All numeric cases below are synthetic teaching examples, not empirical research findings.
Example: conditional delivery
| P(funding)=0.8. P(launch | funding)=0.7. P(launch | no funding)=0.1. |
P(launch)=0.80.7+0.20.1=0.58. Merely multiplying 0.8*0.7 gives the funded-launch pathway, not every launch pathway.
Example: unknown dependence
P(A)=0.7 and P(B)=0.6. P(A and B) is bounded between 0.3 and 0.6. The value 0.42 requires independence or an equivalent supplied conditional.
Example: incompatible conditionals
| P(Y)=0.6, P(S)=0.5, P(Y | S)=0.8 and P(Y | not S)=0.1 imply P(Y)=0.45. Report the inconsistency and ask which estimates should be revised. Do not silently normalize the three values. |