Question 945915
  
Given
An unbalanced die such that
1) P(2)+P(3)+P(5) = {{{1/2}}}
2) P(5)+P(6)+P(4) = {{{1/2}}}
3) P(5) = {{{3/16}}}
Find 
A) P(2)+P(3)+P(1) = ?
B) P(2)+P(3)+P(4)+P(5)+P(6) = ?
  
Solution
Given P(5) = {{{3/16}}}, then
P(2)+P(3)+P(5) = P(2)+P(3)+ {{{3/16}}} = {{{1/2}}}
which means 
P(2)+P(3) = {{{1/2 - 3/16}}} = {{{5/16}}}
Similarly,
using
P(5)+P(6)+P(4) = {{{1/2}}}
we find
P(6)+P(4) = {{{1/2 - 3/16}}} = {{{5/16}}}

Since there are only six faces, we calculate P(1) as follows:
A)
P(1) = 1-( P(2)+P(3) + P(5) + P(6)P(6)+P(4) ) = {{{1-(5/16+3/16+5/16)}}} = {{{3/16}}}
  
B)
P(2)+P(3)+P(5)+P(6)P(6)+P(4) = 1-P(1) = {{{1-3/16}}} = {{{13/16}}}