An architecture company built 200 bridges, 400 hospitals, and 600 hote

1.	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» 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.) = \(\frac{\frac{1}{2} \times 0.03}{\frac{1}{6} \times 0.01 + \frac{1}{3} \times 0.15 +\frac{1}{2} \times 0.03}\) = \(\frac{\frac{1}{2} \times 0.03}{\frac{1}{6} \times (0.01+0.3\times 0.09)}\) =\(\frac{6}{2} \times \frac{0.03}{0.4} = \frac{6\times 3\times 10}{2 \times 4 \times 100}\) = \(\frac{9}{40}\)

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»

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.)

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

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

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

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