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: Name the events:
Let \(E_1, E_2, E_3\) be the events that the contract goes to X, Y, Z. Let \(F\) be the event that a part of the building falls off (poor quality).
We need \(P(E_2 \mid F)+P(E_3 \mid F)\).

Step 2: Write the given probabilities:
\[ P(E_1)=\frac{1}{3},\ P(E_2)=\frac{2}{9},\ P(E_3)=\frac{4}{9} \]
\[ P(F \mid E_1)=\frac{1}{5},\ P(F \mid E_2)=\frac{2}{5},\ P(F \mid E_3)=\frac{3}{10} \]

Step 3: Find the total probability of poor quality:
\[ P(F)=\frac{1}{3}\cdot\frac{1}{5}+\frac{2}{9}\cdot\frac{2}{5}+\frac{4}{9}\cdot\frac{3}{10}=\frac{3}{45}+\frac{4}{45}+\frac{6}{45}=\frac{13}{45} \]

Step 4: Apply Bayes theorem to Y and Z:
\[ P(E_2 \mid F)=\frac{4/45}{13/45}=\frac{4}{13},\qquad P(E_3 \mid F)=\frac{6/45}{13/45}=\frac{6}{13} \]
The events "Y" and "Z" cannot happen together, so we add the two probabilities.
\[ P(E_2 \text{ or } E_3 \mid F)=\frac{4}{13}+\frac{6}{13}=\frac{10}{13} \]

Final Answer:
The probability that the fault came from company Y or Z is \(\frac{10}{13}\). \[ \boxed{\frac{10}{13}} \]