Total number of possible outcomes, n(S) = 5 + 8 + 4 = 17 

(i) Number of favorable outcomes, 

n(E) = 5

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{5}{17}\)

(ii) Number of favorable outcomes, 

n(E) = 8

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{8}{17}\)

(iii) Number of events of drawing a green marble = 4

 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{4}{17}\)

Probability of not drawing green marble = 1 – P(E)

= 1 – \(\frac{4}{17}\) = \(\frac{13}{17}\)

A box contains 5 red marbles, 8 white marbles and 4 green marbles.

∴ Total number of marbles, n(S) 

= 5 + 8 + 4 = 17 

(i) Probability that the 1 red marble drawn is n(A) = 5 

∴ Probability, P(A) = \(\frac{n(A)}{n(S)} = \frac{5}{17}\)

(ii) Possibility that 1 white marble drawn, n(B) = 8 

∴ Probability, P(B) = \(\frac{n(B)}{n(S)} = \frac{8}{17}\)

(iii) Possibility that 1 not green marble ? 

P(C) = 17 – 4 = 13 (∵ Except 4 green marbles) 

∴ Probability, P(C) = \(\frac{n(c)}{n(S)} = \frac{13}{17}\)