Model replacement and independent repeated trials
Explain when branch probabilities stay unchanged and distinguish independence from identical success rates.
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
- Multiply path probabilities and add disjoint paths.
- Test and interpret independence.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
With replacement and independent mixing, does drawing red change the next red probability?
What you will learn
- Use a replacement model with independent mixing.
- Combine independent trials that have different success probabilities.
Replacement restores the available individuals
A bag with four red and three blue is sampled, the selected ball replaced, and the bag independently mixed before the second draw. Each red branch remains 4/7 and each blue branch 3/7. P(RR) = 16/49, P(BB) = 9/49 and P(different colours) = 2(4/7)(3/7) = 24/49.
Same-colour probability is 25/49. Without replacement from the same starting bag, different-colour probability is 4/7, not 24/49. Replacement plus the stipulated independent mixing supports independence. Simply seeing the same marginal probability twice does not prove independence if the drawing mechanism preserves a dependence.
Independent does not require identical probabilities
Three independent trial successes with probabilities 1/2, 2/3 and 3/4 have all-success probability (1/2)(2/3)(3/4) = 1/4. Independence allows multiplication of marginals even though the probabilities differ. An identical success-rate assumption is a separate condition used in some later distribution models.
Exactly two successes requires SSF, SFS or FSS. Their probabilities are (1/2)(2/3)(1/4) = 1/12, (1/2)(1/3)(3/4) = 1/8 and (1/2)(2/3)(3/4) = 1/4. The sum is 11/24. The three paths need not have equal probabilities, so multiplying one path by three would be wrong here.
Pause and explain
Two replaced independent draws from four red and three blue: what is P(RR)?
Worked example
Draw twice from four red and three blue with replacement and independent mixing. Find P(different colours).
Show the worked solution
- Each red branch is 4/7 and each blue branch 3/7.
- RB and BR each have probability 12/49.
- Add them to get 24/49.
Answer 24/49.
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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 from four red and three blue with replacement and independent mixing. Find P(both red).
Give me a hint
Both stages use the original bag proportions.
Compare my reasoning
- First red has probability 4/7.
- Replacement and independent mixing leave second red at 4/7.
- Multiply to obtain 16/49.
16/49.
Look for these in your work
- State the replacement and independence assumptions.
- Use multiplication for both stages succeeding.
Under the same replaced-draw model, find P(different colours) and P(same colour).
Give me a hint
Add the two orders for different colours and take its complement.
Compare my reasoning
- RB and BR each have probability 12/49.
- Different colours has probability 24/49.
- Same colour is 1 − 24/49 = 25/49, also RR + BB.
Different 24/49; same 25/49.
Look for these in your work
- Include both mixed orders.
- Check that complementary probabilities sum to one.
Three independent trials have success probabilities 1/2, 2/3 and 3/4, in that order. Find P(exactly two successes) and explain why three times one path is invalid.
Give me a hint
Write SSF, SFS and FSS using the success or failure probability for each actual trial.
Compare my reasoning
- SSF has probability 1/12; SFS has probability 1/8.
- FSS has probability 1/4, giving total 1/12 + 1/8 + 1/4 = 11/24.
- The path probabilities differ because the trial success rates differ, despite independence.
11/24; the three favourable paths have unequal probabilities.
Look for these in your work
- Use each trial’s own probability.
- Add distinct favourable paths rather than assuming they are equal.
Common mistakes
- Replacement must be part of the stated mechanism.
- Equal marginals alone do not prove independence.
- Independent trials can have different success probabilities and unequal path probabilities.
What can you explain now?
Do independent trials have to share the same success probability?
Compare with the explanation
No; independence concerns joint probability factorisation.
Identical distributions are a separate assumption from independence.
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Make it stick
Check the mechanism and independence; multiply the actual stage probabilities and add paths with their own weights.
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.