Question

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

• A. $\frac{1}{26}$
• B. $\frac{1}{52}$
• C. $\frac{7}{52}$
• D. $noneofthese$

Answer 1

The correct option is B $\frac{1}{52}$

Let $E_1,E_2,E_3$ and A be the events defined as follows:
$E_1$ = construction chosen is a bridge.
$E_2$ = construction chosen is a hospital.
$E_3$ = construction chosen is a hotel.
A = construction gets damaged
Since there are 1200 constructions, therefore
$P\left(E_1\right)=\frac{200}{1200}=\frac{1}{6},P\left(E_2\right)=\frac{400}{1200}=\frac{1}{3}$
and $P\left(E_3\right)=\frac{600}{1200}=\frac{1}{2}$
It is given that $P\left(\frac{A}{E_1}\right)$= Probability that a construction gets damaged is a bridge = 0.01
Similarly, $P\left(\frac{A}{E_2}\right)=0.03$ and $P\left(\frac{A}{E_3}\right)=0.15$
We are required to find $P\left(\frac{E_1}{A}\right)$ by Baye's rule.
$P\left(\frac{E_1}{A}\right)=\frac{P\left(E_1\right)P\left(\frac{A}{E_1}\right)}{P\left(E_1\right)P\left(\frac{A}{E_1}\right)+P\left(E_2\right)P\left(\frac{A}{E_2}\right)+P\left(E_3\right)P\left(\frac{A}{E_3}\right)}=\frac{\frac{1}{6}\times 0.01}{\frac{1}{6}\times 0.01+\frac{1}{3}\times 0.03+\frac{1}{2}\times 0.15}=\frac{1}{1+6+45}=\frac{1}{52}$