2.4 Events, Sample Spaces and Complements — Answers and Rubrics
These solutions explain the new practice tasks in this lesson. For simulation parts, a valid recorded run may differ from another learner’s run; the rubric checks the model, recorded work and interpretation.
Original generated practice for the Fall 2026 course. Return to the lesson and independent checks.
Three Fair Coin Tosses: worked answer
A fair coin is tossed independently three times. Record the tosses in order. Let E be the event “at least two heads.” This is a new complete model; all eight ordered outcomes have equal probability.
Part a
List the complete sample space. Explain why its outcomes are nonoverlapping and exhaustive, and state its total probability.
The space is HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Every ordered three-toss result appears once, so the outcomes are nonoverlapping and exhaustive. Their probabilities sum to 1.
| Evidence for credit | Points |
|---|---|
| Lists all eight outcomes once. | 1 |
| Explains exhaustive/nonoverlapping structure and total probability 1. | 1 |
Part b
Identify every outcome in E and calculate P(E). Explain why a count ratio is justified. Write the complement using E′, an overbar, and E superscript C; describe it in words and calculate its probability in two ways.
E contains HHH, HHT, HTH and THH, so . Fair independent tosses make all eight ordered outcomes equally likely. means fewer than two heads and contains HTT, THT, TTH, TTT. Its probability is , or .
| Evidence for credit | Points |
|---|---|
| Correctly identifies E and obtains 1/2 with the equal-likelihood reason. | 1 |
| Correctly names/describes the complement and all three notations. | 1 |
| Shows both direct counting and subtraction for the complement. | 1 |
Part c
Calculate the probability of at least one head and the probability of four heads. Explain the endpoints/range check, and show that E and its complement account for the whole sample space.
At least one head has probability ; four heads in three tosses is impossible and has probability 0. These and the earlier probabilities lie in [0,1]. E and its complement exhaust the space, and .
| Evidence for credit | Points |
|---|---|
| Obtains 7/8 and 0 with valid reasoning. | 1 |
| Checks [0,1] and the event/complement total 1. | 1 |