probability
posted by Reshma .
An insurer offers a health plan to the employees of a large company. As part of this plan, the individual employees may choose two of the supplementary covaerges A, B and C or they may choose no supplementary coverage. The proportions of the company's employees that choose coverage 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.

Choices are AB, AC, BC, None
Pr(A and B)= Pr(A)Pr(B)= 1/4*1/3= 1/12
Pr(B and C)= Pr(B)Pr(C)= 1/3*5/12=5/36
Pr(A and C)= 1/4 * 5/12=5/48
Pr(None)=1 sum of the three other choices above. 
I've got a different answer from the Actuary exam guide. You need to make a Venn Diagram and make x,y, and z into the middle portions of the diagram between AB, AC, and BC. From the question you know that x+y=1/4, x+z=1/3 and y+z=5/12. Adding these equations gives
(x+y)+(x+z)+(y+z)=1/4+1/3+5/12
then, 2(x+y+z)=1/2
Since you're looking for 1(x+y+z) then 11/2=1/2. The probability is 1/2
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