Why does knowing one event change the odds of another?
Ever wondered why the chance of drawing a red card changes after you’ve already taken a black one? That’s conditional probability in action – and it shows up a lot in exam questions.
💡 In Simple Words: Conditional probability tells you the chance of something happening **after** you already know another thing has happened. It’s like checking the odds of rain **given** that the sky is already dark.
What is Conditional Probability?
In plain language, conditional probability is the probability of an event A **when** another event B is known to have occurred. We write it as P(A|B) – read as “the probability of A given B”.
Key Formula
The core rule is:
P(A|B) = \frac{P(A\cap B)}{P(B)}
Here P(A∩B) means the probability that **both** A and B happen together (the intersection). P(B) is the probability that B happens at all. The fraction simply tells you how much of B’s “world” is also A.
When to Use It
- If the problem says “given that…”, “provided that…”, or “on the condition that…”, you’re looking at conditional probability.
- When two events are linked – like drawing two cards without replacement – the second draw depends on the first. That dependency is handled by the formula.
- Even if events seem unrelated, checking whether they are independent (the outcome of one doesn’t affect the other) is part of the process.
Worked Example 1: Drawing Cards
Suppose you have a standard deck of 52 cards. You pick one card, don’t replace it, and then pick a second card. What is the probability that the second card is a king **given** that the first card was a king?
Step 1: Identify A = “second card is a king”, B = “first card is a king”.
Step 2: Find P(A∩B). There are 4 kings in the deck. For both draws to be kings, you need to pick one of the 4 kings first and then one of the remaining 3 kings. So the number of favorable ordered pairs is 4 × 3 = 12. The total ordered pairs without replacement are 52 × 51 = 2652. Hence P(A∩B) = 12/2652 = 1/221.
Step 3: Find P(B). The chance the first card is a king is 4/52 = 1/13.
Step 4: Apply the formula: P(A|B) = (1/221) ÷ (1/13) = 13/221 ≈ 0.0589. So after you’ve already drawn a king, the chance the next card is also a king drops to about 5.9%.
Worked Example 2: Medical Test
A disease affects 1 in 1,000 people. A test for the disease is 99% accurate – it gives a positive result 99% of the time when the disease is present, and a negative result 99% of the time when the disease is absent. If a person tests positive, what is the probability they actually have the disease?
Let D = “person has disease”, T = “test is positive”. We need P(D|T).
First, compute P(D∩T). That is the chance a random person both has the disease and tests positive: 1/1000 × 0.99 = 0.00099.
Next, compute P(T). Two ways to get a positive result:
- True positive: disease + correct test = 0.00099 (as above).
- False positive: no disease (999/1000) × test mistakenly positive (0.01) = 0.00999.
Now apply the formula:
P(D|T) = 0.00099 ÷ 0.01098 ≈ 0.0902, i.e. about 9%.
Even a very accurate test can give a surprising result when the disease is rare – a classic conditional probability surprise!
Quick Comparison Table
| Scenario | Formula | Interpretation |
|---|---|---|
| Independent events | P(A|B)=P(A) | Knowing B happens doesn’t change the chance of A. |
| Dependent events | P(A|B)=P(A∩B)/P(B) | The chance of A depends on B occurring first. |
Tips to Remember for the ISC Exam
- Always write down what A and B are before plugging numbers.
- If the problem involves “without replacement”, treat the second draw as a new sample space – that’s why we divide by P(B).
- Check if events are independent. If they are, you can skip the fraction and just use P(A).
- When numbers look tiny, convert fractions to decimals to avoid arithmetic slip‑ups.
- Practice the “tree diagram” method – it visualises conditional steps nicely.
📝 Likely Exam Questions
- Question: A bag contains 3 red, 2 blue, and 5 green balls. Two balls are drawn one after another without replacement. Find the probability that the second ball is green given that the first ball was red.
Answer: P(second green | first red) = 5/9 ≈ 0.556. - Question: In a class of 40 students, 25 study maths, 18 study physics, and 10 study both. What is the probability that a randomly chosen student studies physics given that they study maths?
Answer: P(Physics|Maths) = 10/25 = 0.4. - Question: A factory produces 2% defective items. An inspector finds a defective item. The test used correctly identifies a defective item 95% of the time and incorrectly flags a good item 3% of the time. Find the probability the item is actually defective.
Answer: P(defective|positive) ≈ 0.39 (about 39%). - Question: Two fair dice are rolled. What is the probability that the sum is 8 given that the first die shows a 3?
Answer: Only (3,5) works, so P=1/6 ≈ 0.167. - Question: Explain why P(A|B) = P(A) when A and B are independent, using a short example.
Answer: If tossing a coin (A) and rolling a die (B) are independent, knowing the die result doesn’t change the ½ chance of heads, so P(heads|die=4)=½=P(heads).