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IB Maths Probability and Expectation Paper 4: Error Review
IB Maths Probability and Expectation Paper 4: Error Review
Ten original error-review items, each with an answer and a one-line explanation.
Direct answers Items 1-3 revisit the numerator and denominator in basic probability, 4-6 cover the averaging meaning of expectation, 7-9 apply it to expected gain, two dice and drawing without replacement, and item 10 clarifies that an expectation need not be attainable. Cover the answers first, then check each note.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: IB Maths Probability and Expectation |
| Level | IB |
02 Questions (10 items)
- FoundationError review: a fair die is rolled once. Find P(4).
- FoundationError review: a fair coin is tossed twice. Find the probability of two heads.
- FoundationError review: a bag holds 3 red and 5 blue balls. One ball is drawn at random. Find P(red).
- IntermediateError review: a fair die is rolled once. Find the probability of an even score.
- IntermediateError review: a fair die is rolled once. Find the expected score.
- IntermediateError review: four cards are numbered 1, 2, 3 and 4. One is drawn at random. Find the expected value.
- ChallengeError review: a game pays $10 with probability 0.3 and otherwise loses $4. Find the expected gain.
- ChallengeError review: two fair dice are rolled. Find the probability that the total is 7.
- ChallengeError review: a bag holds 2 red and 3 green balls. Two are drawn without replacement. Find P(both red).
- Error redoError review: a pupil says the expected value must be an outcome that can actually occur, yet a die has expected score 3.5. What is the right idea?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 1/6 | The six outcomes are equally likely and only one is a 4, so the probability is 1/6; 4/6 puts the score in the numerator. |
| 2 | 1/4 | The tosses are independent, so 1/2 x 1/2 = 1/4; 1/2 counts only one toss. |
| 3 | 3/8 | There are 3 favourable and 3 + 5 = 8 equally likely outcomes, so 3/8; 3/5 wrongly uses the blue count as the denominator. |
| 4 | 1/2 | The even scores 2, 4 and 6 give 3 out of 6, that is 1/2; 1/3 counts a single even score. |
| 5 | 3.5 | (1 + 2 + 3 + 4 + 5 + 6) / 6 = 3.5; answering 3 confuses the median with the mean. |
| 6 | 2.5 | (1 + 2 + 3 + 4) / 4 = 2.5; answering 2 just takes the middle integer. |
| 7 | $0.2 | 0.3 x 10 – 0.7 x 4 = 3 – 2.8 = 0.2; counting only the win gives $3. |
| 8 | 1/6 | Six of the 36 equally likely pairs total 7, so 6/36 = 1/6; 1/11 wrongly treats the eleven possible totals as equally likely. |
| 9 | 1/10 | The first draw is 2/5 and the second is 1/4 of the four left, giving 2/20 = 1/10; using 2/5 twice forgets there is no replacement. |
| 10 | The expected value is a long-run average, not necessarily an attainable outcome | It is the mean over many trials: 3.5 describes the long-run average and never appears on a single roll; treating it as the most likely score is the root of this slip. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
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