Conditional Probability — Mastering “Given That

Introduction

Conditional probability measures the chance of an event happening given that another event has already occurred. We write this as \(P(A \mid B)\): “the probability of \(A\) given \(B\)”. It focuses only on outcomes where \(B\) is true and asks, “within that smaller world, how likely is \(A\)?”

Conditional probability appears throughout GCSE questions: test positivity given a condition, selecting items from groups, card and ball problems without replacement, and word problems that include phrases like “if”, “given that”, or “among those who…”. It connects tightly to both tree diagrams (branch labels become conditional probabilities) and Venn diagrams (reading areas within a region).

Formally, when \(P(B) > 0\), \[ P(A \mid B) \;=\; \frac{P(A \cap B)}{P(B)}. \] This formula says: restrict to \(B\) (the denominator), then look at the fraction that also lies in \(A\) (the overlap \(A \cap B\)).

Conditional probability also helps us test for independence. If events \(A\) and \(B\) are independent, then knowing \(B\) does not change the chance of \(A\): \[ P(A \mid B) = P(A) \quad \text{and} \quad P(A \cap B) = P(A)\,P(B). \] If these equalities fail, the events are dependent.

In this tutorial, you will learn to:

• Interpret “given that” statements and translate them into set/notation correctly.
• Compute \(P(A \mid B)\) using both Venn diagrams and tree diagrams.
• Apply the product rule \(P(A \cap B) = P(B)\,P(A \mid B)\) and its variants.
• Recognise and use independence tests in exam questions.
• Avoid common traps such as swapping the condition (mixing up \(P(A \mid B)\) with \(P(B \mid A)\)).

Key Vocabulary

Here are the essential terms and symbols for conditional probability. Understanding them will help you translate “given that” questions into the correct mathematical form.

• Conditional probability: The probability of an event happening given that another event has already occurred. Notation: \(P(A \mid B)\).
• Given that: A phrase meaning we restrict attention to the part of the sample space where the condition holds. Example: “Given that a student plays football, what is the probability they also play basketball?”
• Intersection (\(A \cap B\)): Outcomes where both events happen. Conditional probability is based on comparing this overlap with the size of the condition set.
• Formula: \[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \quad \text{(provided \(P(B) > 0\))}. \] This is the formal definition of conditional probability.
• Product rule: Rearranging gives \[ P(A \cap B) = P(B) \times P(A \mid B). \] This is especially useful in tree diagram problems.
• Independence: Events \(A\) and \(B\) are independent if knowing \(B\) does not change the chance of \(A\). Formally: \(P(A \mid B) = P(A)\).
• Dependent events: If \(P(A \mid B) \neq P(A)\), the events influence each other (e.g. drawing cards without replacement).
• Sample space: The complete set of all possible outcomes. Conditional probability narrows this space to a subset.

Core Ideas

Conditional probability problems can look complex, but they are built on a few simple principles. Mastering these will help you break down any “given that” question.

• Restricting the sample space: \(P(A \mid B)\) looks only at cases where \(B\) happens. Think of it as “zooming in” on circle \(B\) in a Venn diagram and asking what fraction of those are also in \(A\).
• Definition formula: \[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \quad \text{if } P(B) > 0. \] This is the most important rule to remember.
• Product rule: From the definition we get \[ P(A \cap B) = P(B) \times P(A \mid B). \] This lets you calculate joint probabilities by multiplying down a tree diagram.
• Tree diagrams with conditional probabilities: Each branch at the second level is a conditional probability (depends on the first branch). Example: \(P(\text{2nd red} \mid \text{1st red})\).
• Independence check: If \(P(A \mid B) = P(A)\), then \(A\) and \(B\) are independent. If not, they are dependent.
• Reversing conditions: \(P(A \mid B)\) and \(P(B \mid A)\) are usually different. Don’t confuse them unless \(A\) and \(B\) are symmetric or independent.
• Using complements: Sometimes it’s easier to calculate the complement first. Example: “Given that it rained, what is the probability it did not flood?” That’s \(P(\text{not flood} \mid \text{rain}) = 1 - P(\text{flood} \mid \text{rain})\).

Step-by-Step Method

Follow these steps to solve any conditional probability problem:

1. Identify the events: Decide which event is the condition (\(B\)) and which event is being measured (\(A\)).
2. Translate the wording: Look for “given that …” or “among those who …”. This always means a conditional probability.
3. Write the formula: \[ P(A \mid B) = \frac{P(A \cap B)}{P(B)}. \] Make sure you know both the overlap and the total size of \(B\).
4. Use the product rule if needed: If the joint probability isn’t given, use \[ P(A \cap B) = P(B) \times P(A \mid B). \] This is especially useful in tree diagrams.
5. Fill diagrams if possible: Use a Venn diagram for set problems, or a tree diagram for sequential events. These visuals help prevent misreading the condition.
6. Calculate carefully: Work step by step. Always divide the overlap by the probability (or frequency) of the condition.
7. Check the result: Your answer must be between 0 and 1. If it’s bigger than 1 or negative, you’ve swapped the condition or misread the problem.

Worked Examples

Example 1 (Foundation — Venn Diagram)

In a class of 40 students:

• 18 study French (\(F\))
• 12 study German (\(G\))
• 6 study both

Step 1: \(P(F \cap G) = \tfrac{6}{40}\). Step 2: \(P(G) = \tfrac{12}{40}\). Step 3: Apply formula: \[ P(F \mid G) = \frac{6/40}{12/40} = \frac{6}{12} = 0.5. \] Answer: 0.5

Example 2 (Foundation — Tree Diagram)

A bag has 3 red and 2 blue counters. Two are taken without replacement. Find \(P(\text{2nd red} \mid \text{1st red})\).

Step 1: After removing one red, the bag has 2 red and 2 blue left (4 total). Step 2: \(P(\text{2nd red} \mid \text{1st red}) = \tfrac{2}{4} = 0.5\). Answer: 0.5

Example 3 (Higher — Using Formula)

In a group, \(P(A) = 0.4\), \(P(B) = 0.5\), and \(P(A \cap B) = 0.2\). Find \(P(A \mid B)\).

\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{0.2}{0.5} = 0.4. \] Answer: 0.4

Example 4 (Higher — Reversing Condition)

In a survey, 35% of people like tea, 20% like both tea and coffee, and 50% like coffee. Find \(P(\text{Tea} \mid \text{Coffee})\).

\[ P(\text{Tea} \mid \text{Coffee}) = \frac{P(\text{Tea} \cap \text{Coffee})}{P(\text{Coffee})} = \frac{0.20}{0.50} = 0.4. \] Answer: 0.4

Example 5 (Higher — Independence Check)

Suppose \(P(A) = 0.3\), \(P(B) = 0.5\), and \(P(A \cap B) = 0.15\). Are \(A\) and \(B\) independent?

If independent, we would have \(P(A \cap B) = P(A)P(B) = 0.3 \times 0.5 = 0.15\). This matches the given \(P(A \cap B)\). Answer: Yes, \(A\) and \(B\) are independent.

Common Mistakes & Fixes

Conditional probability is a common source of errors in exams. Here are the main pitfalls and how to avoid them:

• Swapping the condition: Mistake: Calculating \(P(B \mid A)\) when the question asks for \(P(A \mid B)\). Fix: Always check the order: the event after the vertical bar “\(|\)” is the condition.
• Forgetting to restrict the sample space: Mistake: Using total outcomes instead of only those where the condition holds. Fix: Remember: conditional probability is always relative to the condition set.
• Mixing up probabilities and frequencies: Mistake: Putting counts directly into the formula without dividing. Fix: Convert frequencies to probabilities or cancel ratios consistently.
• Wrong use of complements: Mistake: Thinking \(P(A' \mid B) = 1 - P(A')\). Fix: The complement must also be conditional: \(P(A' \mid B) = 1 - P(A \mid B)\).
• Forgetting “without replacement” changes probabilities: Mistake: Using the same probability for the second event after the first draw is removed. Fix: Update both numerator and denominator when an item is removed.
• Not checking for independence properly: Mistake: Assuming independence just because the events sound unrelated. Fix: Test with the formula: if \(P(A \cap B) = P(A)P(B)\), they are independent.

Practice Questions — Foundation

These questions test basic conditional probability using simple frequencies, Venn diagrams, or tree diagrams. Work step by step and show your calculations.

1. In a group of 30 students:
   • 12 like apples (\(A\))
   • 10 like bananas (\(B\))
   • 4 like both
   Find \(P(A \mid B)\).
2. A bag contains 5 red and 3 blue counters. Two counters are drawn without replacement. Find \(P(\text{2nd red} \mid \text{1st blue})\).
3. In a class, 20 students play football, 15 play basketball, and 5 play both. If a student plays basketball, what is the probability they also play football?
4. A die is rolled. Let \(A=\) “rolling an even number”, \(B=\) “rolling a number greater than 3”. Find \(P(A \mid B)\).
5. A box has 4 pens: 3 working, 1 broken. Two are taken without replacement. Find \(P(\text{2nd working} \mid \text{1st working})\).

Practice Questions — Higher

These problems involve multi-step reasoning, careful use of the definition \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\), and product rule with trees.

1. In a cohort, \(P(M)=0.58\), \(P(P)=0.42\), and \(P(M\cap P)=0.23\). (a) Find \(P(M\mid P)\). (b) Find \(P(P\mid M)\). (c) Are \(M\) and \(P\) independent? Justify.
2. A clinic screens for a condition. \(2\%\) of people have it. The test is \(90\%\) sensitive (positive if condition present) and \(95\%\) specific (negative if condition absent). (a) Find \(P(\text{positive})\). (b) Find \(P(\text{condition}\mid \text{positive})\).
3. A bag has 6 black and 4 white counters. Two counters are drawn without replacement. (a) Find \(P(\text{second white}\mid \text{first black})\). (b) Find \(P(\text{first white}\mid \text{second white})\). (Hint: symmetry or Bayes’ idea.)
4. Cards numbered 1–10 are in a box. One card is drawn at random (then replaced), and a second card is drawn. Let \(A=\{\text{sum is even}\}\) and \(B=\{\text{first card is even}\}\). (a) Compute \(P(A)\). (b) Compute \(P(A\mid B)\). (c) Are \(A\) and \(B\) independent?
5. In a college: \(P(A)=0.6\), \(P(B)=0.5\), and \(P(A\mid B)=0.7\). (a) Find \(P(A\cap B)\). (b) Find \(P(B\mid A)\). (c) Find \(P(A'\mid B)\).

Challenge Questions

These problems combine conditional probability with trees, Venns, complements, and independence tests. Work systematically and justify each step.

1. In a school, 55% of students take Maths (\(M\)), 40% take Physics (\(P\)), and 25% take both. (a) Find \(P(M\mid P)\) and \(P(P\mid M)\). (b) Are \(M\) and \(P\) independent? Explain using a suitable test. (c) Find \(P(M'\mid P)\) and interpret it in words.
2. A factory has two machines: Machine A makes 60% of items, Machine B makes 40%. Defect rates: \(2\%\) for A, \(5\%\) for B. An item is chosen at random and found defective. (a) Find \(P(\text{A}\mid \text{defective})\). (b) Find \(P(\text{B}\mid \text{defective})\). (c) If the defective item is replaced by a good one from A, what is \(P(\text{good})\) for the replacement?
3. A bag has 4 red and 6 blue counters. Two counters are drawn without replacement. (a) Find \(P(\text{2nd red}\mid \text{1st blue})\). (b) Find \(P(\text{1st red}\mid \text{2nd red})\). (c) Hence or otherwise, find \(P(\text{both red})\) using the product rule with a conditional probability.
4. In a town, 30% of households own a dog (\(D\)), 25% own a cat (\(C\)), and 12% own both. (a) Find \(P(D\mid C)\) and \(P(C\mid D)\). (b) A household is known not to own a cat. Find \(P(D\mid C')\). (c) Are pet ownership of dogs and cats independent? Justify.
5. Cards 1–10 are in a box. One card is drawn, kept aside (no replacement), then a second card is drawn. Let \(A=\{\text{sum of the two cards is } \ge 12\}\) and \(B=\{\text{first card is at least }8\}\). (a) Compute \(P(A)\). (b) Compute \(P(A\mid B)\). (c) Comment on whether \(A\) and \(B\) appear independent.
6. A medical test has sensitivity \(0.92\) and specificity \(0.95\). The prevalence of the condition is \(0.04\). (a) Find \(P(\text{positive})\). (b) Find \(P(\text{condition}\mid \text{positive})\). (c) The test is applied twice independently and a person is declared positive if at least one test is positive. Find the new overall sensitivity and specificity, then recompute \(P(\text{condition}\mid \text{positive})\).
7. In a game, a biased coin has \(P(H)=0.6\). It is tossed twice. Let \(A=\{\text{at least one head}\}\) and \(B=\{\text{first toss is head}\}\). (a) Find \(P(A)\). (b) Find \(P(A\mid B)\). (c) Are \(A\) and \(B\) independent? Justify clearly.

Quick Revision Sheet

Use this as a last-minute checklist for conditional probability.

Key Formulas

• Definition: \(P(A \mid B)=\dfrac{P(A \cap B)}{P(B)}\) (provided \(P(B)>0\)).
• Product rule: \(P(A \cap B)=P(B)\,P(A \mid B)=P(A)\,P(B \mid A)\).
• Independence test: \(A\) and \(B\) independent ⇔ \(P(A \mid B)=P(A)\) ⇔ \(P(A \cap B)=P(A)P(B)\).
• Complements (conditional): \(P(A' \mid B)=1-P(A \mid B)\).

Spotting Conditional Probability

• Keywords: “given that…”, “among those who…”, “if we know…”.
• Second-level branches in a tree are conditional (e.g. \(P(\text{2nd red} \mid \text{1st red})\)).
• On a Venn: restrict to circle \(B\), then take the fraction that is also in \(A\).

Method (Fast)

1. Identify condition \(B\) and target \(A\).
2. Write \(P(A \mid B)=\dfrac{P(A\cap B)}{P(B)}\).
3. Get \(P(A\cap B)\) and \(P(B)\) from counts/diagram/tree; use product rule if needed.
4. Compute; check result is between 0 and 1.

Common Traps

• Swapping condition: \(P(A \mid B)\neq P(B \mid A)\).
• Forgetting to restrict to \(B\) (using the whole sample space by mistake).
• Using unconditional complements: use \(1-P(A \mid B)\), not \(1-P(A)\).
• Ignoring “without replacement” updates.

Answers

Foundation

1. \(P(A\cap B)=\tfrac{4}{30},\; P(B)=\tfrac{10}{30}\). \(P(A\mid B)=\dfrac{4/30}{10/30}=\dfrac{4}{10}=0.4\).
2. After 1st blue: bag has 5 red, 2 blue (7 total). \(P(\text{2nd red}\mid \text{1st blue})=5/7\).
3. Overlap = 5. \(P(F\mid B)=5/15=1/3\).
4. \(A=\{2,4,6\},\; B=\{4,5,6\},\; A\cap B=\{4,6\}\). \(P(B)=3/6=0.5,\; P(A\cap B)=2/6=1/3\). \(P(A\mid B)=(1/3)/(1/2)=2/3\).
5. After 1st working: bag has 2 working, 1 broken (3 total). \(P(\text{2nd working}\mid \text{1st working})=2/3\).

Higher

1. (a) \(P(M\mid P)=0.23/0.42\approx0.548\). (b) \(P(P\mid M)=0.23/0.58\approx0.397\). (c) Check independence: \(P(M)P(P)=0.2436\neq0.23\). Not independent.
2. (a) \(P(+)=0.02(0.9)+0.98(0.05)=0.018+0.049=0.067\). (b) \(P(\text{cond}\mid+)=0.018/0.067\approx0.269\).
3. (a) After 1st black: 6B,4W → 5B,4W (9 total). \(P(2W\mid1B)=4/9\). (b) By symmetry, \(P(1W\mid2W)=6/10=0.6\).
4. (a) \(P(A)=0.5\). (Because half the pairs sum even). (b) Given 1st even, 5 choices; check distribution → \(P(A\mid B)=0.5\). (c) Since \(P(A\mid B)=P(A)\), independent.
5. (a) \(P(A\cap B)=P(B)P(A\mid B)=0.5\times0.7=0.35\). (b) \(P(B\mid A)=0.35/0.6\approx0.583\). (c) \(P(A'\mid B)=1-0.7=0.3\).

Challenge

1. (a) \(P(M\mid P)=0.25/0.40=0.625,\; P(P\mid M)=0.25/0.55\approx0.455\). (b) \(P(M)P(P)=0.22\neq0.25\), so not independent. (c) \(P(M'\mid P)=1-0.625=0.375\).
2. (a) Defective prob = \(0.6(0.02)+0.4(0.05)=0.012+0.02=0.032\). \(P(A\mid def)=0.012/0.032=0.375\). (b) \(P(B\mid def)=0.625\). (c) Replacement good = 0.98.
3. (a) After 1B: 4R,5B left. \(P(2R\mid1B)=4/9\). (b) \(P(1R\mid2R)=4/10=0.4\). (c) \(P(\text{both R})=P(1R)P(2R\mid1R)=(4/10)(3/9)=12/90=0.133\).
4. (a) \(P(D\mid C)=0.12/0.25=0.48\). (b) \(P(C\mid D)=0.12/0.30=0.40\). (c) \(P(D\mid C')=(0.30-0.12)/(1-0.25)=0.18/0.75=0.24\). Not independent (\(0.3\times0.25=0.075\neq0.12\)).
5. (a) Pairs sum≥12 out of 45 possible distinct draws = compute ≈0.244. (b) Given 1st≥8, restrict, compute conditional ≈0.5. (c) Since changed, not independent.
6. (a) \(P(+)=0.04(0.92)+0.96(0.05)=0.0368+0.048=0.0848\). (b) Posterior = 0.0368/0.0848≈0.434. (c) Two-test system: sensitivity=1-(0.08^2)=0.9936, specificity=(0.95^2)=0.9025. \(P(+)=0.04(0.9936)+0.96(0.0975)=0.03974+0.0936=0.1333\). Posterior=0.03974/0.1333≈0.298.
7. (a) \(P(A)=1-P(\text{no heads})=1-0.4^2=0.84\). (b) \(P(A\mid B)=1-P(\text{2nd T}\mid 1st H)=1-0.4=0.6\). (c) Not independent: 0.84≠0.6.

Conclusion & Next Steps

Conditional probability unlocks multi-step and “given that” problems by focusing on a restricted sample space. Using the definition \(P(A \mid B)=\dfrac{P(A\cap B)}{P(B)}\) and the product rule \(P(A\cap B)=P(B)\,P(A\mid B)\), you can read values directly from Venn diagrams or multiply along branches in tree diagrams. Always check whether events might be independent by testing \(P(A\cap B)=P(A)P(B)\).

• What you’ve mastered: Translating “given that” into notation, computing conditional probabilities from Venns and trees, using complements, and testing independence.
• Exam habits: Identify the condition first, write the formula before numbers, and sanity-check the result lies between 0 and 1.

Next topics to study:

1. Bayes’ Theorem (Intro): Reversing conditions: \(P(A\mid B)\) from \(P(B\mid A)\) with prior probabilities.
2. Law of Total Probability: Splitting complicated events across partitions to find \(P(B)\).
3. Compound Events: Longer trees (three draws/tosses) and “at least one” via complements within conditions.