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Learn4024Paper 2 · CalculatorProbabilityWithout replacement and Venn diagrams
Probability · Subtopic 3 of 3

Without Replacement and Venn Diagrams

When an item is not put back, the second pick has one fewer item, and possibly one fewer of the colour you took. A Venn diagram sorts a group into "only AA", "both", "only BB" and "neither": fill in the overlap first.

Cambridge 4024 Paper 1Cambridge 4024 Paper 2Edexcel 4MA1 Papers 1H, 2HEdexcel 4MA1 Papers 1F, 2FAQA 8300 Papers 1–3≈ 30 minRevision 5 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Change the second-stage probabilities when items are not replaced.
  • Find "both the same" and "one of each" probabilities without replacement.
  • Complete a Venn diagram from given totals.
  • Read probabilities from a Venn diagram.
Pause and think

A bag has 22 red and 11 blue sweet. You eat one red sweet. What is the probability the next one is red?

Section 2 of 14

Core idea

Start hereSweets from a bag

Take a sweet and eat it, then take another. The bag now has one fewer sweet, so the second probabilities have changed. Events like this are dependent.

The big idea

Without replacement: second-stage totals drop by one. Conditional probability P(A∣B)P(A\mid B): count only inside BB.

1. Without replacement

7 red and 3 blue counters: two taken without replacement
RBRBRB7/103/106/93/97/92/9RR: 42/90RB: 21/90BR: 21/90BB: 6/90
After a red is taken, 6 reds are left out of 9 counters. P(same colour)=42+690=815P(\text{same colour})=\tfrac{42+6}{90}=\tfrac{8}{15}.
Worked example7 red, 3 blue, take two

P(same colour), without and with replacement.

Goal
Compare.
Strategy
Tree; adjust totals.
  1. 710⋅69=4290\tfrac7{10}\cdot\tfrac69=\tfrac{42}{90}
    Both red: one red gone.
  2. 310⋅29=690\tfrac3{10}\cdot\tfrac29=\tfrac6{90}
    Both blue.
  3. 4890=815\tfrac{48}{90}=\tfrac8{15}
    Add.
  4. 0.49+0.09=0.580.49+0.09=0.58
    With replacement.
Answer815≈0.533\frac8{15}\approx0.533; 0.580.58.
✓ CheckWithout replacement, matching is harder ✓.

2. Venn diagrams

40 people: 25 have a cat, 18 a dog, 10 both
151087CatDog
P(dog∣cat)P(\text{dog}\mid\text{cat}): only look inside the cat circle (25 people), of whom 10 have a dog: 1025=25\tfrac{10}{25}=\tfrac25.
P(A∣B)=n(A∩B)n(B)P(A\mid B)=\frac{n(A\cap B)}{n(B)}
where
∣\mid
"given".
Worked examplePets

40 people; 25 cat, 18 dog, 10 both.

Goal
P(dog only); P(dog | cat).
Strategy
Fill the Venn; restrict.
  1. 15, 10, 8, 715,\ 10,\ 8,\ 7
    Cat only, both, dog only, neither.
  2. P(dog only)=840=15P(\text{dog only})=\tfrac8{40}=\tfrac15
    Out of everyone.
  3. P(D∣C)=1025=25P(D\mid C)=\tfrac{10}{25}=\tfrac25
    Only among cat owners.
Answer15\frac15; 25\frac25.
✓ Check15+10+8+7=4015+10+8+7=40 ✓.
Common misconceptionWrong denominator For "given cat", divide by 25, not 40.
Pause and think

P(cat | dog)?

Summary

  • Reduce totals without replacement.
  • Conditional: restrict to the group.
Key words
Conditional probability
Probability given extra information.
HistoryJohn Venn. John Venn, an English logician, introduced his overlapping circles in 1880 to picture logical statements. A stained-glass window at Gonville and Caius College, Cambridge, where he studied and taught, shows a Venn diagram in his honour.
Fun factThe birthday surprise. In a group of just 2323 people, the chance that two share a birthday is over 50%50\%. Working it out uses a long chain of "without replacement" probabilities: 365365×364365×363365×⋯\tfrac{365}{365}\times\tfrac{364}{365}\times\tfrac{363}{365}\times\cdots for no match.
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