In a survey of 1,000 elders, 57% of them have been hospitalized before. Of those who have been hositalized before, 54% have insurance. Of th

Question

In a survey of 1,000 elders, 57% of them have been hospitalized before. Of those who have been hositalized before, 54% have insurance. Of those never hospitalized before, 33% do not have insurance. What is the probability that an elder selected at random has insurance?

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Ayla 1 week 2021-09-15T20:53:21+00:00 1 Answer 0

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    2021-09-15T20:55:10+00:00

    Answer:

    0.5959 or 59%

    Step-by-step explanation:

    The number of total elders who have insurance is the sum of the elders who have been hospitalized and have insurance and the elders who have not been hospitalized and have insurance.

    Hospitalized, have insurance:

    P(H\cap I)=0.57*0.54=0.3078

    Not hospitalized, have insurance:

    P(NH\cap I)=(1-0.57)*(1-0.33)=0.2881

    Therefore, the probability that an elder selected at random has insurance is:

    P(I)=P(H\cap I)+P(NH\cap I)\\P(I)= 0.3078+0.2881\\P(I) = 0.5959\ or\ 59\%

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