📊 Probability Table
How to build, complete and read a probability table
🎯 What Is a Probability Table?
A probability table is a tool that displays all probabilities of two events in an organised format.
The table helps us:
- See all possibilities at a glance
- Calculate conditional probabilities easily
- Check whether events are independent
📐 Table Structure
| \(B\) | \(\bar{B}\) (not B) | Row total | |
|---|---|---|---|
| \(A\) | \(P(A \cap B)\) | \(P(A \cap \bar{B})\) | \(P(A)\) |
| \(\bar{A}\) (not A) | \(P(\bar{A} \cap B)\) | \(P(\bar{A} \cap \bar{B})\) | \(P(\bar{A})\) |
| Column total | \(P(B)\) | \(P(\bar{B})\) | 1 |
✅ Table Rules
1️⃣ Row sum
Sum of every cell in a row = that row's probability
\(P(A \cap B) + P(A \cap \bar{B}) = P(A)\)
2️⃣ Column sum
Sum of every cell in a column = that column's probability
\(P(A \cap B) + P(\bar{A} \cap B) = P(B)\)
3️⃣ Grand total
Sum of all 4 inner cells = 1
Total row/column always sums to 1
4️⃣ Complement
\(P(A) + P(\bar{A}) = 1\)
\(P(B) + P(\bar{B}) = 1\)
✏️ Example 1: Completing a Table from Given Data
Given:
- 40% of students study maths (M)
- 50% study physics (P)
- 20% study both maths and physics
Step 1: Enter what is known
| P | Not P | Total | |
|---|---|---|---|
| M | 0.20 | ? | 0.40 |
| Not M | ? | ? | ? |
| Total | 0.50 | ? | 1 |
Step 2: Complete the missing values
\(P(M \cap \bar{P}) = P(M) - P(M \cap P) = 0.40 - 0.20 = 0.20\)
\(P(\bar{M} \cap P) = P(P) - P(M \cap P) = 0.50 - 0.20 = 0.30\)
\(P(\bar{P}) = 1 - P(P) = 1 - 0.50 = 0.50\)
\(P(\bar{M}) = 1 - P(M) = 1 - 0.40 = 0.60\)
\(P(\bar{M} \cap \bar{P}) = P(\bar{M}) - P(\bar{M} \cap P) = 0.60 - 0.30 = 0.30\)
Step 3: Completed table
| P | Not P | Total | |
|---|---|---|---|
| M | 0.20 | 0.20 | 0.40 |
| Not M | 0.30 | 0.30 | 0.60 |
| Total | 0.50 | 0.50 | 1 |
Check: \(0.20 + 0.20 + 0.30 + 0.30 = 1\) ✓
🔍 Reading Information from the Table
Using the table we completed, answer these questions:
Question 1: What is the probability that a student studies maths or physics (or both)?
\(P(M \cup P) = P(M) + P(P) - P(M \cap P) = 0.40 + 0.50 - 0.20 = 0.70\)
Or: sum of all cells except \(P(\bar{M} \cap \bar{P})\): \(1 - 0.30 = 0.70\)
Question 2: What is the probability that a student studies only maths (and not physics)?
\(P(M \cap \bar{P}) = 0.20\) (read directly from the table)
Question 3: What is the probability that a student studies neither subject?
\(P(\bar{M} \cap \bar{P}) = 0.30\) (read directly from the table)
🎯 Conditional Probability from the Table
\(P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\text{shared cell}}{\text{total of the condition's row/column}}\)
Question 4: What is the probability that a student studies maths, given that they study physics?
\(P(M|P) = \frac{P(M \cap P)}{P(P)} = \frac{0.20}{0.50} = \frac{2}{5} = 0.40\)
Question 5: What is the probability that a student studies physics, given that they study maths?
\(P(P|M) = \frac{P(M \cap P)}{P(M)} = \frac{0.20}{0.40} = \frac{1}{2} = 0.50\)
Question 6: What is the probability that a student does not study physics, given that they do not study maths?
\(P(\bar{P}|\bar{M}) = \frac{P(\bar{M} \cap \bar{P})}{P(\bar{M})} = \frac{0.30}{0.60} = \frac{1}{2} = 0.50\)
🔗 Testing Independence from the Table
Events are independent if:
\(P(A \cap B) = P(A) \cdot P(B)\)
Check in our example:
\(P(M) \cdot P(P) = 0.40 \times 0.50 = 0.20\)
\(P(M \cap P) = 0.20\)
Since \(0.20 = 0.20\) ✓ → the events are independent!
💡 Alternative check:
If \(P(A|B) = P(A)\) → independent
In the example: \(P(M|P) = 0.40 = P(M)\) ✓
✏️ Example 2: Table with Counts (Not Probabilities)
Given: A class of 30 students. 18 girls, 12 boys. 20 wear glasses – of whom 14 are girls.
Frequency table:
| Glasses | No glasses | Total | |
|---|---|---|---|
| Girls | 14 | 4 | 18 |
| Boys | 6 | 6 | 12 |
| Total | 20 | 10 | 30 |
Calculations:
Probability that a random student is a girl with glasses:
\(P(\text{girl} \cap \text{glasses}) = \frac{14}{30} = \frac{7}{15}\)
Probability that a student with glasses is a boy (conditional):
\(P(\text{boy}|\text{glasses}) = \frac{6}{20} = \frac{3}{10}\)
(6 boys with glasses out of 20 with glasses)
Probability that a random girl wears glasses:
\(P(\text{glasses}|\text{girl}) = \frac{14}{18} = \frac{7}{9}\)
(14 girls with glasses out of 18 girls)
💡 Tips for the Exam
1️⃣ Always draw a table
Even if not asked – it helps organise the data
2️⃣ Start with what is known
Fill in the given values first, then complete by subtraction
3️⃣ Check the sums
Rows and columns sum to their totals, and everything sums to 1
4️⃣ Conditional = divide
The required cell divided by the total of the condition
📝 Summary
Row total = \(P(A)\) | Column total = \(P(B)\) | Grand total = 1
\(P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\text{shared cell}}{\text{total of condition}}\)
Independent: \(P(A \cap B) = P(A) \cdot P(B)\)