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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?

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Solution

The correct option is **B** 152

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 1200 constructions, therefore

P(E1)=2001200=16,P(E2)=4001200=13

and P(E3)=6001200=12

It is given that P(AE1)= Probability that a construction gets damaged is a bridge = 0.01

Similarly, P(AE2)=0.03 and P(AE3)=0.15

We are required to find P(E1A) by Baye's rule.

P(E1A)=P(E1)P(AE1)P(E1)P(AE1)+P(E2)P(AE2)+P(E3)P(AE3)=16×0.0116×0.01+13×0.03+12×0.15=11+6+45=152

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 1200 constructions, therefore

P(E1)=2001200=16,P(E2)=4001200=13

and P(E3)=6001200=12

It is given that P(AE1)= Probability that a construction gets damaged is a bridge = 0.01

Similarly, P(AE2)=0.03 and P(AE3)=0.15

We are required to find P(E1A) by Baye's rule.

P(E1A)=P(E1)P(AE1)P(E1)P(AE1)+P(E2)P(AE2)+P(E3)P(AE3)=16×0.0116×0.01+13×0.03+12×0.15=11+6+45=152

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