Teacher page: Venn diagrams, tree diagrams and conditional probability

The one thing to do with the animation

Before you drag anything, ask the class whether two mutually exclusive events are independent.

Most will say yes, because both words sound like "nothing to do with each other". Then drag the circles apart until they no longer overlap and read P(A | B) off the panel: it is zero.

Knowing B happened has told you A definitely did not, which is the strongest possible dependence. Mutually exclusive events are about as far from independent as events can get, and one drag settles it permanently.

Why the conditional bar is a division and not a lookup

Students read P(A | B) as "find B, then find A". Drag the overlap and watch the denominator change: the whole of B becomes the new universe, and the numerator is only the part of A inside it.

Say the sentence “given B, B is now everything” every time. It is the one idea that makes the formula memorable instead of arbitrary.

The answers

1. P(F or T)	(18 + 12 − 7) over 30 = 23/30 = 0.767.
2. P(F \| T)	7 of the 12 tennis players also play football, so 7/12 = 0.583.
3. P(both red)	5/8 × 4/7 = 20/56 = 0.3571.

Where the marks go

1 markSubtracting the overlap once in an “or” question. Adding 18 and 12 to get 1 is the error the mark exists to catch.

1 markDividing by the condition, not by the grand total, on a conditional question.

1 markChanging the denominator on the second branch of a without-replacement tree.

What each wrong answer tells you

They give	What it means
1 (Q1)	Added 18 and 12 and divided by 30 without removing the 7 counted twice. A probability of exactly 1 should have stopped them; worth saying that aloud.
0.233 (Q1)	Found the complement, or answered a different question. Check which, because they are different problems.
0.6 (Q1 or Q2)	Used 18/30 or divided by 30 instead of by 12. This is the conditional error: they looked A up rather than restricting to B.
0.389 (Q2)	7/18, so they conditioned on football rather than tennis. The bar is not symmetric and this is where that bites.
0.3906 (Q3)	(5/8)². They treated the second draw as a fresh one, which is replacement.
0.625 (Q3)	Stopped after the first draw.

Other things they will say

"Is P(A | B) the same as P(B | A)?" No, and this question is worth two minutes. P(ill | positive test) and P(positive test | ill) are wildly different numbers, and confusing them is how people misread medical results.

"When do I add and when do I multiply?" Add across alternatives, multiply along a path. On a tree that is literally down the page versus across it, which is why trees are worth drawing even when a formula would do.

"Do the branches have to add to 1?" Every set of branches leaving one point does. It is the fastest check there is, and it catches a wrong complement immediately.

A possible order

	What is happening
1	The independence question, cold, before any theory. Collect hands.
2	Drag to no overlap. Read P(A \| B) = 0. Let it land before you name anything.
3	The formula, the Venn, and the three questions.
4	Without replacement on the tree, with the denominator change said out loud.
5	Past paper conditional question in context. Medical testing is the one that holds attention.

Two things not to say

Do not say “mutually exclusive means independent” as shorthand for “separate”. The lesson above exists because that sentence gets remembered.

Do not draw the Venn with equal circles and a symmetric overlap every time. It teaches that P(A | B) and P(B | A) are the same, and then you have to unteach it.

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.

Union formulaP(A) = 0.6, P(B) = 0.5 and P(A ∪ B) = 0.8. Find P(A ∩ B).
0.6 + 0.5 − 0.8 = 0.3.
ConditionalFor the same events find P(A | B).
0.3 / 0.5 = 0.6.

2In context

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

Test independenceUsing those values, decide whether A and B are independent, showing the test.
P(A)P(B) = 0.6 × 0.5 = 0.3, which equals P(A ∩ B), so they ARE independent. Note they also overlap, so independent and mutually exclusive are not the same thing.
A dependent pairNow take P(A ∪ B) = 0.85 with the same P(A) and P(B). Test independence again.
P(A ∩ B) = 0.25 but P(A)P(B) = 0.3, so they are NOT independent. Here P(A | B) = 0.5, below P(A) = 0.6, so B makes A less likely.

3Challenge

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

The common confusionA student says that because A and B can both happen they cannot be independent. Correct them using the first set of numbers.
Independence is about whether knowing B changes P(A), not about whether they can co-occur. In the first set they overlap AND are independent. Mutually exclusive events with non-zero probability are in fact always dependent, since one happening makes the other impossible.
Build the treeA box has 5 red and 7 blue. Two are drawn without replacement. Find P(exactly one red), via the tree.
(5/12)(7/11) + (7/12)(5/11) = 70/132 = 35/66 ≈ 0.530. Both branches, because exactly one does not say which draw.

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.