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April 18, 2014

April 18, 2014

Posted by **Reshma** on Monday, February 1, 2010 at 8:31pm.

- probability -
**bobpursley**, Monday, February 1, 2010 at 8:44pmChoices 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.

- probability -
**Tyler**, Wednesday, July 14, 2010 at 1:25pmI'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 1-1/2=1/2. The probability is 1/2

- probability -
**Tyler**, Wednesday, July 14, 2010 at 1:25pmI'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 1-1/2=1/2. The probability is 1/2

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