Question

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. Then, the probability that the construction job will be completed on time is

• A. $0.614$
• B. $0.488$
• C. $0.128$
• D. $0.288$

Answer 1

Answer: (B)

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. We have to find $P(A)$.
We have, $P(B)=0.65$,
$ P(\text { no strike })=P\left(B^{\prime}\right)=1-P(B)=1-0.65=0.35$
$ P(A / B)=0.32, P\left(A / B^{\prime}\right)=0.80$
Since, events $B$ and $B^{\prime}$ form a partition of the sample space $S$, therefore, by theorem on total probability, we have
$P(A) =P(B) P(A / B)+P\left(B^{\prime}\right) P\left(A / B^{\prime}\right)$
$ =0.65 \times 0.32+0.35 \times 0.8$
$ =0.208+0.28$
$ =0.488$
Thus, the probability that the construction job will be completed in time is $0.488$.