01 · All three
Find P(SSS).
Hint
Multiply all three success chances.
Worked solution
(1/2)(3/5)(3/4)=9/40.
Understand · explore · practise
Multi-stage probability
Enumerate three-stage events, distinguish exactly from at least, calculate first-success probabilities and state the independence needed when repeating whole experiments or sampling groups.
Before you startProbability trees, complements and independence.
01 / Define the whole outcome
A three-stage path records three ordered results.
With two possible results at each of three stages, there are eight complete paths.
Multiply the appropriate probabilities along a path, then add the qualifying paths.
Independence lets this model use the stated stage probabilities after any earlier result. Unequal stage probabilities mean different paths need not be equally likely.
Choose a three-stage eventExplore
Three mutually independent stages have success probabilities 1/2,3/5,3/4. S=success,F=failure. The stages do not all have the same success chance.
No successes: FFF, probability(1/2)(2/5)(1/4)=1/20.
02 / Count successes in each path
Exactly two has three alternative orders.
Watch: equal counts can have unequal path probabilities
Pause, replay or seek freely. The notes explain the same idea and stay in view.
Success chances are 1/2,3/5 and3/4.Worked example
SSF: (1/2)(3/5)(1/4)=3/40
Failure is at the third stage.
SFS: (1/2)(2/5)(3/4)=6/40
Failure is at the second stage.
FSS: (1/2)(3/5)(3/4)=9/40
Exactly-two probability=(3+6+9)/40=9/20.
02 · Exactly one
Find the probability of exactly one success.
Hint
Add SFF,FSF,FFS.
Worked solution
SFF=2/40, FSF=3/40, FFS=6/40. Total 11/40.
03 / Use the complementary event
At least one excludes only no successes.
03 · None
Find P(FFF).
Hint
Use all three failure chances.
Worked solution
(1/2)(2/5)(1/4)=2/40=1/20.
04 · At least one
Find P(at least one success).
Hint
Complement none.
Worked solution
1−1/20=19/20.
05 · At least two
Find P(at least two successes).
Hint
Exactly two or exactly three.
Worked solution
18/40+9/40=27/40.
06 · Check the distribution
Verify probabilities for 0,1,2,3 successes sum to 1.
Hint
Use2/40,11/40,18/40,9/40.
Worked solution
(2+11+18+9)/40=1.
04 / First success imposes earlier failures
A later success alone does not identify the first one.
Find the probability the first success occurs at stage 3.Worked example
The path must start with two failures
Only FFS qualifies.
P(first at 3)=(1/2)(2/5)(3/4)=3/20
Multiplying just the third success probability would ignore earlier results.
First at 1:1/2; first at 2:(1/2)(3/5)=3/10
Together with first at 3 and no success, these disjoint cases sum to 1.
07 · First by stage 2
Find the probability of a first success by the end of stage 2.
Hint
Either first at 1 or first at 2.
Worked solution
1/2+3/10=4/5, also1−(1/2)(2/5).
08 · Identical trials
For independent trials with success chance 0.2, find the probability of the first success on trial 3.
Hint
Fail twice then succeed.
Worked solution
0.8×0.8×0.2=0.128. This uses identical independent trial probabilities.
05 / Respect a stopping limit
By a trial limit includes several possible first-success times.
09 · Within four trials
With independent success chance 0.2 per trial, find the probability of at least one success within four trials.
Hint
Complement four consecutive failures.
Worked solution
1−0.8⁴=1−0.4096=0.5904.
10 · Not guaranteed
Why does allowing four trials not guarantee success?
Hint
The failure path has positive probability.
Worked solution
All four can fail, with probability 0.8⁴=0.4096. A larger number of opportunities does not make a finite guarantee.
06 / Repeat a whole experiment
Treat the defined experiment outcome as the next event.
Call the original three-stage experiment a pass if it has at least two successes. Its pass chance is 27/40. Repeat the whole experiment independently twice.Worked example
Both passes: (27/40)²=729/1600
Independence is now stated between whole experiments.
Exactly one pass:2(27/40)(13/40)=351/800
The two pass/fail orders have equal probabilities in this repeated identical model.
At least one pass:1−(13/40)²=1431/1600
Complement two failures of the whole experiment.
11 · Repetition assumption
Can these powers be used if a shared fault links the two experiments?
Hint
The experiment-level outcomes may be dependent.
Worked solution
Not without further information. The simple products rely on independent repetitions; a shared fault may change the conditional chance in the second run.
07 / Dependence within groups can coexist with independence between groups
Choose the event at the level supported by the assumptions.
12 · Two-person groups
In each group, each person succeeds with probability 0.6, but the probability both succeed is 0.4. Two groups are sampled independently from this model. Find the probability all four people succeed.
Hint
Use the probability for both succeeding in one group.
Worked solution
0.4×0.4=0.16. Using 0.6⁴ would incorrectly assume independence of the two people within each group.
13 · At least one complete group
For those two independently sampled groups, find the probability at least one group has both people succeed.
Hint
A group fails the both-success event with probability 0.6.
Worked solution
1−0.6²=0.64. The group event is defined before using the complement.
14 · Finite group sampling
Does choosing two groups without replacement from a small fixed list automatically justify independent group outcomes?
Hint
Selection changes which groups remain.
Worked solution
No. Use the finite-list counts or conditional information. The independent-group calculation requires that assumption or a sampling process that supports it.
08 / Separate paths, counts and repetitions
State the level at which independence applies.
List complete paths for exactly-count events, use complements for at-least questions and require earlier failures when finding a first success. When repeating experiments or sampling groups, define the group event and state the independence assumption before multiplying.
Section 1 of 8 · Define the whole outcome