Question

An architecture company built 200 bridges, 400 hospitals, and 600 hotels. The probability of damage due to an earthquake of a bridge, a hospital, and a hotel are 0.01, 0.15, 0.03 respectively. One of the construction gets damaged in an earthquake. What is the probability that it is a hotel?

(a) \(\frac{1}{26}\)

(b) \(\frac{1}{40}\)

(c) \(\frac{7}{52}\)

(d) \(\frac{9}{40}\)

Answer 1

Answer: (D) = \(\frac{9}{40}\)

Let E₁, E₂, E₃ and A be the events defined as follows: 

E₁ = Construction chosen is a bridge. 

E₂ = Construction chosen is a hospital. 

E₃ = Construction chosen is a hotel. 

A = Construction gets damaged. 

Since there are (200 + 400 + 600) = 1200 constructions,

\(P(E_1) =\frac{200}{1200} =\frac{1}{6}\) , \(P(E_2) =\frac{400}{1200} =\frac{1}{3} , P(E_3) =\frac{600}{1200} =\frac{1}{2}\)

Given, Probability that the construction that gets damaged is a bridge 

= P(A/E₁) = 0.01 

Similarly, P(A/E₂) = 0.15 and P(A/E₃) = 0.03 

\(\therefore\)  Probability that a hotel gets damaged in an earthquake =\(P\big(\frac{E_3}{A}\big)\) 

 = \(\frac{P(E_3) \times P(A/E_3)}{P(E_1) \times P(A/E_1) + P(E_2) \times P(A/E_2)+ P(E_3) \times P(A/E_3)}\)  (Using Bayes Th.)

= \(\frac{\frac{1}{2} \times 0.03}{\frac{1}{6} \times 0.01 + \frac{1}{3} \times 0.15 +\frac{1}{2} \times 0.03}\) 

= \(\frac{\frac{1}{2} \times 0.03}{\frac{1}{6} \times (0.01+0.3\times 0.09)}\)

=\(\frac{6}{2} \times \frac{0.03}{0.4} = \frac{6\times 3\times 10}{2 \times 4 \times 100}\)

= \(\frac{9}{40}\)