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February 28th, 2014, 08:16 PM  #1 
Newbie Joined: Nov 2013 Posts: 3 Thanks: 0  Probability Problem / Insurance problem
Not sure what to title this post. It's from basic probability but it's been about a year and a half. I'm studying for exam P and experienced this problem at the end of the section I'm reading and am having trouble figuring out precisely what to do. I know the answer is 0.5 but not clear on how to get there. I tried 2 ways and got 97/144 once and 23/36 another time. So here is the question. An insurer offers a health plan to the employees of a large company. As part of this plan, the individual employees may choose exactly two of the supplementary coverages A; B; and C; or they may choose no supplementary coverage. The proportions of the company's employees that choose coverages A; B; and C are 1/4, 1/3, and 5/12, respectively. Determine the probability that a randomly chosen employee will choose no supplementary coverage. Any insight will be appreciated. 
March 1st, 2014, 02:00 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,435 Thanks: 562  Re: Probability Problem / Insurance problem
1/4 + 1/3 + 5/12 = 1. Since the employees who chose a plan had to choose 2, divide by 2 to get fraction of employees who chose plans (1/2). The remaining 1/2 chose no plan.

March 1st, 2014, 03:50 PM  #3 
Newbie Joined: Nov 2013 Posts: 3 Thanks: 0  Re: Probability Problem / Insurance problem
Thanks so much. I had considered that but it seemed way too simple. I guess overthinking is gonna be a problem on exam P.


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