Question 394621
if the viruses are truly independent then 
probability that the system gets damaged=
P(AUBUC)=P(A)+P(B)+P(C)-P(AnB)-P(anc)-P(bnc)+P(AnBnC)=
0.4+0.5+0.2-0.4*0.5-0.4*0.2-0.5*0.2+0.4*0.5*0.2=1.1-0.2-0.08-0.1+0.04
=0.76