Suppose that Box-I contains 8 red, 3 blue and 5 green balls, Box-ll contains 24 red, 9 blue and 15 green…

$P\left(\text{Red from Box} \mathrm{I}\right)=\frac{8}{16}=\frac{1}{2}$

$P\left(\text{Blue from Box} \mathrm{I}\right)=\frac{3}{16}$

$P\left(\text{Green from Box} \mathrm{I}\right)=\frac{10}{32}=\frac{5}{16}$

Now,

$P\left(\text{Green from Box} \mathrm{II}\right)=\frac{15}{48}$

$P\left(\text{Green from Box} \mathrm{III}\right)=\frac{12}{16}=\frac{3}{4}$

$P\left(\text{White from Box} \mathrm{IV}\right)=\frac{6}{32}=\frac{3}{16}$

Event $A$ : One of the chosen ball is white
$B:$ At least one of the chosen ball is green

$P\left(\frac{A}{B}\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}$

$=\frac{\frac{5}{16}\times \frac{3}{16}}{\frac{1}{2}\times \frac{15}{48}+\frac{3}{16}\times \frac{3}{4}+\frac{5}{16}\times 1}$

$=\frac{5}{52}$