A telephone call center where people place marketing calls to customers has a probability of success of 0.12 . Find the number of

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A telephone call center where people place marketing calls to customers has a probability of success of 0.12 . Find the number of calls needed to ensure that there is a probability of at least of obtaining or more successful calls.

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  1. According to given,
    p = 0.12 (success)
    q = 1 – 0.12
      = 0.88 (failure)
    x = 7
    By applying binomial probability distribution we get –
    P(x) = nCx.p^x.q^(n-x)
    ⇒ P(x=7) = nC7.(0.12)^7.(0.88)^(n-7)
    Also given,
    Probability of 7 successful trials = 0.77
    Thus,
    nC7.(0.12)^7.(0.88)^(n-7) = 0.77
    ⇒ n!/(7! (n-7)!).(0.12)^7.(0.88)^(n-7) = 0.77
    ⇒ n!/(7! (n-7)!).(0.88)^n = 0.77 * (0.88/0.12)^7
    ⇒ n!/(7! (n-7)!).(0.88)^n = 878214.711
    ⇒ n = 130
    Thus, the number of calls needed to ensure that there is a probability of at least 0.77 of obtaining 7 or more successful calls is 130.
    Learn more about binomial probability distribution : https://brainly.com/question/24756209
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    Question : A telephone call centre where people place marketing calls to customers has a probability of success of 0.12. Find the number of calls needed to ensure that there is a probability of at least 0.77 of obtaining 7 or more successful calls.

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