Question 258838
P(both fail)={{{(0.2)(0.2)=0.04}}}
P(neither fail)={{{(0.8)(0.8)=0.64}}}
P(one fail)=P(first fails,second doesn't)+P(first doesn't, second fails)={{{(0.2)(0.8)+(0.8)(0.2)=0.32}}}
Ptotal={{{0.04+0.64+0.32=1}}} (all possible events are covered)