Discrete Models Beyond the Binomial · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Geometric, hypergeometric and Poisson

Mathematics · Probability · ages 18-19
Name ______________________   Date ____________

  1. A bag holds 11 counters, 3 of them red. One red counter is drawn and kept out. What is the chance the next draw is red?

    • 3 / 11
    • 2 / 11
    • 2 / 10
    • 3 / 10

  2. A hand of 5 cards is dealt from a deck of 52 and the number of aces is counted. Which model fits best?

    • Hypergeometric: two card types, drawn without replacement
    • Geometric, because the count waits for a first ace
    • Poisson, because aces are rare events in a fixed window
    • Binomial, because the draws are independent with a fixed ace chance

  3. A fair six-sided die is rolled independently each time until the first six appears, and the number of rolls is recorded. Which model fits?

    • Poisson
    • Geometric
    • Hypergeometric

  4. Cards are drawn from a deck without replacement. Tom says each draw is independent with the same chances throughout. Is Tom right?

    Circle one:   True   False

  5. Each sales call succeeds with chance 0.25, independently. How many calls are expected in total, counting the successful call, until the first success?

    Answer: ______________

  6. A box holds 5 bulbs, 2 of them faulty. Two are tested without putting any back. What is the chance both are faulty? Give a decimal.

    Answer: ______________

  7. A spinner shows a winner with chance 0.2 on each spin, independently. What is the chance the first winner comes on spin 3? Give a decimal.

    Answer: ______________

  8. A coin with P(heads) = 0.5 is tossed independently until the first head. What is the chance the first head lands on toss 3?

    Answer: ______________

  9. Calls arrive at a constant average rate. Counts in disjoint time intervals are independent, and multiple calls in a very short interval are negligible. Which model fits the count in one hour, and why?

    • Geometric, since the hour waits for a first call
    • Poisson, for arrivals with these assumptions
    • Hypergeometric, since hours are drawn without replacement

  10. Kim draws 4 parts from a large batch, putting each part back before the next draw, and counts the defectives. Kim says the hypergeometric model fits. Is Kim right?

    Circle one:   True   False

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Answer key

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

Geometric, hypergeometric and Poisson W1-mt_9zc5xYjDg_-s1

  1. 2 / 10 · After one red leaves, 2 red remain among 10 counters, so the chance is 2 / 10.
  2. Hypergeometric: two card types, drawn without replacement · Cards leave the deck and there are two types, aces and others, so the sample is hypergeometric.
  3. Geometric · The rolls are independent, each has success chance 1 / 6, and the count stops at the first success. This is a geometric count.
  4. False · Tom's claim is false: drawn cards leave the deck, so later chances change and the draws depend on each other.
  5. 4 · Counting the successful call, the mean geometric wait is 1 divided by p: 1 divided by 0.25 is 4 calls.
  6. 0.1 · P(first faulty) = 2/5. Then 1 faulty bulb remains among 4, so 2/5 times 1/4 = 0.1.
  7. 0.128 · Two misses then a win: 0.8 times 0.8 times 0.2 = 0.128.
  8. 0.125 · Two failures then a success: 0.5 times 0.5 times 0.5 is 0.125.
  9. Poisson, for arrivals with these assumptions · The stated assumptions describe a Poisson arrival process. Its count in a fixed one-hour window is Poisson, rather than a wait until first success or a sample without replacement.
  10. False · Kim is wrong. Putting each part back keeps the chances the same, so the draws are independent. The hypergeometric model is for sampling without replacement.
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