The shape of the lesson
By the end of the lesson, learners can
- The first success: The conditions are the same as for the binomial (independent trials, two outcomes, constant \(p\)) except that there is NO fixed number of trials: you stop at the first success.
- Cumulative probabilities: No long sums are needed. 'More than \(r\) trials are needed' means the first \(r\) trials all fail, which has probability \((1 - p)^{r}\).
- Expectation and the least n: On average you wait \(\tfrac{1}{p}\) trials for the first success: 6 throws for a six, 2 tosses for a head.
- Model: independent trials, constant \(p\), counting up to the first success: \(X \sim \text{Geo}(p)\).
- Exact: \(\text{P}(X = r) = p\,q^{r-1}\).
- Cumulative: \(\text{P}(X > r) = q^{r}\), \(\text{P}(X \le r) = 1 - q^{r}\).
- Mean: \(\text{E}(X) = \dfrac{1}{p}\).
- Least n: solve \(q^{n} < \dots\) by trying values, to 4 s.f.
One picture every idea returns to
If \(X \sim \text{Geo}(p)\), \(X\) is the number of the trial on which the first success occurs: \(\text{P}(X = r) = p(1 - p)^{r - 1}\) for \(r = 1, 2, 3, \dots\). Also \(\text{P}(X > r) = (1 - p)^{r}\) (the first \(r\) trials all fail) and \(\text{E}(X) = \dfrac{1}{p}\).
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, Waiting for the first success.. Keep repeating a trial until it succeeds for the first time, and count the trials. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: The first success | The conditions are the same as for the binomial (independent trials, two outcomes, constant \(p\)) except that there is NO fixed number of trials: you stop at the first success. Model the worked example An exact probability on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Cumulative probabilities | No long sums are needed. 'More than \(r\) trials are needed' means the first \(r\) trials all fail, which has probability \((1 - p)^{r}\). Model the worked example More than, at most on the board, one line at a time. Your turn: \(X \sim \text{Geo}(0.2)\). Find \(\text{P}(X \ge 4)\). | Pairs |
| 21 to 30 min | Teach and check: Expectation and the least n | On average you wait \(\tfrac{1}{p}\) trials for the first success: 6 throws for a six, 2 tosses for a head. Model the worked example Expectation both ways on the board, one line at a time. | Rally Coach |
| 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, 20 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 Expectation and the least n, 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.
Rally Coach
Pairs share one worksheet. One partner solves the first question aloud while the other coaches, checking each line and offering a prompt rather than the answer; they swap roles for the next question.
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 |
|---|---|
| Using \((1 - p)^{r}\,p\). | If the first success is on trial \(r\), only \(r - 1\) trials fail before it, so the power is \(r - 1\). |
| Mixing up \(X > r\) and \(X \ge r\). \(\text{P}(X \ge 4) = q^{3}\), not \(q^{4}\): only the first three trials have to fail. | Return to Cumulative probabilities: ask a pair to state the correct version and show it on the lab before moving on. |
| Taking \(n = 8\) because \(0.75^{8}\) 'rounds to 0.1'. | The inequality is strict, and 0.1001 is not less than 0.1. Work to 4 significant figures. |
The geometric distribution: worksheet
6 questions, 20 marks. Written in the style of the papers; no past paper question is reproduced.
A basketball player scores from a free throw with probability 0.7. She keeps taking free throws until she scores. State a suitable distribution for the number of throws she takes, and its mean.
\(X \sim \text{Geo}(0.15)\). Find (a) \(\text{P}(X = 3)\), (b) \(\text{P}(X \le 4)\).
\(X \sim \text{Geo}(p)\), where \(p > 0.5\), and \(\text{P}(X = 2) = 0.24\). Find \(p\) and \(\text{E}(X)\).
The probability that a randomly chosen person has blood group B is 0.12. People are tested one at a time. Find the least number of people who must be tested for the probability of finding at least one person with group B to be at least 0.95.
A fair die is thrown until a six appears. Show that the probability that the number of throws is even is \(\tfrac{5}{11}\).
Ana and Ben take turns to throw a fair die, Ana first. The first person to throw a six wins.
(a) Find the probability that the game lasts more than 4 throws. (b) Find the probability that Ana wins. (c) Find the expected number of throws in a game.
Answer key and marking notes
- \(X \sim \text{Geo}(0.7)\), assuming throws are independent with constant probability of scoring.
- \(\text{E}(X) = \dfrac{1}{0.7} = 1.43\) (3 s.f.).
- (a) \(0.15 \times 0.85^{2} = 0.108\) (3 s.f.).
- (b) \(1 - 0.85^{4} = 1 - 0.5220 = 0.478\) (3 s.f.).
- \(p(1 - p) = 0.24\), so \(p^{2} - p + 0.24 = 0\), \((p - 0.6)(p - 0.4) = 0\).
- \(p > 0.5\), so \(p = 0.6\).
- \(\text{E}(X) = \dfrac{1}{0.6} = \dfrac{5}{3}\).
- \(X \sim \text{Geo}(0.12)\); need \(\text{P}(X \le n) = 1 - 0.88^{n} \ge 0.95\), i.e. \(0.88^{n} \le 0.05\).
- \(0.88^{23} = 0.0529\): too big. \(0.88^{24} = 0.0465\): works.
- Least \(n = 24\).
- \(\text{P}(X \text{ even}) = \text{P}(2) + \text{P}(4) + \dots = \tfrac{1}{6}\left(\tfrac{5}{6}\right) + \tfrac{1}{6}\left(\tfrac{5}{6}\right)^3 + \dots\)
- This is a geometric series with first term \(\tfrac{5}{36}\) and common ratio \(\left(\tfrac{5}{6}\right)^2 = \tfrac{25}{36}\).
- Sum to infinity \(= \dfrac{5/36}{1 - 25/36} = \dfrac{5/36}{11/36} = \dfrac{5}{11}\).
- (a) The first 4 throws are not sixes: \(\left(\tfrac{5}{6}\right)^4 = 0.482\) (3 s.f.).
- (b) Ana wins on throw 1, 3, 5, ...: \(\tfrac{1}{6} + \tfrac{1}{6}\left(\tfrac{5}{6}\right)^2 + \dots = \dfrac{1/6}{1 - 25/36} = \dfrac{6}{11}\).
- (c) The number of throws \(X \sim \text{Geo}\left(\tfrac{1}{6}\right)\), so \(\text{E}(X) = 6\).
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.
\(X \sim \text{Geo}(0.35)\). Find \(\text{P}(X = 3)\).
- A0.148
- B0.0961
- C0.0796
- D0.444
\(X \sim \text{Geo}(0.15)\). Find \(\text{P}(X > 4)\).
- A0.522
- B0.614
- C0.478
- D0.444
\(X\) has a geometric distribution with \(\text{E}(X) = 5\). Find \(\text{P}(X = 2)\).
- A0.16
- B0.04
- C0.128
- D0.32
\(X \sim \text{Geo}(p)\) and \(q = 1 - p\). Which statement is true?
- A\(\text{P}(X \ge r) = q^{r-1}\)
- B\(\text{P}(X \ge r) = q^{r}\)
- CThe most likely value of \(X\) is \(\dfrac{1}{p}\)
- D\(X\) can take the value 0
\(X \sim \text{Geo}(0.2)\). Find \(\text{P}(2 \le X \le 5)\).
- A0.472
- B0.312
- C0.538
- D0.672
\(X \sim \text{Geo}(0.1)\). Find the least value of \(n\) such that \(\text{P}(X \le n) > 0.75\).
- A14
- B13
- C10
- D15
A fair die is thrown until the first six appears. Find the probability that the first six occurs on an even-numbered throw.
- A\(\dfrac{5}{11}\)
- B\(\dfrac{1}{2}\)
- C\(\dfrac{6}{11}\)
- D\(\dfrac{5}{36}\)
Three friends each throw a fair die repeatedly until they get a six. Find the probability that exactly two of the three friends need more than 4 throws.
- A0.361
- B0.388
- C0.120
- D0.423
Quiz key
Two failures then a success: \(\text{P}(X = 3) = 0.65^2 \times 0.35 = 0.147\,875 = 0.148\) (3 s.f.). The tempting 0.0961 uses \(0.65^3\), one failure too many. 0.0796 swaps \(p\) and \(1 - p\), and 0.444 multiplies by 3 as if it were a binomial count.
\(\text{P}(X > 4) = 0.85^4 = 0.522\) (3 s.f.). The tempting 0.614 is \(0.85^3 = \text{P}(X > 3) = \text{P}(X \ge 4)\), which wrongly includes \(X = 4\). 0.478 is \(\text{P}(X \le 4)\), and 0.444 is \(0.85^5 = \text{P}(X > 5)\).
\(\dfrac{1}{p} = 5\), so \(p = 0.2\). \(\text{P}(X = 2) = 0.8 \times 0.2 = 0.16\). The tempting 0.04 is \(0.2^2\), two successes. 0.128 is \(0.8^2 \times 0.2 = \text{P}(X = 3)\), and 0.32 multiplies by 2 as if the order did not matter.
\(X \ge r\) happens exactly when the first \(r - 1\) trials all fail, so \(\text{P}(X \ge r) = q^{r-1}\). The tempting \(q^{r}\) is \(\text{P}(X > r)\). The most likely value is always 1 (the probabilities shrink by the factor \(q\)), while \(\dfrac{1}{p}\) is the mean. \(X\) counts trials up to and including the first success, so its smallest value is 1.
\(2 \le X \le 5\) is the same as \(1 < X \le 5\), so the probability is \(0.8^1 - 0.8^5 = 0.8 - 0.327\,68 = 0.472\) (3 s.f.). The tempting 0.312 is \(0.8^2 - 0.8^5\), which leaves out \(X = 2\). 0.538 is \(0.8 - 0.8^6\), which wrongly includes \(X = 6\), and 0.672 is \(\text{P}(X \le 5)\), which includes \(X = 1\).
Need \(0.9^{n} < 0.25\). \(0.9^{13} = 0.2542\), too big; \(0.9^{14} = 0.2288 < 0.25\). Least \(n = 14\). The tempting 13 rounds 0.2542 to 0.25, but the inequality is strict and must actually hold. 10 is \(\text{E}(X)\), and 15 works but is not the least.
\(\text{P}(X = 2) = \tfrac{5}{6} \times \tfrac{1}{6} = \tfrac{5}{36}\), and each later even term is \(\left(\tfrac{5}{6}\right)^2 = \tfrac{25}{36}\) times the one before. Sum to infinity \(= \dfrac{5/36}{1 - 25/36} = \dfrac{5/36}{11/36} = \dfrac{5}{11}\). The tempting \(\dfrac{1}{2}\) assumes odd and even are equally likely, but throw 1 is the most likely. \(\dfrac{6}{11}\) is the probability of an odd-numbered throw, and \(\dfrac{5}{36}\) is the first term only.
One friend: \(\text{P}(X > 4) = \left(\tfrac{5}{6}\right)^4 = 0.4823\). The number of friends needing more than 4 throws is \(\text{B}(3, 0.4823)\): \(\text{P}(2) = 3 \times 0.4823^2 \times 0.5177 = 0.361\) (3 s.f.). The tempting 0.423 uses \(\left(\tfrac{5}{6}\right)^3\), which is \(\text{P}(X \ge 4)\), not \(\text{P}(X > 4)\). 0.388 is the probability that exactly two need at most 4 throws, and 0.120 leaves out the coefficient 3.
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.