A box contains identical balls of which 12 are red, 16 white and 8 blue.

Explanation

n(R) = number of red balls = 12;

n(W) = number of white balls = 16;

n(B) = number of blue balls = 8

Total number of balls = 36

(a) P(three are red) = \(\frac{12}{36} \times \frac{11}{35} \times \frac{10}{34}\)

= \(\frac{11}{357}\)

(b) P(first is blue and other two red) = \(\frac{8}{36} \times \frac{12}{35} \times \frac{11}{34}\)

= \(\frac{44}{1785}\)

(c) P(two are white and one is blue) = \(\frac{16}{36} \times \frac{15}{35} \times \frac{8}{34}\)

= \(\frac{16}{357}\)