Nine balls one drawn simultaneously from a bad containing 5 white and 7 black balls. The probability of…

Out of $12$ balls we have to draw $9$ balls.

Therefore, selecting $9$ balls from $12$ balls can be done in $\mathrm{n}\left(\mathrm{S}\right)={}^{12}\mathrm{C}_9 \text{ways}$

$=\frac{12!}{9!\times 3!}=220$

Now, number of ways for selecting $3$ white balls will be $={}^{5}C_3=\frac{5!}{3!\times 2!}=10$

Number of ways for selecting $6$ black balls will be $={}^{7}C_6=\frac{7!}{6!\times 1!}=7$

Therefore, total number of ways for selecting $3$ white and $6$ black balls$\mathrm{n}\left(\mathrm{P}\right)=10\times 7=70$

Therefore, probability of getting $3$ white and $6$ black balls$=\frac{\mathrm{n}\left(\mathrm{P}\right)}{\mathrm{n}\left(\mathrm{S}\right)}=\frac{70}{220}=\frac{7}{22}$