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?

Answer 1

0.5959 or 59%

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\%`