When an item is drawn and put back, probabilities stay the same for each draw — this is with replacement. When the item is kept out, the total decreases and each fraction changes — this is without replacement, producing dependent events that need updated branch fractions on a tree diagram.
What is the difference between with and without replacement?
With replacement: after each selection, the item is returned and the collection resets. Every draw is independent — knowing the outcome of the first draw tells you nothing new about the second.
Without replacement: the item drawn is kept out. The total count decreases by 1 for every subsequent draw. The draws are dependent events — the first outcome changes the probabilities for the second.
This single distinction changes the fractions on a tree diagram and can significantly alter the final probability.
How do you draw a tree diagram for without replacement?
Step 1 — State the total number of items and how many of each type exist.
Step 2 — Draw the first set of branches with fractions based on the original total.
Step 3 — For each branch, update the fractions for the second draw: total decreases by 1, and one specific type also decreases by 1 (the type drawn first).
Step 4 — Multiply along branches for combined probabilities (AND rule).
Step 5 — Add separate paths for the same outcome (OR rule).
Worked example 1: bag of counters, without replacement
A bag contains 4 red and 3 blue counters. Two counters are drawn without replacement. Find the probability that both are red.
First draw: P(Red) = 4/7, P(Blue) = 3/7.
Second draw (given first was red): 3 red and 3 blue remain (total 6).
P(Red | first Red) = 3/6 = 1/2.
P(both Red) = P(Red) × P(Red | first Red) = 4/7 × 1/2 = 4/14 = 2/7
Worked example 2: same bag, find P(one of each colour)
"One of each colour" means Red then Blue OR Blue then Red.
Path 1 — Red then Blue:
P(Red) × P(Blue | first Red) = 4/7 × 3/6 = 12/42 = 2/7.
Path 2 — Blue then Red:
P(Blue) × P(Red | first Blue) = 3/7 × 4/6 = 12/42 = 2/7.
P(one of each) = 2/7 + 2/7 = 4/7
Worked example 3: same scenario, but WITH replacement
Now the counter is replaced after each draw.
P(both Red) = 4/7 × 4/7 = 16/49
P(one of each) = (4/7 × 3/7) + (3/7 × 4/7) = 12/49 + 12/49 = 24/49
Comparing the two: without replacement gives 2/7 (≈ 0.286) vs with replacement 16/49 (≈ 0.327) for "both red". The difference is real and significant — never assume replacement is implied.
How do you update fractions when items are different types?
After a first draw of Red (say from 4 Red + 3 Blue = 7 total):
| Second draw follows: | Red remaining | Blue remaining | Total remaining |
|---|---|---|---|
| First was Red | 3 | 3 | 6 |
| First was Blue | 4 | 2 | 6 |
Only the type actually drawn decreases. The total always decreases by 1 per draw (for without replacement).
Frequently asked questions
How do I know from the question whether it is with or without replacement?
Look for these phrases:
- "without replacement", "not replaced", "kept" → without replacement (dependent).
- "with replacement", "replaced", "put back" → with replacement (independent).
- If neither is stated, consider the context: drawing cards from a deck without shuffling back is normally without replacement; rolling dice is effectively with replacement.
Can I use the formula P(A and B) = P(A) × P(B) for both types?
The formula P(A and B) = P(A) × P(B) applies to independent events — that is, with replacement. For without replacement (dependent events), use the conditional version: P(A and B) = P(A) × P(B|A). The tree diagram automatically handles this if you write the correct fractions on the second set of branches.
What if three items are drawn without replacement?
Extend the tree diagram to three levels. After each draw, decrease the total by 1 and decrease the count of whichever type was drawn. Each path probability is the product of three branch fractions. At GCSE, questions rarely go beyond two draws, but the method scales directly.
Is this the same as conditional probability?
Yes — "without replacement" problems are a specific and very common application of conditional probability. P(second is red | first was red) is a conditional probability, and the tree diagram is the standard tool for calculating it. The Venn diagram approach can also be used for two-event without-replacement problems if the question provides totals rather than asking you to track draws.
For step-by-step GCSE probability practice with personalised hints — visit aitutors.me.