Explanation

(a)
A selection of 3 balls so as to include at least one black ball, can be made in the following 3 mutually exclusive ways
(i) The number of ways in which 1 black balls and 2 others are selected
$=^{3}C_{1} \times^{6}C_{2} = 3 \times 15 = 45$
(ii) The number of ways in which 2 black balls and 1 other are selected
$=^{3}C_{2} \times^{6}C_{1} = 3 \times 6 = 18$
(iii) The number of ways in which 3 black balls and no other are selected $=^{3}C_{3} = 1$
$\therefore$ Total numbers of ways $= 45 + 18 + 1 = 64$