A bag contains 3 red balls, 5 black balls.
Totally there are 8 balls.
∴ n(S) = 8
i) Possibility that the red ball drawn, n(E) = 3
∴ Probability, P(E) = \(\frac{n(E)}{n(S)} = \frac{3}{8}\)
(ii) Possibility that the 1 black ball drwn is n(F) = 5
∴ Probability, P(F) = \(\frac{n(F)}{n(S)} = \frac{5}{8}\)