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18 July, 14:19

A large industrial firm uses three 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 Sheraton, and 30% at the Lakeview Motor Lodge. If the plumbing is faulty in 5% of the rooms at the Ra - mada Inn, in 4% of the rooms at the Sheraton, and in 8% of the rooms at the Lakeview Motor Lodge, what is 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 accommodations at the Lakeview Motor Lodge?

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  1. 18 July, 16:03
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    (a) The probability is 0.054

    (b) The probability is 0.4444

    Step-by-step explanation:

    The probability P that a client will be assigned a room with faulty plumbing can be calculate as the sum of three probabilities: P1, P2 and P3

    P1 is the probability that a client is assigned in Ramada Inn and the plumbing is faulty. This is calculate as:

    P1 = 20%*5% = 0.01

    P2 is The probability that a client is assigned in Sheraton and the plumbing is faulty. This is calculate as:

    P2 = 50%*4% = 0.02

    P3 is The probability that a client is assigned in Lakeview Motor Lodge and the plumbing is faulty. This is calculate as:

    P3 = 30%*8% = 0.024

    Then, the probability P is:

    P = 0.01 + 0.02 + 0.024 = 0.054

    For part b, the probability that a person with a room having faulty plumbing was assigned accommodations at the Lakeview Motor Lodge is calculate as a division between P3 and P.

    Where P3 is the probability that a person was assigned to Lakeview Motor Lodge and has a room with faulty plumbing and P is the probability that a person is going to be assigned a room with faulty plumbing.

    So, the probability that a person with a room having faulty plumbing was assigned accommodations at the Lakeview Motor Lodge is:

    Probability = 0.024/0.054 = 0.4444
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