Teacher page: The binomial distribution

The one thing to do with the animation

Set the inequality to "more than 3" and ask which bars should be shaded.

A good number of students will include the bar at 3. The shading makes the off-by-one visible rather than abstract, and it is worth doing twice with different wordings.

Then switch to "at least one" and watch them try to add nine bars. The single unshaded bar at zero is the whole method: 1 minus P(X = 0), and seeing that saves them a calculator error in every exam they sit.

The four conditions, used as a filter

Fixed number of trials, two outcomes, constant p, independent trials. Students can recite these and still not apply them, so make it a filter: before any calculator work, ask which of the four a question might break.

Without replacement breaks two of them at once, and that is the distractor in Q3 and in most “is this binomial” questions.

The answers

1. P(X = 3) for B(10, 0.3)	120 × 0.027 × 0.0824 = 0.2668.
2. P(X ≥ 4)	1 − 0.6496 = 0.3504. Note the 3 in the subtraction, not the 4.
3. Not binomial	B, drawing without replacement. p changes after each draw, so the trials are not independent and p is not constant. That is a tree diagram problem.

Where the marks go

1 markStating the distribution as B(n, p) with both numbers, before computing anything.

1 markConverting the inequality correctly, which is where the 3 or 4 decision lives.

1 markReading the right calculator function: exactly versus up to and including.

What each wrong answer tells you

They give	What it means
0.6496 (Q1 or Q2)	Gave P(X ≤ 3). They used the cumulative function where they wanted the individual one, or forgot to subtract.
0.3828 (Q1)	Used P(X ≤ 3) minus P(X ≤ 2) incorrectly, or mixed up n and x in the coefficient.
0.027 (Q1)	0.3³, so the combinations factor and the failures are both missing. They have the shape of the formula and none of the structure.
0.6172 (Q2)	Subtracted P(X ≤ 4) instead of P(X ≤ 3). This is the off-by-one the widget is for.
0.2001 (Q2)	Computed P(X = 4) alone, treating “at least” as “exactly”.
"A, tossing a coin" (Q3)	They are looking for something that sounds complicated rather than checking the four conditions. Send them back to the filter.

Other things they will say

"At least or at most?" Make them rewrite every inequality in terms of ≤ before touching the calculator, because that is the only form the calculator offers. Doing it in writing, every time, costs ten seconds and saves the mark.

"What if n is huge?" Still binomial, and still one calculator call. The formula is impractical by hand at n = 200, which is a good moment to say that the formula is for understanding and the calculator is for answers.

"Is 0.5 special?" Only in that it makes the distribution symmetric. Nothing else changes, and asking it is a sign they are thinking about shape, which is worth encouraging.

A possible order

	What is happening
1	The shading exercise with two or three different wordings. Do not move on early; this is the lesson.
2	The four conditions as a filter, with the without-replacement case.
3	The three questions, on calculators, with the inequality rewritten in every one.
4	Mean and variance, np and np(1 − p), and a question that needs both.
5	Set up the link to the normal approximation if your group is going on to Higher Level.

Two things not to say

Do not let them compute without writing B(n, p) first. It is a mark, and it also stops them using the wrong n.

Do not call the four conditions “just the assumptions”. Questions award marks for spotting a broken one, so they are content.

Questions to set

Three tiers, ramping the way practice should: the method on its own, then the method inside something real, then a challenge. Set the tier the class in front of you needs rather than one undifferentiated sheet. Answers are given so these can go straight onto a board.

1Fluency

The method on its own, with friendly numbers. Set these first and move on quickly once they are secure.

Single probabilityX ~ B(12, 0.25). Find P(X = 3).
0.258
Mean and varianceFor the same X state the mean and the standard deviation.
np = 3 and np(1 − p) = 2.25, so the standard deviation is 1.5.

2In context

The same skill inside a real situation, where the first job is working out what is being asked.

At leastA quiz has 12 questions with 4 options each and a student guesses every one. Find the probability of at least 3 correct.
1 − P(X ≤ 2) = 1 − 0.391 = 0.609.
Check the conditionsState the four conditions a binomial model needs, then say which is doubtful for 'number of students late in a week'.
Fixed n, two outcomes, constant p, independent trials. Independence is doubtful: the same traffic or weather makes several latenesses happen together.

3Challenge

Reasoning, working backwards, or spotting an error. These are where the top grades are decided.

More than, carefullyExplain why P(X > 3) is not 1 − P(X ≤ 2), and give the correct statement.
More than 3 starts at 4, so P(X > 3) = 1 − P(X ≤ 3). The expression given is P(X ≥ 3). One step either way changes the answer and this is the most common binomial slip.
Work backwards to nFor p = 0.25, find the smallest n for which P(at least one success) exceeds 0.9.
1 − 0.75ⁿ > 0.9 needs 0.75ⁿ < 0.1, so n > log 0.1 / log 0.75 = 8.00 to 2 dp. Testing: n = 8 gives 0.8999, n = 9 gives 0.9249, so n = 9. This is why the inequality should be checked rather than rounded.

Practicalities

Works on a phone. Nothing is loaded from any other site, so it runs behind a school firewall, and nothing a student does is saved or sent anywhere.