The completion of a construction job may be delayed due to strike. The probability of strike is 0.6. The probability that the construction job gets completed on time if there is no strike is 0.85 and the probability that the construction job gets completed on time if there is a strike is 0.35. What is the probability that the construction job will not be completed on time ?

Concept:

• P(A|B) is the probability of A after B happens.
• P(A ∩ B) + P(A ∩ B̅) = P(A)
• P(A|B) = \(P(A \cap B)\over P(B)\)
• P(A̅ ) = 1 - P(A)

Calculation:

Let A be the event that the construction job is completed on time and B be the event that there is strike.

Given:

The probability of strike is 0.6 ⇒ P(B) = 0.6  __(i)

The probability that the construction job gets completed on time if there is no strike is 0.85

⇒ P(A| B̅) = 0.85  __(ii)

The probability that the construction job gets completed on time if there is a strike is 0.35.

⇒ P(A| B) = 0.35 __(iii)

Find: The probability that the construction job will not be completed on time = 1 - P(A)

Solution:

From (ii), 

 P(A| B̅) = 0.85 

⇒ \({P(A \cap \overline B)\over P(\overline B) }= 0.85\)

⇒ \({P(A \cap \overline B) }= 0.85 P(\overline B)\)

⇒ \({P(A \cap \overline B) }= 0.85 (1 -P(B))\) 

⇒ \({P(A \cap \overline B) }= 0.85 (1 - 0.6)\)  {From (i)}

⇒ \({P(A \cap \overline B) }= 0.34\)  __(iv)

Similarly from (iii),

 P(A| B) = 0.35

⇒ \({P(A \cap B)\over P( B) }= 0.35\)

⇒ \({P(A \cap B) }= 0.35 P( B)\)  

⇒ \({P(A \cap B) }= 0.35 (0.6)\)  {From (i)}

⇒ \({P(A \cap B) }= 0.21\)  __(v)

Adding (iv) and (v)

⇒ \({P(A \cap \overline B) }+ {P(A \cap B) }= 0.34 + 0.21\)

⇒ P(A) = 0.55

⇒ The probability that the construction job is not completed on time = 1 - P(A) = 1 - 0.55

⇒ The probability that the construction job is not completed on time = 0.45

∴ The correct option is (2).