The shape of the lesson
By the end of the lesson, learners can
- When is it binomial?: Check the four conditions every time, and say them in the context of the question when asked to justify the model.
- At most, at least, between: Cumulative questions mean adding several terms.
- Mean, variance and finding n: The mean \(np\) and variance \(np(1-p)\) can be used forwards, or backwards to find \(n\) and \(p\).
- Binomial in context: Longer questions often use one binomial answer as the \(p\) of a second binomial: first find the probability that one pack, week or group 'succeeds', then count how many packs, weeks or groups succeed.
- Model: state \(X\), \(n\) and \(p\), and check the four conditions.
- Exact: \(\text{P}(X = r) = \dbinom{n}{r} p^{r}(1-p)^{n-r}\).
- Cumulative: turn the words into an inequality; use the complement when shorter.
- Mean and variance: \(np\) and \(np(1-p)\).
- Least n: set up the inequality and try values either side.
One picture every idea returns to
If \(X \sim \text{B}(n, p)\), then \(\text{P}(X = r) = \dbinom{n}{r} p^{r} (1 - p)^{n - r}\) for \(r = 0, 1, \dots, n\), with \(\text{E}(X) = np\) and \(\text{Var}(X) = np(1 - p)\).
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, Count the successes.. Repeat the same independent trial a fixed number of times and count how often it succeeds: that count follows a binomial distribution. 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: When is it binomial? | Check the four conditions every time, and say them in the context of the question when asked to justify the model. Model the worked example One exact probability on the board, one line at a time. | Whole class, then pairs |
| 11 to 17 min | Teach and check: At most, at least, between | Cumulative questions mean adding several terms. Model the worked example At most on the board, one line at a time. Your turn: \(X \sim \text{B}(12, 0.4)\). Find \(\text{P}(X > 2)\). | Pairs |
| 17 to 23 min | Teach and check: Mean, variance and finding n | The mean \(np\) and variance \(np(1-p)\) can be used forwards, or backwards to find \(n\) and \(p\). Model the worked example Mean and variance on the board, one line at a time. | Pairs |
| 23 to 30 min | Teach and check: Binomial in context | Longer questions often use one binomial answer as the \(p\) of a second binomial: first find the probability that one pack, week or group 'succeeds', then count how many packs, weeks or groups succeed. Model the worked example Two layers on the board, one line at a time. | Round Table |
| 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, 24 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 Binomial in context, 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.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
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 |
|---|---|
| Leaving out \(\dbinom{n}{r}\). \(0.3^3 \times 0.7^7\) is the probability of ONE particular order, such as SSSFFFFFFF; there are 120 orders. | Return to When is it binomial?: ask a pair to state the correct version and show it on the lab before moving on. |
| Treating 'more than 2' as \(X \ge 2\). | For whole numbers 'more than 2' starts at 3. |
| Stopping at \(n = 20\) because \(0.0115\) 'is about 0.01'. | The inequality must actually hold, so check the value either side. |
| Rounding the first answer to 0.74 before using it again. | Keep at least 4 significant figures in intermediate values so the final 3 s.f. is right. |
The binomial distribution: worksheet
6 questions, 24 marks. Written in the style of the papers; no past paper question is reproduced.
A fair die is thrown 15 times and \(X\) is the number of sixes. State the distribution of \(X\) and give two conditions needed for it to be valid.
\(X \sim \text{B}(15, 0.2)\). Find (a) \(\text{P}(X = 4)\), (b) \(\text{P}(X < 3)\).
\(X \sim \text{B}(n, p)\) has mean 4.8 and variance 2.88. Find \(n\) and \(p\), and hence find \(\text{P}(X = 5)\).
A fair die is thrown \(n\) times. Find the least value of \(n\) for which the probability of at least one six is greater than 0.9.
A multiple-choice test has 20 questions, each with 4 options, only one of which is correct. A student guesses every answer.
(a) Find the expected number of correct answers and the variance. (b) The pass mark is 8. Find the probability that the student passes.
A light bulb passes a quality check with probability 0.9, independently of other bulbs. Bulbs are packed in boxes of 10, and a box is 'good' if at least 9 of its bulbs pass.
(a) Find the probability that a box is good. (b) A shop orders 6 boxes. Find the probability that at least 5 of them are good.
Answer key and marking notes
- \(X \sim \text{B}(15, \tfrac{1}{6})\).
- The throws are independent, and the probability of a six is the same (\(\tfrac{1}{6}\)) on every throw.
- (a) \(\dbinom{15}{4}(0.2)^4(0.8)^{11} = 1365 \times 0.0016 \times 0.085\,90 = 0.188\) (3 s.f.).
- (b) \(\text{P}(0) + \text{P}(1) + \text{P}(2) = 0.035\,18 + 0.131\,94 + 0.230\,89 = 0.398\) (3 s.f.).
- \(1 - p = 2.88 \div 4.8 = 0.6\), so \(p = 0.4\) and \(n = 4.8 \div 0.4 = 12\).
- \(\text{P}(X = 5) = \dbinom{12}{5}(0.4)^5(0.6)^7 = 792 \times 0.010\,24 \times 0.027\,99 = 0.227\) (3 s.f.).
- Need \(1 - \left(\tfrac{5}{6}\right)^n > 0.9\), i.e. \(\left(\tfrac{5}{6}\right)^n < 0.1\).
- \(\left(\tfrac{5}{6}\right)^{12} = 0.112\): too big. \(\left(\tfrac{5}{6}\right)^{13} = 0.0935 < 0.1\).
- Least \(n = 13\).
- (a) \(X \sim \text{B}(20, 0.25)\): \(\text{E}(X) = 5\), \(\text{Var}(X) = 20 \times 0.25 \times 0.75 = 3.75\).
- (b) \(\text{P}(X \ge 8) = 1 - \text{P}(X \le 7)\).
- \(\text{P}(X \le 7) = 0.8982\) (adding the eight terms from 0 to 7).
- \(\text{P}(X \ge 8) = 0.102\) (3 s.f.).
- (a) \(X \sim \text{B}(10, 0.9)\): \(\text{P}(9) + \text{P}(10) = 10(0.9)^9(0.1) + 0.9^{10} = 0.3874 + 0.3487 = 0.7361\), so 0.736 (3 s.f.).
- (b) \(Y \sim \text{B}(6, 0.7361)\).
- \(\text{P}(Y = 5) = 6(0.7361)^5(0.2639) = 0.3422\); \(\text{P}(Y = 6) = (0.7361)^6 = 0.1591\).
- \(\text{P}(Y \ge 5) = 0.501\) (3 s.f.).
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.
\(X \sim \text{B}(9, 0.35)\). Find \(\text{P}(X = 4)\).
- A0.219
- B0.001\,74
- C0.828
- D0.118
\(X \sim \text{B}(8, 0.4)\). Find \(\text{P}(X < 2)\).
- A0.106
- B0.315
- C0.0896
- D0.894
Which of these random variables has a binomial distribution?
- AThe number of sixes when a fair die is thrown 20 times
- BThe number of red cards when 5 cards are dealt from a pack without replacement
- CThe number of throws of a fair die needed to get the first six
- DThe total score when a fair die is thrown 20 times
\(X \sim \text{B}(25, 0.12)\). Find \(\text{E}(X)\) and \(\text{Var}(X)\).
- A\(\text{E}(X) = 3\), \(\text{Var}(X) = 2.64\)
- B\(\text{E}(X) = 3\), \(\text{Var}(X) = 1.62\)
- C\(\text{E}(X) = 3\), \(\text{Var}(X) = 0.36\)
- D\(\text{E}(X) = 2.64\), \(\text{Var}(X) = 3\)
\(X \sim \text{B}(15, 0.3)\). Find \(\text{P}(X > 3)\).
- A0.703
- B0.873
- C0.297
- D0.485
\(X \sim \text{B}(n, p)\) has mean 4.8 and variance 2.88. Find \(n\) and \(p\).
- A\(n = 12\), \(p = 0.4\)
- B\(n = 8\), \(p = 0.6\)
- C\(n = 20\), \(p = 0.24\)
- D\(n = 6\), \(p = 0.8\)
Each time Kai enters a prize draw, the probability that he wins is 0.15, independently. Find the least number of draws he must enter for the probability that he wins at least once to be greater than 0.95.
- A19
- B18
- C7
- D20
8% of the items made by a machine are defective, independently. Items are packed in boxes of 10, and a box passes inspection if it contains at most 1 defective item. Four boxes are inspected. Find the probability that exactly 3 of them pass.
- A0.403
- B0.101
- C0.536
- D0.838
Quiz key
\(\text{P}(X = 4) = \dbinom{9}{4}(0.35)^4(0.65)^5 = 126 \times 0.015\,006 \times 0.116\,03 = 0.219\) (3 s.f.). The tempting 0.001 74 leaves out \(\dbinom{9}{4}\), giving the probability of one particular order. 0.828 is \(\text{P}(X \le 4)\), and 0.118 swaps \(p\) and \(1 - p\).
\(\text{P}(X = 0) = 0.6^8 = 0.016\,80\) and \(\text{P}(X = 1) = 8 \times 0.4 \times 0.6^7 = 0.089\,58\). Sum \(= 0.106\) (3 s.f.). The tempting 0.315 is \(\text{P}(X \le 2)\), treating "fewer than 2" as including 2. 0.0896 is \(\text{P}(X = 1)\) alone, and 0.894 is \(\text{P}(X \ge 2)\).
Twenty independent throws, each a six (success) or not with constant \(p = \tfrac{1}{6}\): \(\text{B}(20, \tfrac{1}{6})\). The tempting card example fails because dealing without replacement changes \(p\) and makes the trials dependent. The number of throws to the first six has no fixed \(n\) (it is geometric), and a total score is not a count of successes.
\(\text{E}(X) = 25 \times 0.12 = 3\) and \(\text{Var}(X) = 25 \times 0.12 \times 0.88 = 2.64\). The tempting 1.62 is the standard deviation \(\sqrt{2.64}\). 0.36 is \(np^2\), and the last option swaps the mean and the variance.
\(\text{P}(X \le 3) = 0.004\,75 + 0.030\,52 + 0.091\,56 + 0.170\,04 = 0.296\,87\), so \(\text{P}(X > 3) = 1 - 0.296\,87 = 0.703\) (3 s.f.). The tempting 0.873 is \(1 - \text{P}(X \le 2) = \text{P}(X \ge 3)\), which wrongly includes 3. 0.297 is \(\text{P}(X \le 3)\), and 0.485 also subtracts \(\text{P}(X = 4)\).
\(np = 4.8\) and \(np(1 - p) = 2.88\), so \(1 - p = \dfrac{2.88}{4.8} = 0.6\), \(p = 0.4\) and \(n = 4.8 \div 0.4 = 12\). The tempting \(n = 8\), \(p = 0.6\) takes 0.6 as \(p\) instead of \(1 - p\); its variance is 1.92. The other options also have mean 4.8, but variances 3.648 and 0.96.
Need \(1 - 0.85^{n} > 0.95\), so \(0.85^{n} < 0.05\). \(0.85^{18} = 0.053\,65\), too big; \(0.85^{19} = 0.045\,60 < 0.05\). Least \(n = 19\). The tempting 18 comes from rounding \(0.0536\) to 0.05; it does not satisfy the inequality. 7 is about \(\tfrac{1}{0.15}\), the expected wait for one win, and 20 works but is not the least.
One box: \(\text{P}(\text{pass}) = 0.92^{10} + 10 \times 0.08 \times 0.92^{9} = 0.4344 + 0.3777 = 0.8121\). Boxes: \(Y \sim \text{B}(4, 0.8121)\), \(\text{P}(Y = 3) = 4 \times 0.8121^3 \times 0.1879 = 0.403\) (3 s.f.). The tempting 0.101 leaves out the coefficient 4. 0.536 is \(0.8121^3\), ignoring the box that fails, and 0.838 is \(\text{P}(Y \ge 3)\).
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.