Given probability of first examination P(I)=0.8
Probability of atleast (I or II)=$P(I \cup II)=0.95$
$\therefore P(I \cup II)=P(I)+P(II)-P(I \cap II)$
$0.95=0.8+0.7-P(I \cap II)$
$\therefore P(I \cap II)=1.5-0.95$
$\therefore$ Probability of passing both is 0.55
Hence (A) is the correct answer.