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Thread: Conditional Probability

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    Conditional Probability




    Dear all Please help in solving the following problem.
    A large industrial firm uses 3 local motels to provide overnight accommodations for its clients. from past experience, it is known that 20% of the clients are assigned rooms at the Ramada Inn, 50% at the sheeraton and 30% at the Lakeview Motor Lodge. if the plumbing is faulty in 5% of the rooms at the Ramada Inn, in 4% of the rooms at the Sheraton and in 8% of the rooms at the Lakeview Motor Lodge, what the probability that
    a) A client will be assigned a room with faulty plumbing?
    b) A person with a room having faulty plumbing was assigned accommodation at the Lakeview Motor Lodge?
    I have come with the following solution.
    Let L: a client is assigned accommodation at Lakeview Motor Lodge
    R: a client is assigned accommodation at Ramada Inn
    S: a client is assigned accommodation at Sheraton
    P(Lf): Probability of faulty plumbing rooms in Lakeview=0.08
    P(Rf): Probability of faulty plumbing rooms in Ramada=0.05
    P(Sf): Probability of faulty plumbing rooms in Sheraton=0.04
    Now
    a) P(client assigned with faulty plumbing room)=P(Lf/L)+P(Rf/R)+P(Sf/S)
    =0.3*0.08+0.2*0.05+0.5*0.04
    b) P(faulty plumbing room in Lakeview)=P(L/Lf)
    =0.08*0.3

    Needs help

  2. #2
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    Re: Conditional Probability


    b) is looking for probability of Lakeview room given faulty plumbing. L: Lakeview room f: faulty plumbing

    P(L|f)= [P(f|L)P(L)]/P(f) This is Baye's Rule

    The denominator P(f) is what you found in a)
    Your answer for part b) is the numerator
    "I have discovered a truly remarkable proof of this theorem which this margin is too narrow to contain." Pierre de Fermat

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