Use the following information to answer the question. Suppose that the probability that a person books a hotel using an online travel websit

Question

Use the following information to answer the question. Suppose that the probability that a person books a hotel using an online travel website is 0.68. For the questions that follow, consider a sample of fifteen randomly selected people who recently booked a hotel. What is the probability that no more than four out of fifteen people used an online travel website when they booked their hotel

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Amity 5 years 2021-07-20T19:26:25+00:00 1 Answers 38 views 0

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  1. Edana
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    2021-07-20T19:28:13+00:00
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    Answer:

    0.0013 = 0.13% probability that no more than four out of fifteen people used an online travel website when they booked their hotel

    Step-by-step explanation:

    For each person, there are only two possible outcomes. Either they booked the hotel using an online travel website, or they did not. Each person is independent of each other. So we use the binomial probability distribution to solve this question.

    Binomial probability distribution

    The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

    P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

    In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

    C_{n,x} = \frac{n!}{x!(n-x)!}

    And p is the probability of X happening.

    Suppose that the probability that a person books a hotel using an online travel website is 0.68.

    This means that p = 0.68

    Sample of fifteen:

    This means that n = 15

    What is the probability that no more than four out of fifteen people used an online travel website when they booked their hotel

    P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

    So

    P(X = 0) = C_{15,0}.(0.68)^{0}.(0.32)^{15} = 0

    P(X = 1) = C_{15,1}.(0.68)^{1}.(0.32)^{14} = 0

    P(X = 2) = C_{15,2}.(0.68)^{2}.(0.32)^{13} = 0

    P(X = 3) = C_{15,3}.(0.68)^{3}.(0.32)^{12} = 0.0002

    P(X = 4) = C_{15,4}.(0.68)^{4}.(0.32)^{11} = 0.0011

    P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0 + 0 + 0 + 0.0002 + 0.0011 = 0.0013

    0.0013 = 0.13% probability that no more than four out of fifteen people used an online travel website when they booked their hotel

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