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 the building constructed was found to be strong, then what is the probability that the construction contract was given to the company X ?

Hint:

Use Bayes theorem with P(strong given X) = 4/5, P(strong given Y) = 3/5 and P(strong given Z) = 7/10.

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 \(S\) be the event that the building is strong.
We want \(P(E_1 \mid S)\), so Bayes theorem applies.

Step 2: Write the given probabilities:
The selection probabilities are \(P(E_1)=\frac{1}{3}\), \(P(E_2)=\frac{2}{9}\), \(P(E_3)=\frac{4}{9}\).
Poor quality has chances 20%, 40%, 30%. Strong quality is the complement of poor quality.
\[ P(S \mid E_1)=1-0.2=\frac{4}{5},\quad P(S \mid E_2)=1-0.4=\frac{3}{5},\quad P(S \mid E_3)=1-0.3=\frac{7}{10} \]

Step 3: Find the total probability of a strong building:
Add the three cases using the total probability rule.
\[ P(S)=\frac{1}{3}\cdot\frac{4}{5}+\frac{2}{9}\cdot\frac{3}{5}+\frac{4}{9}\cdot\frac{7}{10}=\frac{4}{15}+\frac{2}{15}+\frac{14}{45} \]
\[ P(S)=\frac{12}{45}+\frac{6}{45}+\frac{14}{45}=\frac{32}{45} \]

Step 4: Apply Bayes theorem:
\[ P(E_1 \mid S)=\frac{P(E_1)\,P(S \mid E_1)}{P(S)}=\frac{12/45}{32/45}=\frac{12}{32}=\frac{3}{8} \]

Final Answer:
The probability that the contract went to company X, given a strong building, is \(\frac{3}{8}\). \[ \boxed{\frac{3}{8}} \]