Free printable Halloween activity

Trick-or-Treat Probability Lab

A 20-problem Halloween probability activity with candy bags, spinners, dice, compound events, expected values, and experimental analysis.

Grades 5–820 problemsAbout 60 minutesAnswers included
Download the free activity PDF

Why this activity works

Probability becomes clearer when prediction meets data. Students calculate theoretical outcomes, perform short trials, compare results, and decide whether differences are surprising or simply the natural variation of a small experiment.

Teaching note

Experimental answers will vary. Credit a correct fraction based on the student's actual results and focus discussion on why short experiments rarely match theory exactly.

Materials

  • Pencil
  • One coin
  • One six-sided die
  • Optional small bag of colored counters

How to use it

Predict first, run the small experiments where directed, then compare theoretical and experimental probability.

Complete activity preview

All prompts below are included in the printable version. Keep the answer key closed until the activity is complete.

Candy bag predictions

Find probabilities from a known collection.

  1. 1A bag has 5 chocolate, 4 fruit, and 3 mint candies. Find P(chocolate).
  2. 2Find P(not mint) from the same bag.
  3. 3Which is more likely: fruit or mint? By how much?
  4. 4Two candies are selected with replacement. Find the probability of mint, then chocolate.

Spinner trials

Use equally likely sectors and expected counts.

  1. 5A spinner has 8 equal sectors: 3 bats, 2 ghosts, 2 pumpkins, 1 cat. Find P(ghost).
  2. 6In 80 spins, about how many bats would you expect?
  3. 7What is the probability of landing on a pumpkin or cat?
  4. 8How many spins are expected before the spinner lands on a cat 12\displaystyle 12 times?

Coin-and-die experiment

Build a compound sample space and compare it with trials.

  1. 9Flip a coin and roll a die. How many equally likely outcomes are possible?
  2. 10Find P(heads and an even number).
  3. 11Run 12 trials. Report your experimental probability of heads and an even number.
  4. 12Find the probability of tails or a roll of 6\displaystyle 6.

Without replacement

Update probabilities after an outcome changes the collection.

  1. 13A bowl has 3 orange and 2 black tokens. Find P(orange, then orange) without replacement.
  2. 14From the same bowl, find P(black, then orange) without replacement.
  3. 15Find P(two tokens of different colors).
  4. 16Find the probability of at least one black token in two draws without replacement.

Carnival challenge

Analyze fairness, complements, and repeated events.

  1. 17A game wins on a die roll of 5 or 6. Is a $2 prize on a $1 play favorable to the player on average?
  2. 18If P(rain) = 0.18, find P(no rain).
  3. 19A player has a 14\displaystyle \frac{1}{4} chance to win each round. Find the probability of losing two rounds in a row.
  4. 20A game pays $5\displaystyle 5 with probability 14\displaystyle \frac{1}{4} and $1\displaystyle 1 otherwise. Find the expected payout.
Open the complete solutions

Open-ended and experimental responses may differ when the reasoning meets the stated conditions.

Candy bag predictions

1. 512\displaystyle \frac{5}{12}

2. 912\displaystyle \frac{9}{12}, or 34\displaystyle \frac{3}{4}

3. Fruit by 112\displaystyle \frac{1}{12}

4. 312⋅512=548\displaystyle \frac{3}{12}\cdot\frac{5}{12}=\frac{5}{48}

Spinner trials

5. 28\displaystyle \frac{2}{8}, or 14\displaystyle \frac{1}{4}

6. 30 bats

7. 38\displaystyle \frac{3}{8}

8. 96\displaystyle 96 spins

Coin-and-die experiment

9. 12 outcomes

10. 312\displaystyle \frac{3}{12}, or 14\displaystyle \frac{1}{4}

11. Answers vary; successful trials / 12

12. 712\displaystyle \frac{7}{12}

Without replacement

13. 35\displaystyle \frac{3}{5} x 24\displaystyle \frac{2}{4} = 310\displaystyle \frac{3}{10}

14. 25\displaystyle \frac{2}{5} x 34\displaystyle \frac{3}{4} = 310\displaystyle \frac{3}{10}

15. 35\displaystyle \frac{3}{5}

16. 1−35⋅24=710\displaystyle 1-\frac{3}{5}\cdot\frac{2}{4}=\frac{7}{10}

Carnival challenge

17. No; expected prize is $2(13\displaystyle \frac{1}{3}) = $0.67, less than the $1 cost

18. 0.82

19. 916\displaystyle \frac{9}{16}

20. $2\displaystyle 2

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