Question

An organisation needs to select a building construction company to make houses in cities. Three companies X, Y and Z are short-listed with the probability of their selection being \(\frac{1}{3}\), \(\frac{2}{9}\) and \(\frac{4}{9}\) respectively. It was found that there were 20%, 40% and 30% chances respectively, of delivering poor construction quality.

Based on the above information, answer the following questions :

If a part of the newly constructed building falls off, then what is the probability that it was due to the selection of company Y or Z ?

Hint:

Find P(poor) = 13/45, then P(X given poor) = 3/13. The answer for Y or Z is 1 minus that.

Answer 1

Step 1: Use the complement idea:
There are only three companies. So "Y or Z" is the opposite of "X".
Let $B$ be the event that part of the building falls off. Then $P(Y \cup Z \mid B)=1-P(X \mid B)$.

Step 2: Find the chance of poor quality overall:
Use the poor-quality chances 0.2, 0.4 and 0.3 with the selection probabilities.
\[ P(B)=\frac{1}{3}(0.2)+\frac{2}{9}(0.4)+\frac{4}{9}(0.3)=\frac{3}{45}+\frac{4}{45}+\frac{6}{45}=\frac{13}{45} \]

Step 3: Find the part for X:
The joint probability for X is $P(X \cap B)=\frac{1}{3}(0.2)=\frac{3}{45}$.
By Bayes theorem:
\[ P(X \mid B)=\frac{3/45}{13/45}=\frac{3}{13} \]

Step 4: Subtract from 1:
\[ P(Y \cup Z \mid B)=1-\frac{3}{13}=\frac{10}{13} \]
We can check by adding the direct parts: $\frac{4}{13}+\frac{6}{13}=\frac{10}{13}$.

Final Answer:
If part of the building falls off, the contract was with Y or Z with probability $\frac{10}{13}$. \[ \boxed{\frac{10}{13}} \]