SOLUTION: In a survey of 1,000 elders, 56% of them have been hospitalized before. Of those who have been hositalized before, 54% have insurance. Of those never hospitalized before, 32% do n

Question 604777: In a survey of 1,000 elders, 56% of them have been hospitalized before. Of those who have been
hositalized before, 54% have insurance. Of those never hospitalized before, 32% do not have
insurance. What is the probability that an elder selected at random has insurance?
Answer by alicealc(293) (Show Source): You can put this solution on YOUR website!
Elders who have been hospitalized = 56%
Elders who have never been hospitalized = 100% - 56% = 44%

Elders who have been hospitalized and have insurance = 54% * 56% = 54/100 * 56% = 30.24%

Elders who have never been hospitalized and do not have insurance = 32% * 44% = 32/100 * 44% = 14.08%
Elders who have never been hospitalized and have insurance = 44% - 14.08% = 29.92%

so, the probability that an elder has insurance = 30.24% + 29.92% = 60.16%

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