Conditional Probability with Anti Set Ethnic Group    O    A    B    AB
1             .082 .106 .008 .004
2             .135 .141 .018 .006
3             .215 .200 .065 .020


A given population is in one of the four above blood types and one of the three above ethnicities.
Above is given the "joint" probabilities.
Have to calculate probability of a randomly selected individual being from ethnicity 1 given  that their blood type is NOT B.

My Solution:
$P(1|B') = \dfrac{P(1 \cap (O \cup A \cup AB))}{P(O \cup A \cup AB)}$

But this yielded the wrong answer ... Any idea how to approach this?
Question on your answer is given by the Bayes' theorem. I will try to write a complete solution.

Let $B$ denote that given person has a blood type B and denote $C_{i}$ that given person belongs to the i-th ethnicity group. I assume that $P(C_{1}) = P(C_{2}) = P(C_{3}) = \frac{1}{3}$ (which should have been definitely given in the assignment).

According to the Bayes' theorem we have 

$P(C_{1}|B) = \frac{P(B|C_{1}) \cdot P(C_{1})}{\sum^{3}_{i=1} P(B|C_{i}) \cdot P(C_{i})}$.

All the terms on the right hand side of the previous equations are given by your table, so you can easily compute the desired probability $P(C_{1}|B)$.
Question on your answer is given by the Bayes' theorem. I will try to write a complete solution.

Let $B$ denote that given person has a blood type B and denote $C_{i}$ that given person belongs to the i-th ethnicity group. I assume that $P(C_{1}) = P(C_{2}) = P(C_{3}) = \frac{1}{3}$ (which should have been definitely given in the assignment).

According to the Bayes' theorem we have 

$P(C_{1}|B) = \frac{P(B|C_{1}) \cdot P(C_{1})}{\sum^{3}_{i=1} P(B|C_{i}) \cdot P(C_{i})}$.

All the terms on the right hand side of the previous equations are given by your table, so you can easily compute the desired probability $P(C_{1}|B)$.