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7 September, 07:13

The probability that a randomly chosen sales prospect will make a purchase is 20%. What is the probability (to three decimal places) that the salesperson will make four or more sales if six sales calls are made on a given day

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  1. 7 September, 07:56
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    1.7%

    Step-by-step explanation:

    We have to calculate the probability that the salesperson will make four or more sales if six sales calls are made on a given day, that is:

    P (x = > 4)

    Therefore, we must calculate when x = 4, when x = 5, and when x = 6 and add. p = 0.2, n = 6

    P (x = r) = nCr * p ^ r * (1 - p) ^ (n-r)

    Also, nCr = n! / (r! * (n-r) !, now replacing:

    P (x = 4) = 6! / (4! * (6-4) ! * 0.20 ^ 4 * 0.80 ^ (6-4)

    P (x = 4) = 15 * 0.001024 = 0.01536

    P (x = 5) = 6! / (5! * (6-5) ! * 0.20 ^ 5 * 0.80 ^ (6-5)

    P (x = 5) = 6 * 0.000256 = 0.001536

    P (x = 6) = 6! / (6! * (6-6) ! * 0.20 ^ 6 * 0.80 ^ (6-6)

    P (x = 6) = 1 * 0.000064 = 0.000064

    now,

    P (x = > 4) = P (x = 4) + P (x = 5) + P (x = 6)

    P (x = > 4) = 0.01536) + 0.001536 + 0.000064

    P (x = > 4) = 0.01696 = 0.017

    It means that the probability is 1.7%
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