Question

A person has a contract of construction. The probability of being a strike is 0.65. The probabilities of completing the construction work on time in both conditions are 0.80 and 0.32 whether the strike is not happened and it is happened respectively. Then find the probability of completing the construction work in due time.

Hint:

Drawing a probability tree can be very helpful for visualizing problems involving conditional probabilities and the law of total probability. The branches would split first into 'Strike' / 'No Strike', and then each of those would split into 'Complete on Time' / 'Not on Time'.

Answer 1

Step 1: Understanding the Concept:
This problem can be solved using the Law of Total Probability. This law is used to find the probability of an event when it depends on two or more mutually exclusive and exhaustive events.
Step 2: Key Formula or Approach:
Let C be the event of completing the work on time.
Let S be the event that a strike happens.
Let S' be the event that a strike does not happen.
The Law of Total Probability states: \[ P(C) = P(C \cap S) + P(C \cap S') \] Using the conditional probability formula \( P(A \cap B) = P(A|B)P(B) \), this becomes: \[ P(C) = P(C|S)P(S) + P(C|S')P(S') \] Step 3: Detailed Explanation:
From the problem statement, we can extract the following probabilities:
- Probability of a strike: \( P(S) = 0.65 \) - Probability of no strike: \( P(S') = 1 - P(S) = 1 - 0.65 = 0.35 \) - Probability of completing on time, given no strike: \( P(C|S') = 0.80 \) - Probability of completing on time, given a strike happened: \( P(C|S) = 0.32 \) Now, we apply the Law of Total Probability to find \( P(C) \): \[ P(C) = P(C|S) \cdot P(S) + P(C|S') \cdot P(S') \] \[ P(C) = (0.32)(0.65) + (0.80)(0.35) \] Calculate the individual terms: \[ (0.32)(0.65) = 0.208 \] \[ (0.80)(0.35) = 0.280 \] Add them together: \[ P(C) = 0.208 + 0.280 = 0.488 \] Step 4: Final Answer:
The probability of completing the construction work in due time is 0.488.