01 · A and no delay
Find the complete path probability.
Hint
The no-delay probability on A is 0.9.
Worked solution
0.6×0.9=0.54.
Understand · explore · practise
Conditional branches and total probability
Interpret second-stage probabilities conditional on a route or group, form weighted totals across disjoint branches, and check what independence would require.
Before you startTree diagrams, complements and path probabilities.
01 / Read a branch in its context
The probability after a route is chosen is conditional on that route.
P(A and delay)=P(A)P(delay given A).
Path multiplication remains valid with conditional branch values; independence is not required.
Here, 10% is the delay chance among route-A parcels. It is not the probability that a randomly chosen parcel both uses A and is delayed. The first branch determines how much weight that group receives.
Change the route mixtureExplore
A parcel uses route A with probability w and route B otherwise. Delay chance: 10% on A,30% on B. These are model probabilities conditional on route.
At w=0, all parcels use route B, so the delay probability is 0.30.
02 / Calculate each complete path
Use the route proportion before the delay proportion.
Watch: weight each route before adding
Pause, replay or seek freely. The notes explain the same idea and stay in view.
Take w=0.6: route A 60%, route B 40%.Worked example
A and delay: 0.6×0.1=0.06
Six percent of all parcels follow this path.
B and delay: 0.4×0.3=0.12
Twelve percent of all parcels follow this path.
Total delay: 0.06+0.12=0.18
The two delayed paths are disjoint and cover every delayed parcel.
02 · B and no delay
Find the complete path probability.
Hint
The no-delay probability on B is 0.7.
Worked solution
0.4×0.7=0.28.
03 / Add weighted conditional probabilities
A simple mean assumes equal route weights.
P(delay)=w×0.10+(1−w)×0.30.
The groups must be disjoint and exhaustive for this two-branch sum.
03 · Unweighted error
Why is(0.10+0.30)/2=0.20 wrong when w=0.6?
Hint
The routes are not equally likely.
Worked solution
A has 60% weight and B 40%. The correct weighted result is 0.18. A simple average works here only when each route has 50% weight.
04 · Whole tree
Check the four complete path probabilities sum to1.
Hint
Include both delay outcomes for each route.
Worked solution
0.06+0.54+0.12+0.28=1.
05 · No delay
Find P(no delay) in two ways.
Hint
Add the two no-delay paths or complement delay.
Worked solution
0.54+0.28=0.82, also 1−0.18=0.82.
04 / Change the weights, not the branch meanings
A group may have a high rate but a small population share.
06 · Equal routes
Find P(delay) if A and B are equally likely.
Hint
Set w=0.5.
Worked solution
0.5×0.1+0.5×0.3=0.20.
07 · Mostly route A
Find P(delay) for w=0.75.
Hint
The route-B weight is 0.25.
Worked solution
0.75 × 0.1 + 0.25 × 0.3 = 0.075 + 0.075 = 0.15.
08 · Which delayed path is larger?
For w=0.75, does B contribute more delayed parcels because its delay rate is larger?
Hint
Compare the complete path probabilities.
Worked solution
No. A and delay and B and delay each have probability 0.075. The larger B conditional rate is balanced by its smaller route share.
05 / Solve for an unknown mixture
The target total must lie between the two group rates.
Find w if the overall delay probability is 0.14.Worked example
0.10w+0.30(1−w)=0.14
Use the weighted total.
0.30−0.20w=0.14
Collect like terms.
w=0.8
This is a valid probability, so 80% use route A.
09 · Impossible target
Can these same two route rates produce overall delay 0.05 by changing w?
Hint
A weighted average cannot fall below the smaller rate.
Worked solution
No. For 0≤w≤1, the total lies in[0.10,0.30]. Solving formally gives w=1.25, an invalid weight.
10 · Route given delay
At w=0.6, among delayed parcels, what fraction came from B?
Hint
Restrict the denominator to delayed parcels.
Worked solution
P(B given delay)=0.12/0.18=2/3. This differs from P(delay given B)=0.30; the conditioning direction matters.
06 / Compare the group-specific chances
The route and delay events are not independent here.
11 · Product test
At w=0.6, test independence of route A and delay.
Hint
Compare P(A and delay) with P(A)P(delay).
Worked solution
0.06 differs from0.6×0.18=0.108. They are dependent; also P(delay given A)=0.10 differs from the overall 0.18.
12 · Equal branch chances
If both routes instead had delay probability 0.2, what would the overall delay probability be?
Hint
Factor the common conditional rate.
Worked solution
0.2w+0.2(1−w)=0.2. With both route probabilities positive, the unchanged conditional chance shows independence of route and delay.
07 / State where the probabilities came from
A tree can represent an assumption or estimates from records.
13 · Historical rates
The 10% and30% rates were estimated from last month. Is0.18 necessarily exact for next month?
Hint
The future process may differ.
Worked solution
No. It is a model estimate using those conditional rates and the assumed 60/40 route mix. Changes in traffic, procedures or selection can change the result.
14 · Missing route
A third route handles some parcels but is omitted. Can A and B still be assigned probabilities 0.6 and0.4 for all parcels?
Hint
Their probabilities already sum to1.
Worked solution
Not if a positive share uses the third route. The first-stage branches must describe every eligible parcel exactly once, or the experiment must explicitly restrict selection to A/B parcels.
08 / Weight, multiply and add
Conditional branch labels support dependent calculations.
Multiply each group weight by the event chance within that group. Add across disjoint exhaustive groups. Keep the conditioning direction clear, check totals against the range of group rates and state assumptions when using estimated probabilities.
Section 1 of 8 · Read a branch in its context