Conditional Probability
If you know it’s already cloudy, your prediction of rain changes. That’s conditional probability — the probability of one event given that another has already happened. It’s used in healthcare, sports, weather forecasting, and your Higher GCSE exam.
The Core Concept
What is Conditional Probability?
Conditional probability is the probability of event A happening, given that event B has already happened. The notation is $P(A|B)$, read as “the probability of A given B”.
The key idea: knowing B has happened changes the total outcomes you consider — you now focus on a smaller reduced sample space.
In practice: numerator = favourable outcomes in the “given” group; denominator = total in the “given” group.
$P(A|B)$ vs $P(A \text{ and } B)$
These are easily confused but very different:
- $P(A \text{ and } B)$ — probability both A and B happen, out of the entire original group.
- $P(A|B)$ — probability A happens only among the outcomes where B has occurred.
Example: “What’s the chance a student likes both football and rugby from all 30 students?” vs “Out of students who like rugby, what’s the chance they also like football?” — the total changes.
Worked Examples
Example 1: Two-way Table
120 students were asked whether they walk to school and which year group they’re in. Find $P(\text{Walks} | \text{Year 8})$.
| Walks | Other transport | Total | |
|---|---|---|---|
| Year 7 | 35 | 25 | 60 |
| Year 8 ← focus | 20 | 40 | 60 |
| Total | 55 | 65 | 120 |
- Condition is “Year 8” → look only at Year 8 row.
- Total Year 8 students = 60 (denominator).
- Year 8 students who walk = 20 (numerator).
$P(\text{Walks} | \text{Year 8}) = \dfrac{20}{60} = \mathbf{\dfrac{1}{3}}$
Example 2: Venn Diagram
In a class of 30: 18 play football (F), 10 play basketball (B), 6 play both. Find $P(\text{Basketball} | \text{Football})$.
- Condition is “Football” → focus on F circle total = 18 (denominator).
- Students who play both = 6 (numerator).
$P(\text{Basketball} | \text{Football}) = \dfrac{6}{18} = \mathbf{\dfrac{1}{3}}$
Example 3: Tree Diagram (Without Replacement)
A bag has 6 red and 4 blue balls. One is drawn and not replaced. Then a second is drawn. Find $P(\text{2nd is Red} | \text{1st is Blue})$.
- Condition is “1st is Blue” → follow only the Blue branch at the first pick.
- After removing one blue ball, there are 9 balls left: 6 red and 3 blue.
- The probability of picking Red second, given the first was Blue, is shown directly on the highlighted branch: $\dfrac{6}{9}$.
$P(\text{2nd Red} | \text{1st Blue}) = \dfrac{6}{9} = \mathbf{\dfrac{2}{3}}$
Key insight: The numbers on the second set of branches are the conditional probabilities — the tree diagram does the work for you.
Tutor Insights
🤔 Common Misunderstandings
- Mixing up $P(A|B)$ and $P(A \text{ and } B)$. Remember: $P(A \text{ and } B)$ is from the whole group; $P(A|B)$ narrows your focus to the subset where B has occurred.
- Forgetting to adjust in “without replacement” problems. If the first ball isn’t put back, both the total count and the specific colour count change for the second pick.
📝 Common Exam Mistakes
- Wrong denominator — using the overall total instead of the “given” group total. This is the most common error.
- Misreading tree diagram branches — using the joint probability (multiplying branches) instead of reading the conditional directly from the second set of branches.
- Not simplifying fractions in the final answer.
Practice Questions
- Two-way table: 150 students surveyed on favourite sport.
(a) Find $P(\text{Girl})$. (b) Find $P(\text{Netball} | \text{Girl})$.Football Netball Total Boys 45 15 60 Girls 30 60 90 Total 75 75 150 - Tree diagram: A bag has 5 green and 3 yellow sweets. Sara picks one and eats it, then picks another.
(a) Draw a tree diagram. (b) Find $P(\text{2nd is Yellow} | \text{1st is Green})$. - Venn diagram: 50 people: 25 read paper A, 20 read paper B, 8 read both.
(a) Draw a Venn diagram. (b) Find $P(\text{Reads A} | \text{Reads B})$. - Tree diagram challenge: $P(\text{Rain}) = 0.4$. If it rains, $P(\text{Heavy traffic}) = 0.7$. If no rain, $P(\text{Heavy traffic}) = 0.2$.
(a) Draw a tree diagram. (b) Find $P(\text{Rain} | \text{Heavy traffic})$. - Medical test: 1000 people tested for a rare disease. 10 have the disease; 9 of these test positive. Of the 990 without the disease, 90 test positive.
(a) Complete a two-way table. (b) Find $P(\text{Has disease} | \text{Tests positive})$ and explain what it means.
Show Answers
- (a) $P(\text{Girl}) = \frac{90}{150} = \mathbf{\frac{3}{5}}$. (b) Reduced sample space = 90 girls; 60 prefer Netball. $P(\text{Netball}|\text{Girl}) = \frac{60}{90} = \mathbf{\frac{2}{3}}$.
- (b) After picking Green first, 4G and 3Y remain (7 total). $P(\text{Yellow 2nd}|\text{Green 1st}) = \mathbf{\frac{3}{7}}$.
- A only = 17, B only = 12, Both = 8, Neither = 13. (b) $P(\text{Reads A}|\text{Reads B}) = \frac{8}{20} = \mathbf{\frac{2}{5}}$.
- (b) $P(\text{Rain} \cap \text{HT}) = 0.4 \times 0.7 = 0.28$. $P(\text{No Rain} \cap \text{HT}) = 0.6 \times 0.2 = 0.12$. $P(\text{HT}) = 0.28 + 0.12 = 0.40$. $P(\text{Rain}|\text{HT}) = \frac{0.28}{0.40} = \mathbf{\frac{7}{10}}$.
- (a) Disease+Positive=9, Disease+Negative=1; No Disease+Positive=90, No Disease+Negative=900. Total Positive=99. (b) $P(\text{Disease}|\text{Positive}) = \frac{9}{99} = \mathbf{\frac{1}{11}} \approx 9\%$. This means even a positive test result only gives about a 9% chance of actually having the disease — because the disease is rare and the false positive rate is relatively high.
FAQs
Q: What’s the difference between $P(A \text{ and } B)$ and $P(A|B)$?
A: $P(A \text{ and } B)$ is the probability that both happen out of the entire original set. $P(A|B)$ is the probability of A specifically among the outcomes where B has occurred — a smaller, reduced sample space.
Q: Which diagram should I use for which type of problem?
A: Two-way tables are best for organised frequency counts of two categories. Tree diagrams are ideal for sequences of events, especially without-replacement problems. Venn diagrams are great for showing overlapping groups visually.
Q: Is conditional probability always harder than regular probability?
A: Not once you understand the “reduced sample space” idea. The calculation is often just a simple fraction — the challenge is identifying which numbers to use as numerator and denominator.
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