GCSE · Probability

Conditional Probability Higher

"Given that it's raining, what's the chance the bus is late?" Once you know something has happened, some outcomes are ruled out and the chances change. Conditional probability is about shrinking the whole to just the outcomes that are still possible, then counting again.

GCSE subject content (Higher tier): calculate and interpret conditional probabilities through representation using expected frequencies with two-way tables, tree diagrams and Venn diagrams

Before you start

P(event) = outcomes you want ÷ all outcomes, and two-way tables and Venn diagrams with frequencies: sample spaces and Venn diagrams. Notice the KS3 trap there: "a girl is picked" means divide by the girls' total. That was conditional probability in disguise.
Tree diagrams: multiply along, add the ends; without replacement the second stage changes. See tree diagrams and combined events.
Set notation: A ∩ B (A and B), A ∪ B (A or B), A′ (not A). See Venn diagrams and set notation.

1 · "Given that" shrinks the whole Higher

Your friend rolls a dice, hides it, and tells you "it's even". What's the probability it's a 6? Before she spoke it was 1/6. But now 1, 3 and 5 are impossible. Only 2, 4 and 6 are left, all equally likely, so it's 1/3. The extra information shrank the "whole" from 6 outcomes to 3.

We write this P(A | B), said "the probability of A given B". The bar | means "given that". B is the thing you know happened, the condition. To find it: throw away everything outside B, then count how much of what's left is also in A:

P(A | B) = n(A ∩ B) ÷ n(B)

A number from 1 to 20 is picked at random. Choose what you're told (the condition) and what you want. Watch the tiles outside the condition fade away.

given B:

want A:

P(A), no information

P(A | B)

2 · From a two-way table Higher

A two-way table already sorts people into groups, so conditional probability is just choosing the right total to divide by. "Given that the student is a girl" means you only look at the girls' row: its total becomes the bottom of the fraction. The top is the cell where that row meets what you want.

Here are 200 students and their favourite subject. Choose the condition and the event. The table and the dot picture (one dot per student) fade out everyone ruled out by the condition.

given want

P =

Read who is being picked. "A student is picked. Find P(girl and Art)" divides by everyone (200). "A girl is picked. Find P(Art)" divides by the girls (115). "A student who likes Art is picked. Find P(girl)" divides by the Art column (70). Same top number, three different bottoms.

3 · From a Venn diagram Higher

Exactly the same idea. "Given B" means you only look inside circle B. Add up the numbers inside B: that's the bottom. The top is the part of B that is also in A, the overlap. Choose a conditional probability and the diagram shades the new whole in teal and the part you want in orange.

P =

4 · From a tree diagram Higher

You've been using conditional probabilities on trees all along. The second-stage branches after R are probabilities given that R happened: they are P(L | R) and P(L′ | R). When the second stage depends on the first, the branches after R and after R′ are different.

Here R is "it rains" and L is "the bus is late". Set the probabilities. The AND rule now reads P(R and L) = P(R) × P(L | R).

The harder exam question goes backwards: "Given that the bus was late, what's the probability it rained?" That's P(R | L), and it is not on any branch. Shrink the whole to the paths where L happened (teal), then ask what fraction of that is the rain path (orange).

P(R) P(L | R) P(L | R′)

P =

5 · The formula P(A | B) = P(A ∩ B) ÷ P(B) Higher

Sometimes you only have probabilities, not counts. Counting gave P(A | B) = n(A ∩ B) ÷ n(B). Divide the top and the bottom by the total N, and nothing changes, but each part becomes a probability:

P(A | B) = P(A ∩ B) ÷ P(B)

Picture the whole sample space as a bar of length 1. B is the teal part; A ∩ B is the orange part inside it. "Given B" means zoom in until B fills the whole bar. The orange part is now P(A | B).

P(B) P(A ∩ B)

answer =

Rearranged, the formula is the AND rule for dependent events: P(A ∩ B) = P(B) × P(A | B). That's exactly "multiply along the branches" on a tree.

6 · Testing for independence Higher

If knowing B tells you nothing about A, the condition doesn't change anything: P(A | B) = P(A). That is the precise meaning of independent. (Equivalently, P(A ∩ B) = P(A) × P(B).) If P(A | B) is different from P(A), B is giving you information, so the events are dependent.

Below, A is "wears glasses" and B is "is left-handed". Type any numbers in the table. The bars compare P(A) for everyone with P(A | B) for the left-handers and P(A | B′) for everyone else.

7 · The classic trap: P(A | B) ≠ P(B | A) Higher

"The probability a footballer is a girl" and "the probability a girl is a footballer" sound alike but divide by different groups. Mixing them up is the most common conditional-probability mistake, and it has real consequences. The famous example is medical testing.

A disease affects a small percentage of people. A test is good at spotting it, but sometimes says "positive" for healthy people too (a false positive). You test positive. How worried should you be? Don't use percentages; use natural frequencies: imagine 10 000 people and count.

have the disease positive if ill positive if healthy

P(positive | ill)

P(ill | positive)

With the starting numbers, the test catches 90% of ill people, yet a positive result means only about a 15% chance you're ill. The healthy group is so much bigger that its 5% of false positives outnumber all the true positives. Doctors confirm with a second test for exactly this reason.

Now practise telling the two apart. For each sentence, which probability is it?

Common mistakes and exam traps

Dividing by the grand total. That's P(A ∩ B), not P(A | B). "Given", "of those who", "if the person is" all mean: divide by the condition's total.
Swapping A and B. P(A | B) divides by B; P(B | A) divides by A. Say it in words before you calculate.
Reading the "backwards" answer off a tree. P(R | L) is not on a branch. Find P(L) from all L paths, then divide.
Forgetting regions in a Venn diagram. n(B) is everything inside circle B, including the overlap. n(B′) includes the outside.
Assuming independence. Only call events independent if P(A | B) = P(A) (or the question says so).

Worked examples

Use the 200-student table: find P(Art | girl) and P(girl | Art).

The table: girls: Maths 32, Art 48, Music 35, total 115. Art column: boys 22, girls 48, total 70.

P(Art | girl): only girls are possible, so divide by the girls' total. P = 48/115. (It doesn't simplify: 48 = 2⁴ × 3 and 115 = 5 × 23.)

P(girl | Art): only Art-lovers are possible, so divide by the Art total: 48/70 = 24/35.

Notice: same top, different bottoms, different answers: 48/115 ≈ 0.417 but 24/35 ≈ 0.686.

50 people were asked: 23 drink tea (T), 18 drink coffee (C) and 7 drink both. Find P(T | C) and P(C | T′).

Fill the Venn diagram from the middle: T ∩ C = 7; T only = 23 − 7 = 16; C only = 18 − 7 = 11; neither = 50 − (16 + 7 + 11) = 50 − 34 = 16.

P(T | C): inside circle C there are 7 + 11 = 18 people. Of these, 7 also drink tea. P = 7/18.

P(C | T′): T′ is everyone outside circle T: C only (11) and neither (16), 27 people. Of these, 11 drink coffee. P = 11/27.

P(rain) = 0.3. If it rains, P(bus late) = 0.6; if not, P(bus late) = 0.2. The bus was late. Find the probability that it rained.

Tree paths with L: R and L: 0.3 × 0.6 = 0.18. R′ and L: 0.7 × 0.2 = 0.14.

P(L): add every path where the bus is late: 0.18 + 0.14 = 0.32.

P(R | L) = P(R ∩ L) ÷ P(L) = 0.18 ÷ 0.32 = 18/32 = 9/16 (= 0.5625).

Sense check: rain only happens 30% of the time, but late buses are mostly rainy-day buses, so the chance goes up to over a half.

P(A) = 0.5, P(B) = 0.4 and P(A ∩ B) = 0.2. Find P(A | B) and decide whether A and B are independent.

Formula: P(A | B) = P(A ∩ B) ÷ P(B) = 0.2 ÷ 0.4 = 0.5.

Compare: P(A | B) = 0.5 = P(A). Knowing B doesn't change the probability of A.

So A and B are independent. Check the other way: P(A) × P(B) = 0.5 × 0.4 = 0.2 = P(A ∩ B) ✓.

A bag has 5 red and 3 blue counters. Two are taken without replacement. Given that at least one is red, find the probability that both are red.

Tree: RR = 5/8 × 4/7 = 20/56; RB = 5/8 × 3/7 = 15/56; BR = 3/8 × 5/7 = 15/56; BB = 3/8 × 2/7 = 6/56.

The condition, "at least one red": RR, RB and BR. P = 20/56 + 15/56 + 15/56 = 50/56. (Or 1 − 6/56.)

Both red and at least one red is just RR: 20/56.

Divide: P(both red | at least one red) = (20/56) ÷ (50/56) = 20/50 = 2/5. The 56ths cancel, so you can divide the numerators.

1% of people have a disease. A test is positive for 90% of people with it, and for 5% of people without it. Someone tests positive. Find the probability that they have the disease.

Imagine 10 000 people. 1% = 100 are ill; 9900 are healthy.

Ill and positive: 90% of 100 = 90. Healthy and positive: 5% of 9900 = 495.

Everyone positive: 90 + 495 = 585.

P(ill | positive) = 90/585 = 2/13 ≈ 0.154. Not 0.9: that's P(positive | ill).

Quick check

What you've learned

P(A | B) means "the probability of A, given that B has happened". The condition B becomes the new whole.
From counts: P(A | B) = n(A ∩ B) ÷ n(B). In a two-way table, divide by the condition's row or column total; in a Venn diagram, by everything inside the condition's region.
From probabilities: P(A | B) = P(A ∩ B) ÷ P(B), and so P(A ∩ B) = P(B) × P(A | B).
On a tree, second-stage branches are conditional probabilities. To go backwards, add every path for the condition, then divide.
A and B are independent exactly when P(A | B) = P(A).
P(A | B) is usually not P(B | A). Natural frequencies (imagine 10 000 people) make problems like false positives easy to reason about.