The shape of the lesson
By the end of the lesson, learners can
- Probabilities add to one: A probability distribution lists every value a discrete random variable \(X\) can take, with its probability.
- Drawing up the table: Most questions describe a situation and ask you to draw up the distribution.
- Expectation and variance: \(\text{E}(X)\) is what the average value of \(X\) would settle to over many trials.
- Games and profit: In a game, let \(X\) be the profit (winnings minus the cost to play). \(\text{E}(X) < 0\) means the player loses on average; \(\text{E}(X) = 0\) is a fair game.
- Values: list every value \(X\) can take.
- Probabilities: trees, sample spaces or combinations; check they add to 1.
- Expectation: \(\text{E}(X) = \sum x p\).
- Variance: \(\text{E}(X^2) - [\text{E}(X)]^2\) with \(\text{E}(X^2) = \sum x^2 p\).
- Unknowns: one equation from \(\sum p = 1\), another from the given \(\text{E}(X)\).
One picture every idea returns to
The probabilities in a distribution add up to 1. \(\text{E}(X) = \sum x\,p\) is the long-run mean, and \(\text{Var}(X) = \sum x^2 p - \left[\text{E}(X)\right]^2\) measures the spread: the same 'mean of the squares minus the square of the mean' as for data.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Every value, every probability.. A discrete random variable takes separate values, each with a probability. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 11 min | Teach and check: Probabilities add to one | A probability distribution lists every value a discrete random variable \(X\) can take, with its probability. Model the worked example Finding a constant on the board, one line at a time. Your turn: \(\text{P}(X = x) = \dfrac{x^2}{c}\) for \(x = 1, 2, 3\). Find \(c\). | Whole class, then pairs |
| 11 to 17 min | Teach and check: Drawing up the table | Most questions describe a situation and ask you to draw up the distribution. Model the worked example Counters without replacement on the board, one line at a time. | Pairs |
| 17 to 23 min | Teach and check: Expectation and variance | \(\text{E}(X)\) is what the average value of \(X\) would settle to over many trials. Model the worked example From a formula on the board, one line at a time. | Pairs |
| 23 to 30 min | Teach and check: Games and profit | In a game, let \(X\) be the profit (winnings minus the cost to play). \(\text{E}(X) < 0\) means the player loses on average; \(\text{E}(X) = 0\) is a fair game. Model the worked example Is it worth playing? on the board, one line at a time. | Numbered Heads Together |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 25 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Games and profit, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Forgetting that \(\text{P}(X \ge 3)\) includes 3. 'At least 3' and 'more than 2' are the same event for whole-number values; 'more than 3' is not. | Return to Probabilities add to one: ask a pair to state the correct version and show it on the lab before moving on. |
| Missing a route in \(\text{P}(X = 1)\). | One green can come first OR second: red-green and green-red are both needed. |
| Stopping at \(\text{E}(X^2)\), or writing \(\text{Var}(X) = \text{E}(X^2) - \text{E}(X)\). | The expectation must be SQUARED before subtracting. |
| Using the winnings instead of the profit. | Expected winnings are \(\tfrac{7}{6} = \$1.17\); you must still subtract the $1.50 stake. |
Discrete probability distributions, E(X) and Var(X): worksheet
6 questions, 25 marks. Written in the style of the papers; no past paper question is reproduced.
\(\text{P}(X = x) = k(x + 1)\) for \(x = 0, 1, 2, 3\). Find \(k\) and \(\text{P}(X \ge 2)\).
\(X\) takes values 1, 2, 3, 4 with probabilities 0.2, 0.3, 0.4, 0.1. Find \(\text{E}(X)\) and \(\text{Var}(X)\).
Two fair dice are thrown. \(X\) is the difference between the scores (larger minus smaller, or 0 if equal).
(a) Draw up the probability distribution of \(X\). (b) Find \(\text{E}(X)\) and \(\text{Var}(X)\).
\(X\) takes values \(-1, 0, 1, 2\) with probabilities \(a\), 0.3, \(b\), 0.2. Given that \(\text{E}(X) = 0.5\), find \(a\), \(b\) and \(\text{Var}(X)\).
Three people are chosen at random from 4 women and 3 men. \(X\) is the number of women chosen.
(a) Show that \(\text{P}(X = 2) = \tfrac{18}{35}\). (b) Draw up the probability distribution of \(X\). (c) Find \(\text{E}(X)\) and \(\text{Var}(X)\).
A bag contains 3 red and 2 blue discs. A player takes discs one at a time, without replacement, until a red disc is taken. \(X\) is the number of discs taken.
(a) Find the probability distribution of \(X\) and \(\text{E}(X)\). (b) It costs $2.50 to play. The player receives $4 if \(X = 1\), $1 if \(X = 2\) and nothing if \(X = 3\). Find the expected profit per game for the player, and comment.
Answer key and marking notes
- \(k(1 + 2 + 3 + 4) = 1\), so \(k = 0.1\).
- \(\text{P}(X \ge 2) = 0.3 + 0.4 = 0.7\).
- \(\text{E}(X) = 0.2 + 0.6 + 1.2 + 0.4 = 2.4\).
- \(\text{E}(X^2) = 0.2 + 1.2 + 3.6 + 1.6 = 6.6\).
- \(\text{Var}(X) = 6.6 - 2.4^2 = 6.6 - 5.76 = 0.84\).
- (a) Counting the 36 outcomes: \(x = 0\): \(\tfrac{6}{36}\); 1: \(\tfrac{10}{36}\); 2: \(\tfrac{8}{36}\); 3: \(\tfrac{6}{36}\); 4: \(\tfrac{4}{36}\); 5: \(\tfrac{2}{36}\).
- (b) \(\text{E}(X) = \dfrac{0 + 10 + 16 + 18 + 16 + 10}{36} = \dfrac{70}{36} = \dfrac{35}{18} = 1.94\) (3 s.f.).
- \(\text{E}(X^2) = \dfrac{10 + 32 + 54 + 64 + 50}{36} = \dfrac{210}{36} = \dfrac{35}{6}\).
- \(\text{Var}(X) = \dfrac{35}{6} - \left(\dfrac{35}{18}\right)^2 = 2.05\) (3 s.f.).
- \(a + b = 0.5\) and \(-a + b + 0.4 = 0.5\), so \(b - a = 0.1\).
- \(b = 0.3\), \(a = 0.2\).
- \(\text{E}(X^2) = 0.2 + 0 + 0.3 + 0.8 = 1.3\); \(\text{Var}(X) = 1.3 - 0.25 = 1.05\).
- (a) \(\dfrac{{}^{4}C_{2} \times {}^{3}C_{1}}{{}^{7}C_{3}} = \dfrac{6 \times 3}{35} = \dfrac{18}{35}\).
- (b) \(\text{P}(X = 0) = \tfrac{1}{35}\), \(\text{P}(X = 1) = \tfrac{4 \times 3}{35} = \tfrac{12}{35}\), \(\text{P}(X = 2) = \tfrac{18}{35}\), \(\text{P}(X = 3) = \tfrac{4}{35}\).
- (c) \(\text{E}(X) = \dfrac{12 + 36 + 12}{35} = \dfrac{60}{35} = \dfrac{12}{7}\).
- \(\text{E}(X^2) = \dfrac{12 + 72 + 36}{35} = \dfrac{24}{7}\); \(\text{Var}(X) = \dfrac{24}{7} - \dfrac{144}{49} = \dfrac{24}{49}\).
- (a) \(\text{P}(X = 1) = \tfrac{3}{5}\); \(\text{P}(X = 2) = \tfrac{2}{5} \times \tfrac{3}{4} = \tfrac{3}{10}\); \(\text{P}(X = 3) = \tfrac{2}{5} \times \tfrac{1}{4} \times 1 = \tfrac{1}{10}\).
- \(\text{E}(X) = 0.6 + 0.6 + 0.3 = 1.5\).
- (b) Expected receipts \(= 0.6 \times 4 + 0.3 \times 1 + 0.1 \times 0 = 2.7\).
- Expected profit \(= 2.7 - 2.5 = \$0.20\) per game. The player gains on average, so the game is not fair, and the organiser loses money in the long run.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
The random variable \(X\) has \(\text{P}(X = x) = k(6 - x)\) for \(x = 1, 2, 3, 4, 5\). Find \(\text{P}(X > 2)\).
- A\(\dfrac{2}{5}\)
- B\(\dfrac{3}{5}\)
- C\(\dfrac{2}{3}\)
- D6
\(X\) takes the values \(-1, 0, 2, 5\) with probabilities 0.3, 0.25, 0.3, 0.15 respectively. Find \(\text{E}(X)\).
- A1.05
- B1.5
- C1.65
- D0.263
\(X\) takes the values 1, 2, 3 with probabilities 0.2, 0.5, 0.3 respectively. Find \(\text{Var}(X)\).
- A0.49
- B2.8
- C4.9
- D0.7
A bag contains 3 red and 5 blue discs. Two discs are taken at random without replacement, and \(X\) is the number of red discs taken. Find \(\text{P}(X = 1)\).
- A\(\dfrac{15}{28}\)
- B\(\dfrac{15}{56}\)
- C\(\dfrac{15}{32}\)
- D\(\dfrac{3}{28}\)
\(X\) takes the values 1, 2, 3, 4 with probabilities 0.15, \(a\), \(b\), 0.25 respectively, and \(\text{E}(X) = 2.7\). Find \(\text{Var}(X)\).
- A1.01
- B8.3
- C5.6
- D0.51
In a game, a player pays $2 and tosses two fair coins. One head wins $1.50, two heads win \(w\) dollars, and no heads wins nothing. Find the value of \(w\) that makes the game fair (expected profit zero).
- A5
- B6.5
- C3
- D8
Two fair dice are thrown and \(X\) is the positive difference between the two scores (0 if they are equal). Find \(\text{E}(X)\).
- A\(\dfrac{35}{18}\)
- B\(\dfrac{5}{2}\)
- C\(\dfrac{7}{3}\)
- D\(\dfrac{35}{36}\)
\(X\) takes the values 1, 2, 3, 4 with probabilities 0.2, \(k\), \(k^2\), 0.24 respectively. Find \(\text{E}(X)\).
- A2.44
- B2.68
- C0.4
- D2.5
Quiz key
\(k(5 + 4 + 3 + 2 + 1) = 15k = 1\), so \(k = \dfrac{1}{15}\). \(\text{P}(X > 2) = \text{P}(3) + \text{P}(4) + \text{P}(5) = \dfrac{3 + 2 + 1}{15} = \dfrac{2}{5}\). The tempting \(\dfrac{3}{5}\) is \(\text{P}(X \le 2)\), the complement. \(\dfrac{2}{3}\) is \(\text{P}(X \ge 2)\), which wrongly includes 2, and 6 forgets to multiply by \(k\).
\(\text{E}(X) = (-1)(0.3) + 0(0.25) + 2(0.3) + 5(0.15) = -0.3 + 0 + 0.6 + 0.75 = 1.05\). The tempting 1.5 is the plain mean of the four values, ignoring the probabilities. 1.65 drops the minus sign on \(-1\), and 0.263 divides the correct sum by 4.
\(\text{E}(X) = 0.2 + 1 + 0.9 = 2.1\). \(\text{E}(X^2) = 0.2 + 2 + 2.7 = 4.9\). \(\text{Var}(X) = 4.9 - 2.1^2 = 4.9 - 4.41 = 0.49\). The tempting 2.8 subtracts \(\text{E}(X)\) without squaring it. 4.9 is \(\text{E}(X^2)\), and 0.7 is the standard deviation.
Red then blue: \(\dfrac{3}{8} \times \dfrac{5}{7} = \dfrac{15}{56}\). Blue then red: \(\dfrac{5}{8} \times \dfrac{3}{7} = \dfrac{15}{56}\). \(\text{P}(X = 1) = \dfrac{30}{56} = \dfrac{15}{28}\). The tempting \(\dfrac{15}{56}\) counts only one order. \(\dfrac{15}{32}\) treats the draws as with replacement, and \(\dfrac{3}{28}\) is \(\text{P}(X = 2)\).
\(a + b = 0.6\) and \(0.15 + 2a + 3b + 1 = 2.7\), so \(2a + 3b = 1.55\). Then \(b = 1.55 - 1.2 = 0.35\) and \(a = 0.25\). \(\text{E}(X^2) = 0.15 + 4(0.25) + 9(0.35) + 16(0.25) = 8.3\), so \(\text{Var}(X) = 8.3 - 2.7^2 = 8.3 - 7.29 = 1.01\). 8.3 is \(\text{E}(X^2)\), and 5.6 subtracts 2.7 without squaring. The tempting 0.51 swaps the values of \(a\) and \(b\).
P(two heads) \(= \tfrac{1}{4}\), P(one head) \(= \tfrac{1}{2}\) (HT or TH). Fair: \(\tfrac{1}{4}w + \tfrac{1}{2}(1.5) = 2\), so \(\tfrac{1}{4}w = 1.25\) and \(w = 5\). The tempting 6.5 uses \(\tfrac{1}{4}\) for one head, forgetting that it can happen in two ways. 3 is the profit when \(w = 5\), not the prize, and 8 ignores the $1.50 prize.
Frequencies out of 36: difference 0: 6, 1: 10, 2: 8, 3: 6, 4: 4, 5: 2. \(\text{E}(X) = \dfrac{0 + 10 + 16 + 18 + 16 + 10}{36} = \dfrac{70}{36} = \dfrac{35}{18}\). The tempting \(\dfrac{5}{2}\) averages the values 0 to 5 as if they were equally likely. \(\dfrac{7}{3}\) divides by 30, leaving out the 6 doubles, and \(\dfrac{35}{36}\) counts each unequal pair once instead of twice.
\(0.2 + k + k^2 + 0.24 = 1\), so \(k^2 + k - 0.56 = 0\), \((k + 1.4)(k - 0.4) = 0\) and \(k = 0.4\) (\(-1.4\) is not a probability). Probabilities 0.2, 0.4, 0.16, 0.24. \(\text{E}(X) = 0.2 + 0.8 + 0.48 + 0.96 = 2.44\). The tempting 2.68 attaches 0.16 to \(x = 2\) and 0.4 to \(x = 3\), mixing up \(k\) and \(k^2\). 0.4 is \(k\), and 2.5 is the mean of the values.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.