RRB NTPC Mains or Stage 2 Model Paper 5 with solutions

(b) There are 7 + 5 = 12 balls in the bag and the number of ways in which 4 balls can be drawn is ${}_{}^{12}{\mathit{C}}_4^{}$ and the number of ways of drawing 4 black balls (out of seven) is ${}_{}^7{\mathit{C}}_4^{}$.Hence, P (4 black balls)
$\frac{{}_{}^7{\mathit{C}}_4^{}}{{}_{}^{12}{\mathit{C}}_4^{}}=\frac{7.6.5.4}{1.2.3.4}\times \frac{1.2.3.4}{12.11.10.9}=\frac{7}{99}$
Thus the odds against the event ‘all blackballs’ are
$\left(1-\frac{7}{99}\right):\frac{7}{99}\mathit{i},\mathit{e}.,\frac{92}{99}:\frac{7}{99}\mathit{o}\mathit{r}92:7$