11.2 Probability, Counting Techniques & Simulations

Key Takeaways

  • Theoretical probability equals the number of favorable outcomes divided by the number of possible outcomes; experimental probability comes from trials.

  • As the number of trials increases, experimental probability tends to approach theoretical probability (the Law of Large Numbers).

  • Permutations count arrangements where order matters, and combinations count selections where order does not.

  • Geometric probability is a ratio of areas or lengths.

  • Independent events do not affect each other, so past coin flips do not change the probability of the next flip.

Last updated: October 2026

Overview & Exam Relevance

The probability half of Competency 005 asks you to explore probability through data collection, experiments, and simulations, and to describe simple and compound events. It also covers constructing sample spaces and solving problems with combinations and geometric probability, which is probability as a ratio of two areas. Elementary students build intuition with coins, number cubes, spinners, and bags of counters. The 391 expects you to handle the mathematics behind those activities.


Foundations of Probability

Probability measures the quantitative likelihood that a particular event will occur, expressed on a scale from 0 (impossible event) to 1 (certain event) (or 0%0\% to 100%100\%).

Theoretical versus Experimental Probability

  • Theoretical Probability: The ratio of favorable outcomes to total possible equally likely outcomes in a sample space (SS): P(E)=Number of favorable outcomesTotal number of possible outcomesP(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} Example: Rolling a fair six-sided die has a theoretical probability of rolling a 4 of P(4)=16≈16.7%P(4) = \frac{1}{6} \approx 16.7\%.
  • Experimental (Empirical) Probability: The ratio of observed occurrences to the total number of trials conducted in an experiment: P(E)=Number of observed occurrencesTotal number of experimental trialsP(E) = \frac{\text{Number of observed occurrences}}{\text{Total number of experimental trials}} Example: If a student rolls a die 30 times and rolls a 4 eight times, the experimental probability is 830≈26.7%\frac{8}{30} \approx 26.7\%.
  • The Law of Large Numbers: As the number of experimental trials increases, the experimental relative frequency systematically converges toward the theoretical probability. A small sample of 10 coin flips may yield 8 heads (80%80\%), but 10,000 flips will reliably approach 50%50\%.

Compound Probability: Independent versus Dependent Events

  • Independent Events: The occurrence of the first event has no effect on the probability of subsequent events. Calculated by multiplying individual probabilities: P(A and B)=P(A)⋅P(B)P(A \text{ and } B) = P(A) \cdot P(B). Example: Flipping heads on a coin and then rolling a 6 on a die: 12×16=112\frac{1}{2} \times \frac{1}{6} = \frac{1}{12}. Similarly, sampling items with replacement preserves independence.
  • Dependent Events: The occurrence of the first event alters the sample space and changes the probability of subsequent events. Calculated using conditional probability: P(A and B)=P(A)⋅P(B∣A)P(A \text{ and } B) = P(A) \cdot P(B|A). Example: A bag contains 4 red and 6 blue marbles (10 total). Sampling two red marbles without replacement: The first draw is P(R1)=410P(R_1) = \frac{4}{10}. The second draw has only 3 red marbles left out of 9 total marbles: P(R2)=39P(R_2) = \frac{3}{9}. The compound probability is 410×39=1290=215\frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}.

Counting Principles and Sample Spaces

  • Fundamental Counting Principle: If a first choice can be made in mm ways and a second choice in nn ways, the total number of combined sequential outcomes is m×nm \times n. Example: Selecting an outfit from 4 shirts, 3 pants, and 2 pairs of shoes yields 4×3×2=244 \times 3 \times 2 = 24 unique outfit combinations.
  • Tree Diagrams and Systematic Lists: Visual branching organizers used in elementary grades to enumerate sample spaces without overlooking outcomes.

Combinations and Permutations

  • Permutations count arrangements where order matters: P(n,r)=n!(n−r)!P(n, r) = \frac{n!}{(n - r)!}. The number of ways to award first, second, and third place among 5 runners is 5×4×3=605 \times 4 \times 3 = 60.
  • Combinations count selections where order does not matter: C(n,r)=n!r! (n−r)!C(n, r) = \frac{n!}{r!\,(n - r)!}. The number of 2-student committees from 5 students is 5×42×1=10\frac{5 \times 4}{2 \times 1} = 10. The number of 2-topping pizzas from 6 toppings is 6×52=15\frac{6 \times 5}{2} = 15.

Ask: Does switching the order create a different outcome? A committee of Ana and Ben is the same as Ben and Ana (combination); first place Ana, second place Ben differs from the reverse (permutation).


Geometric Probability

When outcomes are points in a region, probability is a ratio of areas. If a dart lands at a random point on a 10 in by 10 in board that contains a circle of radius 2 in, then P(circle)=π(2)2100≈12.57100≈0.13P(\text{circle}) = \frac{\pi (2)^2}{100} \approx \frac{12.57}{100} \approx 0.13


Compound Events, Expected Outcomes, and Simulations

  • "Or" events: P(A or B)=P(A)+P(B)−P(A and B)P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B). On a standard die, P(even or greater than 4)=36+26−16=46P(\text{even or greater than } 4) = \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6}, because 6 is counted in both events.
  • Expected outcomes: Multiply the probability by the number of trials. With a fair coin, about 12×200=100\frac{1}{2} \times 200 = 100 heads are expected in 200 flips.
  • Simulations model real situations with chance devices. To estimate how many cereal boxes a family must buy to collect all 6 prizes, roll a number cube until every number appears, record the rolls, and repeat many times. Simulations and probability models let students test predictions and see the Law of Large Numbers in action.

Classroom Scenario Application

Classroom Context: A fourth-grade class flips a coin and records the results on the board. After five heads in a row, a student, Dylan, announces, "The next flip has to be tails. Tails is due!" Several classmates agree.

Diagnostic Error Analysis: Dylan is showing the gambler's fallacy, the belief that past outcomes of independent events change future probabilities. A coin has no memory, so the probability of tails on the next flip is still 12\frac{1}{2}.

Targeted Instructional Response:

  1. Test the prediction: Each pair flips a coin 20 times and records every time a run of three heads is followed by another flip. The class pools the results and finds that heads and tails follow a run about equally often.
  2. Connect to the Law of Large Numbers: Over 10 flips, results can look lopsided. When the class combines several hundred flips, the experimental probability of heads moves close to the theoretical probability of 12\frac{1}{2}.
  3. Generalize: Students explain in writing why each flip is an independent event and compare it with drawing marbles from a bag without replacement, where earlier draws do change later probabilities (dependent events).
Test Your Knowledge

A prize box contains 6 red bouncy balls, 4 blue bouncy balls, and 2 green bouncy balls (12 total). A student randomly selects one bouncy ball from the box, keeps it on their desk without returning it, and then randomly selects a second bouncy ball. What is the probability that both bouncy balls selected are red?

A

1/4

B

1/6

C

5/24

D

5/22

Test Your Knowledge

A pizza shop offers 6 toppings. How many different pizzas with exactly 2 different toppings are possible, and which counting idea applies?

A

30, because order matters, so it is a permutation

B

12, because 6×2=126 \times 2 = 12

C

15, because order does not matter, so it is a combination

D

36, because 6×6=366 \times 6 = 36

Sections you finish are checked off in the contents.