A person has undertaken a construction job. The probabilities are 0.65 that there will be strike, 0.80 that the construction job will be completed on time if there is no strike, and 0.32 that the construction job will be completed on time if there is a strike. Determine the probability that the construction job will be completed on time.

Let A be the event that the construction job will be completed on time, and B be the event that there will be a strike.
Now    P(B) = 0.65
P(no strike) = P(B’) = 1 - P(B) = 1 - 0.65 = 0.35
P(A | B) = 0.32, P(A | B') = 0.80
Since events B and B' form a partition of the sample space S,
∴  By theorem on total probability, we have
P(A) = P(B) P(A | B) + P(B') P(A | B’)
= 0.65 × 0.32 + 0.35 × 0.8
= 0.208 + 0.28 = 0.488
∴  the probability that the construction job will be completed in time is 0.488.