Welcome to the Chapter on Conditional Probability!
Hello! This chapter is where probability gets really interesting, dealing with events that influence each other. Don't worry if the name sounds complicated—the core idea is very intuitive.
Conditional probability asks: “What is the likelihood of something happening, given that we already know something else has occurred?”
This topic is particularly important when you deal with situations where objects are drawn without replacement, like picking cards or taking marbles out of a bag.
1. Prerequisite Review: Independent vs. Dependent Events
1.1 Independent Events
Independent Events are two or more events where the outcome of one event does not affect the probability of the other event(s).
Example: Flipping a coin twice. Getting a Head on the first flip does not change the probability of getting a Head on the second flip (it’s still 0.5 or \(\frac{1}{2}\)).
To find the probability of two independent events A and B both happening, you multiply their individual probabilities:
Formula (Independent Events):
$$P(A \text{ and } B) = P(A) \times P(B)$$
1.2 Dependent Events (The Need for Conditional Probability)
Dependent Events (or Combined Events without Replacement) are two or more events where the outcome of the first event changes the probability of the subsequent event(s).
This change in probability is exactly what Conditional Probability addresses!
Analogy: Imagine a box of 10 chocolates. If you take out your favourite chocolate (dark chocolate) first, the chance of your friend also getting a dark chocolate is now lower, because there are fewer total chocolates and fewer dark chocolates left. The second choice is conditional on the first.
Key Takeaway
If items are drawn without replacement, the events are dependent, and you must use conditional probability. If items are drawn with replacement, the events are independent.
2. Understanding Conditional Probability
2.1 Defining the Condition
When we talk about conditional probability, we are calculating a probability based on a reduced or modified sample space.
We want to find the probability of Event A happening, given that Event B has already occurred.
Notation:
We write this as: \(P(A | B)\).
Read aloud as: “The Probability of A, given B.”
2.2 The "Universe Shrinks" Analogy
Think of the entire group of possible outcomes as your "universe". When you are given a condition (Event B), your new "universe" shrinks down to only the outcomes in B.
- The original sample space (U) is ignored.
- The new sample space is just Event B.
- We are only interested in the part of Event A that overlaps with Event B.
Don't worry about memorising a complex formula here; in IGCSE, you mostly calculate conditional probability naturally by adjusting the numbers in the numerator and denominator, especially when using tree diagrams or tables.
3. Calculating Conditional Probability using Tree Diagrams
Tree diagrams are the most common tool used to solve problems involving dependent events (without replacement). The conditional probabilities are written directly on the second set of branches.
Step-by-Step Example: Marbles (Without Replacement)
A bag contains 5 Red (R) marbles and 3 Blue (B) marbles. Two marbles are drawn without replacement.
Step 1: Probabilities for the First Draw
Total marbles = 8.
- \(P(\text{1st is R}) = \frac{5}{8}\)
- \(P(\text{1st is B}) = \frac{3}{8}\)
Step 2: Probabilities for the Second Draw (The Conditional Step)
Now, the total has dropped to 7. The exact probabilities for the second draw depend entirely on what happened first.
Scenario 1: Given the 1st marble was Red (R)
The bag now has 4 Red and 3 Blue marbles (Total 7).
- \(P(\text{2nd is R } | \text{ 1st is R}) = \frac{4}{7}\) (One less Red)
- \(P(\text{2nd is B } | \text{ 1st is R}) = \frac{3}{7}\) (Same number of Blue)
Scenario 2: Given the 1st marble was Blue (B)
The bag now has 5 Red and 2 Blue marbles (Total 7).
- \(P(\text{2nd is R } | \text{ 1st is B}) = \frac{5}{7}\) (Same number of Red)
- \(P(\text{2nd is B } | \text{ 1st is B}) = \frac{2}{7}\) (One less Blue)
The probabilities written on the second set of branches are conditional probabilities.
2.3 Calculating Combined Probability (Pathways)
To find the probability of a whole path (e.g., Red followed by Blue), you multiply the probabilities along the branches:
$$P(R \text{ then } B) = P(1\text{st R}) \times P(2\text{nd B} | 1\text{st R})$$
$$P(R \text{ then } B) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}$$
⚠ Common Mistake Alert!
Always check the denominator! If it’s “without replacement,” the total number of items available for the second draw must decrease by one. If you forget to reduce the denominator (and the numerator for the item drawn), your probabilities will be wrong!
Did you know?
If you add the probabilities on any set of branches radiating from a single point, they must always add up to 1! This is a great way to check your work.
4. Conditional Probability using Tables and Venn Diagrams
Conditional probability is not just for "without replacement" scenarios; it is often used when analyzing survey data presented in tables or Venn diagrams. The principle remains the same: The condition restricts the total outcomes.
4.1 Using Two-Way Tables (Frequency Tables)
Suppose 50 students were surveyed about whether they own a Dog (D) or a Cat (C).
| Dog (D) | No Dog (D') | Total | |
| Cat (C) | 15 | 10 | 25 |
| No Cat (C') | 5 | 20 | 25 |
| Total | 20 | 30 | 50 |
Question: Find the probability that a student owns a Cat, given that they own a Dog.
We are looking for \(P(C | D)\).
Method:
-
Identify the condition (new denominator): The student owns a Dog (D).
Look at the ‘Dog (D)’ column total: 20 students. This is your new total sample space. -
Identify the favorable outcome (numerator): Out of those 20 Dog owners, how many own a Cat?
Look at the cell (C and D): 15 students. -
Calculate:
$$P(C | D) = \frac{\text{Number of students who own C AND D}}{\text{Total number of students who own D}}$$ $$P(C | D) = \frac{15}{20} = \frac{3}{4}$$
4.2 Using Venn Diagrams
If the same data above were displayed in a Venn diagram, the key is still to recognize that the conditional event limits your denominator.
If you are asked for \(P(A|B)\), you only look at the numbers inside the circle B.
The calculation is:
$$P(A | B) = \frac{n(A \cap B)}{n(B)}$$
Where:
- \(n(A \cap B)\) is the number of elements in the intersection (A AND B).
- \(n(B)\) is the number of elements in the restricting event (B).
Don't worry if this seems tricky at first. Practice focusing on isolating the group defined by the "given that" phrase—that group becomes your new total!
Quick Review: Key Takeaways for Conditional Probability
- Conditional probability deals with situations where knowing one event has happened affects the probability of another.
- The critical phrase is “given that” or “without replacement.”
- The key calculation technique is restricting the sample space (the denominator changes).
- In Tree Diagrams for "without replacement" problems, the probabilities on the second set of branches are the conditional probabilities (the total count and the count of the item drawn both decrease by 1).
- In Tables/Venn Diagrams, the denominator becomes the total count of the event that is given to have occurred.