Follow probability trees without replacement
Update every second-stage branch after removal, multiply along a path and add distinct paths for an event.
Exam scope and assessment notes
Probability & Statistics 1 (Paper 5, AS & A Level) · Mathematics 9709, 2026–2027, section 5.3. Define the sample space and state its probability assumptions before counting. Fair independent dice give equiprobable ordered pairs; their sums are not equiprobable. Random selection means each specified individual is equally likely under the stated sampling model. Without replacement, later branch probabilities change; with replacement and independent mixing, they remain the same. A′ is a complement, A ∩ B means both, A ∪ B means either or both, and P(A|B) means A given B with P(B) > 0. Mutually exclusive and independent are different properties. The explicit general addition formula is not required by this section; explain overlap through disjoint regions. Conditional probabilities may use P(A|B) = P(A ∩ B)/P(B). Numerical models describe the stipulated experiment, not automatically a real population. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Calculate fractions and conditional probabilities.
- Use combinations for random subsets.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
After removing one red from four red and three blue, what remains?
What you will learn
- Construct a without-replacement tree with changing conditional probabilities.
- Combine ordered paths and check against unordered selection counts.
Every branch refers to its history
A bag contains four red and three blue balls, all individually equally likely to be drawn. Draw two without replacement. First branches are R with 4/7 and B with 3/7. After R, the bag has three red and three blue, so the second branches are 3/6 and 3/6. After B, they are 4/6 and 2/6.
At each node the outgoing probabilities sum to one. Multiply along a path: P(RR) = (4/7)(3/6) = 2/7, P(RB) = 2/7, P(BR) = 2/7 and P(BB) = 1/7. These joint path probabilities add to one. The second-stage branch after R is P(second R | first R), not the unconditional initial red probability.
An event may contain several paths
Different colours means RB or BR, two disjoint paths. Add their joint probabilities to obtain 4/7. Do not use only one order unless the event specifically requests that order. The combination check gives 4 × 3/7C2 = 12/21 = 4/7.
For three draws from five red and three blue without replacement, exactly two red means RRB, RBR or BRR. Each path has probability 5/28, so the total is 15/28. The order-independent subset count gives 5C2 × 3C1/8C3 = 15/28. Use the changing numerator and denominator at each draw, including the third stage.
Pause and explain
Four red, three blue; two draws without replacement. What is P(RR)?
Worked example
Draw twice without replacement from four red and three blue. Find P(different colours).
Show the worked solution
- P(RB) = (4/7)(3/6) = 2/7.
- P(BR) = (3/7)(4/6) = 2/7.
- Add the two mutually exclusive paths to get 4/7.
Answer 4/7.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Draw twice without replacement from four red and three blue. Find P(both red).
Give me a hint
The second red branch follows a first-red removal.
Compare my reasoning
- First red has probability 4/7.
- After red, three of six remaining balls are red, giving 3/6.
- Multiply to obtain (4/7)(3/6) = 2/7.
2/7.
Look for these in your work
- Use the changing bag contents.
- Multiply successive conditional branch probabilities.
For that bag and two draws without replacement, find P(different colours).
Give me a hint
The event has two possible colour orders.
Compare my reasoning
- P(RB) = (4/7)(3/6) = 2/7.
- P(BR) = (3/7)(4/6) = 2/7.
- The paths are disjoint, so their sum is 4/7.
4/7.
Look for these in your work
- Include both colour orders.
- Distinguish path multiplication from event addition.
Draw three times without replacement from five red and three blue. Find P(exactly two red) using ordered paths, then check with combinations.
Give me a hint
The favourable paths are RRB, RBR and BRR.
Compare my reasoning
- Each path has numerator product 5 × 4 × 3 and denominator 8 × 7 × 6, giving 5/28.
- The three disjoint paths total 15/28.
- The subset check gives 5C2 × 3C1/8C3 = 30/56 = 15/28.
15/28 by both methods.
Look for these in your work
- Update all three stages without replacement.
- List every order belonging to the exact composition.
Common mistakes
- Reduce the total after every removal.
- Change the colour count only when that colour was removed.
- Add all permitted orders after multiplying each path.
What can you explain now?
Is the second red branch after a first red still 4/7?
Compare with the explanation
No; it is 3/6 because one red was removed from the seven-ball bag.
Without replacement changes both the relevant remaining count and total.
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Make it stick
Write the remaining bag at each node, multiply down paths, and add the distinct paths belonging to the event.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.