2 ms·p(h)p(a|h)=p(h|a)p(a) therefore if p(a)=p(a|h) then p(h)=p(h|a) too.by d0mine 3y agop(h)p(a|h)=p(h|a)p(a) therefore if p(a)=p(a|h) then p(h)=p(h|a) too.