  • bag of 5 red, 3 blue and 2 green.
  • Green wins. In 50 plays, 13 won.
  • Give the model, experimental and true.
Worked example

Q1: three probabilities

  • Model: from the bag.
  • Experimental: from plays.
  1. model: P(green) = 2 ÷ 10assumes equally likely
  2. = 0.2
  3. experimental: 13 ÷ 50from 50 real plays
  4. = 0.26
  5. true: unknownthe long-run value
  6. both estimate the true value
Your turn

Question 2 · Achieved

  • A nursery sows 6 mānuka seeds. Each
  • fails, independently, with P = 0.15.
  • Find P(all 6 fail) as a decimal.
  • Careful: read the screen properly.
Worked example

Q2: standard form

  • Independent: multiply.
  • E−5 means × 10⁻⁵.
  1. P(all fail) = 0.15⁶independent
  2. screen: 1.1390625E−5
  3. = 1.139 × 10⁻⁵
  4. = 0.00001139point moves 5 left
  5. 1.139over 1: impossible
  6. as a percent: 0.001139%
Your turn

Question 3 · Merit

  • A netball shooter scores with p = 0.7.
  • A simulation: 200 sets of 5 shots,
  • all 5 scored in 34 sets. Compare with
  • the model. What can it not tell you?
Worked example

Q3: a simulation of a model

  • Close to the model.
  • It repeats its assumptions.
  1. simulation: 34 ÷ 200 = 0.17
  2. model: 0.7⁵ = 0.16807independent shots
  3. ≈ 0.16814 d.p.
  4. 0.17 is close to 0.1681
  5. simulation assumes p = 0.7on every shot
  6. and independent shots
  7. it cannot test those
Your turn

Question 4 · Merit, in context

  • A drawing pin lands point up or down.
  • Which estimate should predict the next
  • 200 tosses, and how many point up?
Tossed byTossesPoint up
Ana4026
Whole class600357
Worked example

Q4: no model to use

  • Not symmetric: no model.
  • Trust the bigger experiment.
  1. Ana: 26 ÷ 40 = 0.65
  2. class: 357 ÷ 600 = 0.595
  3. model: nonepin is not symmetric
  4. use 0.595600 tosses vary less
  5. expected = 200 × 0.595
  6. = 119
  7. about 119 of 200 point up
Your turn

Question 5 · Merit

  • A game says 1 in 5 loot boxes holds a
  • rare item. Rawiri opens 40 and gets 4.
  • Is this strong evidence that the
  • game's claim is false?
Worked example

Q5: how unusual is 4?

  • Assume the claim is true.
  • Then 4 or fewer: how likely?
  1. experimental: 4 ÷ 40 = 0.1
  2. expected: 40 × 0.2 = 8if the claim holds
  3. X ~ Bin(40, 0.2)model: 1 in 5
  4. P(X ≤ 4) = 0.075914binomial
  5. ≈ 0.07594 d.p.
  6. about 1 time in 13
  7. not strong evidence40 boxes is small
Your turn

Question 6 · Excellence

  • A 50c coin is spun on a table, not
  • tossed. 1000 spins gave 431 heads.
  • Which estimate of P(heads) should be
  • used to predict 200 more spins?
Worked example

Q6: test the model

  • Suppose the model is right.
  • Read standard form.
  1. model: P(heads) = 0.5two equal faces
  2. experimental: 431 ÷ 1000
  3. = 0.431
  4. if 0.5: X ~ Bin(1000, 0.5)
  5. P(X ≤ 431) = 7.164 × 10⁻⁶binomial
  6. = 0.000007164not 7.164
  7. almost never if fair
Worked example

Q6: which to trust

  • The assumption fails.
  • 1000 spins: a large sample.
  1. model 0.5, experiment 0.431
  2. equal faces assumption failsspun, not tossed
  3. 1000 spins: varies little
  4. predict with 0.431
  5. true value still unknown
  6. expected = 200 × 0.431
  7. = 86.2
  8. about 86 heads in 200 spins
Your turn

Question 7 · Excellence

  • Hemi says P(rain on an October day) is
  • 1/2: it either rains or it does not.
  • Judge his model, choose an estimate,
  • and predict rainy days next October.
RecordsDaysRainy
Last 10 Octobers310186
Last October3121
Worked example

Q7: a model that fails

  • Two outcomes, not equal.
  • More days, more trust.
  1. Hemi: 1 ÷ 2 = 0.5not equally likely
  2. 10 years: 186 ÷ 310 = 0.6experimental
  3. last year: 21 ÷ 31 = 0.6774only 31 days
  4. use 0.6310 days, 10 years
  5. assumes Octobers stay alike
  6. expected = 31 × 0.6
  7. = 18.6
  8. about 19 rainy days
Summary

The method that solved all seven

  • Name each estimate, check the model, trust size, then predict.
StepHow
Truelong run, never known
Modelname its assumptions
Simulationrepeats the model
Experimentmore trials, less wobble
Check fithow likely if the model holds?
Every Pbetween 0 and 1

More in AS 91585

Tables, trees and Venn diagrams from written information6 minConditional probability from tables, trees and Venn diagrams6 minIndependent or mutually exclusive? The two tests6 minRelative risk: times as likely, and when you need a ratio6 minTrue probability, model estimates and experimental estimates7 minOrganising information: 7 exam questions solved step by step11 minConditional probability: 7 exam questions solved step by step10 minIndependent or exclusive: 7 exam questions solved step by step10 minRelative risk: 7 exam questions solved step by step10 minProbability estimates: 7 exam questions solved step by step9 min

Is your child on track?

Take the free placement test