A die is thrown once. If E is the event that ‘the number appearing is a multiple of 3’ and F be the event ‘the number appearing is even’, then the incorrect option is

E and F are dependent events

S = {1, 2, 3, 4, 5, 6}

E = {3, 6}, F = {2, 4, 6}, E ∩ F = {6}

$∴ P(E) =\frac{2}{6}=\frac{1}{3}$,  $P(F) =\frac{3}{6}=\frac{1}{2}$, $P(E ∩ F)=\frac{1}{6}$

Since P(E and F) = $\frac{1}{6}=\frac{1}{3}×\frac{1}{2}=P(E) ×P(F)$

∴ the events E and F are independent variable or events. So option no. 4 is incorrect.