I have 2 problems I am having issues with. 1. At a local hospital, 73% of the ER

Question

I have 2 problems I am having issues with.

1. At a local hospital, 73% of the ER patients have some form of insurance. Furthermore, 80% of the patients have a hospital bill of at least $1500. Finally, 70% of those with a bill in excess of $1500 have some form of insurance. What is the probability that a randomly chosen patient has a bill over than $1500 and does not have insurance? (please round your answer to 2 decimal places)

2. At a local hospital, 73% of the ER patients have some form of insurance. Furthermore, 80% of the patients have a hospital bill of at least $1500. Finally, 70% of those with a bill in excess of $1500 have some form of insurance. Of the ER patients with insurance, what is the probability that a randomly chosen patient has a bill less than $1500? (please round your answer to 2 decimal places)

Explanation / Answer

1)P(having insurance)=0.73

P(hospital bill>1500)=0.8

P(have insurance|hospital bill>1500)=0.7 gives P(no insurance|hospital bill>1500)=0.3

P(hospital bill>1500 & no insrance)=P(no insurance|hospital bill>1500)*P(hospital bill>1500)=0.3*0.8=0.24

2)P(have insurance|hospital bill>1500)*P(hospital bill>1500)=P(hospitalbill>1500|have insurance)*P(have insurance)

Hence 0.7*0.8=P(hospital bill>1500|have insurance)*0.73

gives P(hospital bill>1500|have insurance)=56/73

Hence P(hospital bill<1500|have insurance)=1-56/73=17/73=0.23

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