Question

19. A businessman goes to hotels X, Y and Z 20%, 50%, 30% of the time respectively.
It is known that 5%, 4% and 8% of the rooms in X, Y and Z hotels have faulty
plumbings. What is the probability that businessman's room having faulty is
assigned to hotel Z?​

Answer 1

The probability that the businessman goes to hotel X is P(X) = 0.20.2

The probability that the businessman goes to hotel Y is P(Y) = 0.50.5

The probability that the businessman goes to hotel Z is P(Z) = 0.3

Suppose AA is the event of all the faulty plumbings, then it results in

P(A/X)=0.05 P(A/Y)=0.04 P(A/Z)=0.08

The Probability that faulty plumbing is assigned to hotel Z is P(Z/A)

P(Z/A)=(P(Z)P(A/Z))/P(X)P(A/X)+P(Y)P(A/Y)+P(Z)P(A/Z)

P(Z/A)=(0.30.08)/(0.20.05+0.50.04+0.30.08)

= 0.024/0.1+0.02+0.024

=0.024/0.054

=24/54

=4/9

Hence, the Probability that faulty plumbing is assigned to hotel Z is

4/9