Why Two Success Probabilities Are Needed
A development program can have several sequential stages, each with its own probability of success. A probability attached to one stage may describe success given that the program has reached that stage. It is not automatically the probability that a candidate succeeds across the entire program. To answer the latter question, the probabilities of passing all required stages must be combined in a way that respects their conditions.
This distinction matters when comparing programs or summarizing technical risk. A phase-conditional probability uses a particular group as its denominator: programs that reached that phase and met the stated prior conditions. A cumulative probability instead concerns completing a sequence of required transitions from a specified starting point. The starting point must be explicit; otherwise, even a correctly calculated product can answer the wrong question.
Because \(A_i\) includes all the earlier required successes, \(q_i\) conditions on having reached stage \(i\) by passing the preceding stages. In contrast, the marginal probability \(P(S_i)\) refers to the event of stage \(i\) success without that conditioning. These quantities can differ, particularly when success outcomes across stages are associated.
The Chain Rule for Sequential Success
The cumulative probability follows from the probability chain rule. Each step asks for the probability of adding the next required success, conditional on the successful history so far. Multiplication is appropriate because these are conditional probabilities; independence is not required.
Specify \(A_0\), the state from which the program-level probability is to be calculated. It may represent all candidates under consideration or only those that have already entered development.
For each stage, define \(S_i\) as success under the program’s stated criterion, and set \(A_i=A_{i-1}\cap S_i\).
Set \(q_i=P(S_i\mid A_{i-1})\), making sure that each conditioning event has positive probability.
The product gives \(P(A_K\mid A_0)\); multiply by \(P(A_0)\) if the desired probability is unconditional on entering the starting state.
For the first equality, \(A_K\) is the event that \(A_0\) occurs and all the required stage successes occur. Repeated application of the chain rule gives the conditional probability of that sequence, one transition at a time. Since \(A_{i-1}\) is contained in \(A_0\), conditioning on \(A_{i-1}\) already implies \(A_0\). The second equation then follows from \(A_K\subseteq A_0\): \(P(A_K)=P(A_K\mid A_0)P(A_0)\).
This result requires the events and conditioning history to be defined consistently, and each \(q_i\) to be defined. It does not require independent stage outcomes. Dependence is accommodated by the conditional probabilities along the successful history. Independence is one possible assumption under which those conditional probabilities can be replaced by corresponding marginal probabilities.
Worked Example: Including the Probability of Reaching the Starting State
Suppose a hypothetical program has a 0.80 probability of reaching its specified starting state, \(A_0\). Conditional on reaching that state, the probabilities of passing three required stages are \(q_1=0.70\), \(q_2=0.60\), and \(q_3=0.50\). Each probability is conditional on success through the previous stage. Calculate both the probability of completing all three stages given \(A_0\), and the unconditional probability.
First calculate conditional cumulative success:
Thus, among candidates that reach the defined starting state, the probability of passing all three stages is 0.210, or 21.0%. For the unconditional probability, include the chance of reaching \(A_0\):
The unconditional probability is 0.168, or 16.8%. Reporting 0.210 as the unconditional probability would omit the 20.0% of candidates that do not reach the specified starting state.
Conditional Probabilities and Marginal Probabilities
A frequent shortcut is to multiply the marginal stage probabilities, \(P(S_1),\ldots,P(S_K)\). That product equals the chain-rule product if, along the required successful history, each conditional probability equals the corresponding marginal probability. Full independence of all stage-success events is sufficient for this condition, but it is not necessary. Conversely, if the conditional probabilities differ from the marginals, a product of marginals may give a different result.
Equality of the two products is not, by itself, proof that the stages are independent. The products could be equal by coincidence, including because differences between individual factors offset one another. The correct calculation for sequential success remains the chain-rule product of the probabilities conditional on reaching each stage. A product of marginals is justified only when its relationship to those conditional factors has been established.
Worked Example: Independent Stage Outcomes
Consider a hypothetical three-stage program with \(A_0\) equal to the sure event. Suppose the stage-success events are mutually independent, with marginal probabilities 0.70, 0.60, and 0.50. Calculate the cumulative probability and check the independence assumption’s role.
Under mutual independence, conditioning on earlier stage outcomes does not change the probability of success at the next stage. Therefore, \(q_1=0.70\), \(q_2=P(S_2\mid S_1)=0.60\), and \(q_3=P(S_3\mid S_1\cap S_2)=0.50\). The chain rule gives:
The product of the marginals gives the same value here because independence makes each conditional factor equal to its marginal counterpart. The result is a 21.0% probability of passing all three stages, starting from \(A_0\). The multiplication is not justified merely because there are three phase-specific probabilities; it is justified by the stated independence assumption, or by other information showing that the relevant conditional factors equal those marginals.
Worked Example: Dependent Transitions Change the Cumulative Probability
Now consider a hypothetical three-stage program with \(A_0\) again equal to the sure event. Suppose \(P(S_1)=0.60\), \(P(S_2\mid S_1)=0.50\), and \(P(S_3\mid S_1\cap S_2)=0.80\). To see why marginal probabilities can differ, assume \(P(S_2\mid S_1^c)=0.625\), and that \(P(S_3\mid S_1,S_2)\) is 0.80 when both stages 1 and 2 succeed and 0.40 for each of the other three stage-1/stage-2 histories.
The chain-rule factors along the path of required successes are 0.60, 0.50, and 0.80. Therefore:
The marginal probability of stage 2 success is \(P(S_2)=(0.60)(0.50)+(0.40)(0.625)=0.550\). The probability of the joint stage-1/stage-2 history \(S_1\cap S_2\) is 0.300, leaving probability 0.700 for the other histories. Thus:
The product of marginal probabilities is \((0.60)(0.550)(0.520)=0.1716\), not 0.240. The difference arises because the marginal probabilities do not represent the successive conditional transitions on the all-success path. The specified conditional probabilities are coherent: the four stage-1/stage-2 histories have probabilities 0.300, 0.300, 0.250, and 0.150, which sum to 1, and the stated stage-3 probabilities are between 0 and 1.
Worked Example: Equal Products Do Not Prove Independence
Consider three hypothetical success events. Let \(S_1\) and \(S_2\) be independent, each with probability 0.50. Each of the four joint histories for \(S_1,S_2\) then has probability 0.25. Suppose the conditional probabilities of \(S_3\) for histories \((S_1,S_2)=(1,1),(1,0),(0,1),(0,0)\) are, respectively, 0.50, 1.00, 0, and 0.50.
The cumulative success probability is calculated along the all-success history:
The marginal probability of \(S_3\) is \(P(S_3)=(0.25)(0.50)+(0.25)(1.00)+(0.25)(0)+(0.25)(0.50)=0.50\). Hence the product of the three marginals is also \((0.50)(0.50)(0.50)=0.125\). Yet \(S_3\) is not independent of \(S_1\): \(P(S_3\mid S_1)=(0.50+1.00)/2=0.75\), which is not \(P(S_3)=0.50\). The matching products in this example do not establish independence; they happen to agree for the cumulative event.
Reading and Checking Program Probabilities
A useful comparison is to distinguish the probability’s denominator and conditioning history, not just the stage label. A reported phase-conditional probability describes transition success among candidates who meet its conditioning conditions. A cumulative probability describes passing every required transition from a stated starting point. Marginal phase probabilities, if used without conditioning, describe events over their specified population and need not correspond to the sequential pathway.
| Quantity | What it describes | Use in a sequential program |
|---|---|---|
| \(q_i=P(S_i\mid A_{i-1})\) | Stage \(i\) success given prior required successes | Use as the \(i\)th factor in the chain-rule product |
| \(P(S_i)\) | Stage \(i\) success without conditioning on the successful history | Do not substitute automatically for \(q_i\) |
| \(P(A_K\mid A_0)\) | Success through all \(K\) required stages among those in the starting state | Summarizes cumulative success from that starting state |
| \(P(A_K)\) | Success through all stages, including whether the starting state is reached | Use \(P(A_0)P(A_K\mid A_0)\) |
Before multiplying probabilities, check that the events use the same definition of success, the same starting population, and a compatible sequence of stages. Check that each phase probability has the stated denominator and that every conditioning event has positive probability. If a stage can be reached through different histories, one single transition probability may not describe all candidates at that stage; the pathway and conditioning must be made explicit. These calculations also depend on the chosen program structure: if success can occur through alternative pathways rather than requiring every listed stage, the all-stages product does not represent the program’s success event.
Common Mistakes and Limitations
- Using conditional probabilities as though they were unconditional. A transition probability among candidates that reached a phase is not automatically the chance of getting there from program entry. Multiply along the required transitions, and include \(P(A_0)\) when the target probability is unconditional.
- Multiplying marginals without checking the conditioning. The chain rule uses conditional factors. Independence is sufficient to replace them with marginal probabilities, but it is not required for the chain rule and is not the only way two products can agree.
- Leaving the starting point unstated. \(P(A_K\mid A_0)\) and \(P(A_K)\) answer different questions when \(P(A_0)<1\). Name the population or starting event for the probability being reported.
- Interpreting a product as a guarantee about an individual. A cumulative probability summarizes uncertainty across the defined population and events; it does not establish that a particular candidate will succeed or fail.
The chain-rule calculation is exact for the specified events and conditional probabilities, but its practical value depends on those inputs being appropriate for the target program and population. It does not by itself explain why stages are associated, determine how probabilities should be estimated, or account for uncertainty in the probabilities themselves. Those are separate questions from the arithmetic of cumulative success.
Check Your Understanding
Try each problem, then compare your reasoning with the solutions.
- A hypothetical program starts in \(A_0\) with probability 0.90 and has two transition probabilities, conditional on reaching each stage, of 0.80 and 0.50. Find \(P(A_2\mid A_0)\) and \(P(A_2)\).
- For three sequential stages, explain why multiplying the conditional transition probabilities does not require independence.
- Suppose two stage-success events each have marginal probability 0.60. Is their joint success probability necessarily 0.36? State a condition that would make it 0.36.
- A report gives a probability of 0.40 for success at a stage among programs that reached it. Explain why this alone does not give the probability of success from program entry.
- Can a product of marginal success probabilities equal the cumulative probability even when the stage events are dependent? Explain briefly.
Solutions. (1) \(P(A_2\mid A_0)=(0.80)(0.50)=0.400\); \(P(A_2)=(0.90)(0.400)=0.360\). (2) The chain rule factors the joint successful history into probabilities conditional on the preceding history; it is valid whether or not the stage outcomes are independent. (3) No. The joint probability is \(P(S_1)P(S_2\mid S_1)\), which is 0.36 if, for example, the events are independent so that \(P(S_2\mid S_1)=0.60\). (4) The value is conditional on reaching the stage; the probability of reaching it and the earlier transitions are also needed to calculate success from entry. (5) Yes. The products can agree by coincidence even when events are dependent; equality of the products does not establish independence.