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 $0.488$.