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Complete IGCSE Edexcel Higher-tier guide to conditional probability, without replacement, two-way tables, and Venn diagrams. Worked examples for grades 7-9.
Conditional probability is the probability of an event occurring given that another event has already happened. It changes the sample space — the denominator is no longer the total of all outcomes but the total within the given condition. In IGCSE, conditional probability appears in 'without replacement' tree diagrams, two-way tables, and Venn diagram problems. It is Higher-tier only and typically worth 4-6 marks.
Source: Edexcel IGCSE Mathematics (9-1) Specification 4MA1
Two events are independent if the outcome of one does not affect the probability of the other. For independent events A and B, the probability of both occurring is:
P(A and B) = P(A) x P(B)
Common examples of independent events include tossing a coin multiple times, rolling a dice multiple times, or spinning a spinner and picking a card — as long as one event does not change the conditions for the other.
Two events are dependent if the outcome of the first event changes the probability of the second. The classic IGCSE example is picking items from a bag without replacement: the first pick changes both the total number of items and the composition for the second pick.
For dependent events, the multiplication rule still applies, but the second probability is adjusted to reflect the new conditions: P(A and B) = P(A) x P(B given A).
In "without replacement" problems, once an item is selected it is not put back. This means the total decreases by 1 for the second selection, and the number of the type selected also decreases by 1.
Example setup:
Bag contains 4 red and 6 blue balls. Two are picked without replacement.
| First Pick | Second Pick (if first was red) | Second Pick (if first was blue) |
|---|---|---|
| P(R) = 4/10 | P(R) = 3/9, P(B) = 6/9 | P(R) = 4/9, P(B) = 5/9 |
| P(B) = 6/10 | — | — |
Notice that the denominators change from 10 to 9, and the numerators adjust based on what was picked first. This is the hallmark of dependent events.
A two-way table displays data classified by two categories. To find a conditional probability such as P(A given B), look only at the row or column for B and find the proportion that also satisfies A.
The key insight is that the denominator changes: instead of dividing by the grand total, you divide by the total for the given condition. This is what makes it "conditional" — the sample space is restricted.
For mutually exclusive events (events that cannot happen at the same time), the probability of either occurring is the sum of their individual probabilities:
If mutually exclusive: P(A or B) = P(A) + P(B)
If not mutually exclusive: P(A or B) = P(A) + P(B) - P(A and B)
The second formula accounts for double-counting the overlap. In IGCSE, this is often tested using Venn diagrams where the overlapping region must be subtracted.
A bag contains 5 red and 3 blue marbles. Two are picked without replacement. Find the probability that both are red.
Step 1: P(first red) = 5/8.
Step 2: If first is red: 4 red and 3 blue remain (7 total). P(second red) = 4/7.
Step 3: P(both red) = 5/8 x 4/7 = 20/56 = 5/14.
Answer: P(both red) = 5/14
A two-way table shows: 30 students play football, of whom 18 are boys. 20 students do not play football, of whom 8 are boys. Find the probability that a randomly chosen student plays football given that they are a boy.
Step 1: Total boys = 18 + 8 = 26. Boys who play football = 18.
Step 2: P(football given boy) = 18/26 = 9/13.
Note: The denominator is 26 (total boys), not 50 (total students). This is what makes it conditional.
Answer: P(football | boy) = 9/13
A bag has 6 red and 4 blue counters. Three are taken without replacement. Find the probability of getting exactly two red counters.
Step 1: Three paths give exactly 2 red: RRB, RBR, BRR.
Step 2: P(RRB) = 6/10 x 5/9 x 4/8 = 120/720.
Step 3: P(RBR) = 6/10 x 4/9 x 5/8 = 120/720.
Step 4: P(BRR) = 4/10 x 6/9 x 5/8 = 120/720.
Step 5: P(exactly 2 red) = 120/720 + 120/720 + 120/720 = 360/720 = 1/2.
Answer: P(exactly 2 red) = 1/2
Not adjusting for "without replacement": Keeping the same denominator for the second event when items are not replaced. The total decreases by 1 each time an item is removed.
Using the wrong denominator in conditional probability: For P(A given B), the denominator is the total for B, not the grand total. This is the most common conceptual error.
Missing branches in "exactly" questions: For "exactly 2 red from 3 picks", there are three possible orderings (RRB, RBR, BRR). Missing any one of them gives an incorrect total.
Confusing "and" with "or": P(A and B) means both events happen (multiply). P(A or B) means at least one happens (add, subtracting overlap if not mutually exclusive).
Always draw the tree diagram: Even if the question does not ask for one, drawing a tree diagram organises your thinking and prevents errors. Write adjusted probabilities on each branch.
Check "with" or "without" replacement: Read the question carefully. "With replacement" means independent events (probabilities stay the same). "Without replacement" means dependent events (probabilities change).
Simplify fractions at the end: Do all multiplication with unsimplified fractions first, then simplify the final answer. Simplifying too early often introduces arithmetic errors.
List all orderings for "exactly" questions: For "exactly k successes from n trials", systematically list all orderings. The number of orderings equals "n choose k" — for example, exactly 2 from 3 gives 3 orderings.
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