Question

A box contains $2$ white balls, $3$ black balls & $4$ red balls.In how many ways can $3$ balls be drawn from the box if atleast $1$ black is to be included in the draw?

Answer 1

Solution:-

Probability of not getting a black ball $=\frac{{{}^{6}C}_3}{{{}^{9}C}_3}$

${{}^{6}C}_3=\frac{6!}{3!\times (6-3)!}=\frac{6\times 5\times 4\times 3!}{3\times 2\times 1\times (3!)}=20$

${{}^{9}C}_3=\frac{9!}{3!\times (9-3)!}=\frac{9\times 8\times 7\times 6!}{3\times 2\times 1\times (6!)}=84$

$\therefore$ Probability of not getting a black ball $=\frac{{{}^{6}C}_3}{{{}^{9}C}_3}=\frac{20}{84}$

$\therefore$ Probability of getting atleast 1 black ball $=1-\frac{20}{84}=\frac{84-20}{84}=\frac{64}{84}=\frac{16}{21}$