By the definition of conditional probability, you have
\(\mathbb P(B|A)=\dfrac{\mathbb P(B\cap A)}{\mathbb P(A)}\implies\mathbb P(A\cap B)=0.2\times0.5=0.1\)
\(\mathbb P(A|B)=\dfrac{\mathbb P(A\cap B)}{\mathbb P(B)}\implies\mathbb P(B)=\dfrac{0.1}{0.25}=1\)
\(\mathbb P(
eg B|
eg A)=\dfrac{\mathbb P(
eg B\cap
eg A)}{\mathbb P(
eg A)}\)
Recall that
\(\mathbb P(
eg B\cap
eg A)=\mathbb P(
eg(B\cup A))=1-\mathbb P(B\cup A)\)
and that
\(\mathbb P(B\cup A)=\mathbb P(B)+\mathbb P(A)-\mathbb P(B\cap A)\)
This means
\(\mathbb P(
eg B|
eg A)=\dfrac{1-\mathbb P(B)-\mathbb P(A)+\mathbb P(B\cap A)}{1-\mathbb P(A)}=\dfrac{1-0.4-0.5+0.1}{1-0.5}=0.1\)
