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10 June, 19:09

In one town, 67% of adults have health insurance. What is the probability the 5 adults selected at random from the town all do not have health insurance?

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Answers (2)
  1. 10 June, 19:29
    0
    Answer: P = 0.0039135

    To percentage

    P = 0.0039135 * 100%

    P = 0.39%

    Therefore the probability that the 5 adults selected at random from the town all do not have health insurance is

    P = 0.0039 or 0.39%

    Step-by-step explanation:

    Given;

    Percentage of adult with health insurance = 67%

    Percentage of adult without health insurance = 100% - 67% = 33%

    The probability that 5 adults selected at random don't have health insurance is P.

    For each selection the chances of selecting an adult without health insurance is f' = 0.33

    f' = fraction of adults without health insurance.

    For the five selection, the probability of selecting adults without health insurance five times is

    P = (f') ^5 = (0.33) ^5

    P = 0.0039135

    To percentage

    P = 0.0039135 * 100%

    P = 0.39%

    Therefore the probability that the 5 adults selected at random from the town all do not have health insurance is

    P = 0.0039 or 0.39%
  2. 10 June, 19:31
    0
    0.00391

    Step-by-step explanation:

    67% of adults have health insurance.

    Adults that do not have health insurance = 100 - 67 = 33% = 0.33

    Number of adults selected at random = 5

    Probability that 5 adults selected at random all do not have insurance = 0.33^5

    = 0.00391
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