Probbility question


#1

a businessman goes to hotels x,y,z 20%,50%,30% of the time respectively.it is known that 5%,4%,8% of the rooms in x,y,z hotels have faculty plumbings.what is the probability that businessman’s room having faulty plumbing is assigned to hotel Z?


#2

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
=5/9
Hence, the Probability that faulty plumbing is assigned to hotel Z is
4/9