A contractor is planning a construction project to be completed in three stages. The contractor figures that 1. the chance that the first stage will be completed on time is 0.8.
2. given that the first stage is completed on time, the chance that the second stage will be completed on time is 0.85.
3. given that both the first and the second stages are completed on time, the chance that the third stage will be
completed on time is 0.7.
The schedule is tight in the sense that delay in any stage will necessarily cause delay in later stage(s). That is, the
delay in stage 2 causes delay in stage 3, while delay in stage 1 causes delays in both stages 2 and 3. Find
probabilities of the events
A = the contractor will finish the job on time.
B= the contractor will not finish the 2nd stage on time.
C= The contractor will not finish the 3rd stage on time.
A) P(A) = 0.8 x 0.85 x 0.7 = 0.51
B) P(B) = 1 - 0.85 = 0.15
C) P(C) = 1 - 0.7 = 0.3