Relative Frequency and Expected Outcomes · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Estimating a Chance by Trying It

Mathematics · Probability · ages 15-16
Name ______________________   Date ____________

  1. What is a relative frequency?

    • the number of successes divided by the number of trials
    • the number of trials divided by the number of successes, giving trials per win
    • the difference between the successes and the failures in the whole experiment
    • the number of trials still to be carried out before an estimate can be trusted

  2. Two runs use independent trials with the same success probability: one has 20 trials and one has 500. Why does the 500-trial run usually give a steadier estimate?

    • because the experiment itself becomes fairer and more reliable the longer it is left running for
    • because one odd result is a twentieth of a short run and a five-hundredth of a long one
    • because the person running the experiment gets better at throwing and makes fewer mistakes as the trials go on
    • because the probability being estimated slowly changes towards the answer you keep getting

  3. The estimated probability of rain is 0.3. Predict how many rainy days to expect in 200 days.

    Answer: ______________

  4. A drawing pin is dropped 100 times and lands point up 37 times. Estimate the probability that it lands point up.

    Answer: ______________

  5. A spinner lands on green 7 times in 20 spins. Estimate the probability of green as a decimal.

    Answer: ______________

  6. Each go at a stall is independent and has the same chance of winning. After a long run of losses, the next go is more likely to win.

    Circle one:   True   False

  7. Two people estimate the same probability using independent trials under the same conditions. One did 40 trials and got 0.30; the other did 900 and got 0.26. Which estimate would you usually rely on?

    • the 0.30, because a smaller experiment is easier to run carefully
    • neither of them, since two different answers from the same experiment mean it was run badly
    • the average of the two, because both people did the work properly
    • the 0.26, because 900 trials leave far less room for the fraction to wobble

  8. A stall is tested with the targets 2 metres away, giving a probability of 0.25. On the day the targets are moved to 4 metres. What happens to the estimate?

    • it holds, because the probability of winning a game does not depend on how far away the targets are
    • it holds if the same person is throwing, and not otherwise
    • it should be doubled, since the distance has been doubled
    • it no longer applies, because the trials were run under conditions that have changed

  9. Sam expects 24 winners in 120 independent goes, each with win probability 0.2. He wants to reduce the risk of running out. Sam orders 24 prizes; Mia suggests a few more. Whose plan reduces that risk?

    • Sam, because ordering more than the expected number is a waste of the fair's money
    • Mia, because more than 24 can win and spare prizes reduce shortage risk
    • Sam, because the expected number already allows for the busier afternoons
    • Mia, but only if the probability came from fewer than a hundred practice trials

  10. A fair coin is tossed 500 times and lands heads 261 times. What should you conclude about the coin?

    • the coin is biased towards heads and should not be used for anything fair
    • the relative frequency of 0.522 must be the coin's true probability of landing on heads
    • the coin is unfair, since a fair coin would land on exactly 250 heads
    • nothing much, because 261 heads is an ordinary result in 500 fair tosses
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Answer key

For grown-ups. Fold this page away before handing over the rest.

Estimating a Chance by Trying It W1-mt_OQCnzkcHUH-s1

  1. the number of successes divided by the number of trials · It is the share of trials that succeeded, so successes over trials. Dividing the other way answers how many goes per win, which is a different question.
  2. because one odd result is a twentieth of a short run and a five-hundredth of a long one · Each result is one twentieth of the 20-trial sample, but one five-hundredth of the 500-trial sample. Under this model, the larger sample usually has less sampling variation.
  3. 60 · 0.3 x 200 = 60.
  4. 0.37 · The relative frequency is successes divided by trials: 37 / 100 = 0.37.
  5. 0.35 · 7 divided by 20 is 0.35.
  6. False · The goes are independent with the same chance each time, so the hoop remembers nothing. A long losing run is worth noticing as evidence about that chance, and it changes nothing about the next go.
  7. the 0.26, because 900 trials leave far less room for the fraction to wobble · Different estimates are possible without mistakes. With comparable independent trials, the 900-trial estimate usually has less sampling variation. Neither estimate is guaranteed closer in this particular pair.
  8. it no longer applies, because the trials were run under conditions that have changed · The 0.25 describes throws from 2 metres. It cannot be assumed to describe throws from 4 metres. New trials at that distance can give a new estimate.
  9. Mia, because more than 24 can win and spare prizes reduce shortage risk · Twenty-four is the expected count, not a cap. More than 24 can win, so spare prizes reduce the risk of running out. They do not remove that risk.
  10. nothing much, because 261 heads is an ordinary result in 500 fair tosses · Expecting 250 does not mean getting 250. A gap of 11 in 500 tosses is ordinary, and the relative frequency of 0.522 is an estimate that happens to sit a little above the true 0.5.
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