For the U.S. population in 2012 the probability that an adult between the ages o
ID: 3302068 • Letter: F
Question
For the U.S. population in 2012 the probability that an adult between the ages or 18 and 44 does not have health insurance coverage of any kind is 0.248. The corresponding probability for a person aged between 45 and 64 is 0.156. (a) Suppose that you randomly select a 29-year-old man and an unrelated 36-year-old woman from this population. What is the probability that both individuals are uninsured? (b) What is the probability that both adults of part (a) have health insurance coverage? (c) Now suppose that you randomly select a 53-year-old man and an unrelated 19-year-old woman. What is the probability that both individuals are uninsured? (d) What is the probability that both adults of part (c) have health insurance coverage? (e) If ten unrelated adults between the ages of 45 and 64 are chosen from the population, what is the probability that all ten are insured?Explanation / Answer
a) As both lies in the range of 18 years to 44 years old of age, probability that both individuals are uninsured is computed as = 0.248*0.248 = 0.061504
Therefore 0.061504 is the required probability here.
b) Probability that both insurers in a) have insurance is computed as:
= (1 - 0.248)*(1-0.248) = 0.7522 = 0.565504
Therefore 0.565504 is the required probability here.
c) Here both the adults are from different age buckets, therefore probability that both are uninsured here is computed as: = 0.248*0.156 = 0.038688
Therefore 0.038688 is the required probability here.
d) Probability that both adults of part c) have health insurance coverage is computed as:
= (1 - 0.248)*(1- 0.156)
= 0.634688
Therefore 0.634688 is the required probability here.
e) Probability that all 10 are insured is computed as:
= (1 - 0.156)*(1 - 0.156)*..........10 times
= ( 1- 0.156 )10
= 0.1834
Therefore 0.1834 is the required probability here.