events A and B are such that
p(A) = 1/2
p(A'|b) = 1/3
p(A$\displaystyle \cup$ B) = 3/5
determine P(B|A').
totally have no idea..
how can i define p(A$\displaystyle \cup$B) = 3/5 btw?
I'd suggest you draw a Karnaugh table. From the given data:
$\displaystyle \begin{tabular}{l | c | c | c} & A & A$\, '$ & \\ \hline B & a & b & \\ \hline B$\, '$ & c & (1/2) - b & \\ \hline & 1/2 & 1/2 & 1 \\ \end{tabular}
$
where:
$\displaystyle a + b + c = \frac{3}{5}$ .... (1)
$\displaystyle \frac{b}{a+b} = \frac{1}{3} \Rightarrow a = 2b$ .... (2)
Substitute (2) into (1): $\displaystyle c = \frac{3}{5} - 3b$.
Update the Karnaugh table:
$\displaystyle \begin{tabular}{l | c | c | c} & A & A$\, '$ & \\ \hline B & 2b & b & 3b \\ \hline B$\, '$ & (3/5) - 3b & (1/2) - b & (11/10) - 4b \\ \hline & 1/2 & 1/2 & 1 \\ \end{tabular}
$
Therefore $\displaystyle 3b + \frac{11}{10} - 4b = 1 \Rightarrow b = \frac{1}{10}$.
Update the Karnaugh table:
$\displaystyle \begin{tabular}{l | c | c | c} & A & A$\, '$ & \\ \hline B & 2/10 & 1/10 & 3/10 \\ \hline B$\, '$ & 3/10 & 4/10 & 7/10 \\ \hline & 1/2 & 1/2 & 1 \\ \end{tabular}
$
Now it's easy to get $\displaystyle \Pr(B | A') = \frac{\Pr(B \cap A')}{\Pr(A')} = \frac{1/10}{1/2} = \frac{1}{5}$.