A coin whose faces are marked 3 and 5 is tossed 4 times. The odds against the sum of the numbers thrown being less than 15 are

$E<15$

$\overline{E}\ge 15$

$\left(3355\right)\left(5553\right)\left(5555\right)$

$P\left(\overline{E}\right)=\frac{\frac{4!}{2!2!}+\frac{4!}{3!}+1}{16}$

            =$\frac{11}{16}$

$P\left(E\right)=\frac{5}{16}$

The odds against the sum of the numbers thrown being less than 15 are  11:5