Resultado inciso (a): La probabilidad de obtener un número par es de 0.5.

Solución inciso (b)

E= {1}
F= {1, 2, 3, 4, 5, 6}
P{E}= \(\frac{1}{6}\)
P{F}= \(\frac{1}{6}\)+\(\frac{1}{6}\)+\(\frac{1}{6}\)+\(\frac{1}{6}\)+\(\frac{1}{6}\)+\(\frac{1}{6}\)= \(\frac{6}{6}\)= 1
\(\frac{P\left\{E\right\}}{P\left\{F\right\}}\)= \(\frac{\frac{1}{6}}{\frac{6}{6}}\)= \(\frac{1}{6}\)= 0.16666