Question

Sejam A e B eventos de um mesmo espaço amostral, com p(A∪B)=0,64. Calcule p(A), admitindo que P(B)= 24% e P(A∩B)= 8% *

Answer 1

$\boxed{P(A\cap B)=P(A)+P(B)-P(A\cup B)}$

$P(A\cup B)=0.64\\ P(B)=0.24\\ P(A\cap B)=0.08\\\\ 0.08=P(A)+0.24-0.64\ \therefore\ 0.08=P(A)-0.4\ \therefore\\\\ P(A)=0.08+0.4=0.48\ \therefore\ \boxed{P(A)=0.48=48\%}$