here P(B)=P(A)*P(B|A)+P(Ac)*P(B|Ac)=0.5*(1-0.01)+0.5*(0.01)=0.5
hence P(Bc)=1-P(B)=1-0.5=0.5
hence
P(A|B)=P(A)*P(B|A)/P(B)=0.5*(1-0.01)/0.5 =0.99
P(Ac|B)=P(Ac)*P(B|Ac)/P(B)=0.5*(0.01)/0.5=0.01
P(A|Bc)=P(A)*P(Bc|A)/P(Bc)=0.5*0.01/0.5=0.01
P(Ac|Bc)=P(Ac)*P(Bc|Ac)/P(Bc)=0.5*0.99/0.5=0.99
16. In a noisy communication system transmitting binary digits (zeros and ones) the sender may send...