Ball selection probability

There are five whites and nine blacks in the destiny. We will choose three balls at random. What is the probability that
a) the selected balls will not be the same color,
b) will there be at least two blacks between them?

Final Answer:

p₁ =  0.7418
p₂ =  0.7253

Step-by-step explanation:

$$\begin{gathered} C_3(14)=\left(\genfrac{}{}{0ex}{}{14}{3}\right)=\frac{14!}{3!(14-3)!}=\frac{14\cdot 13\cdot 12}{3\cdot 2\cdot 1}=364 \\ {C}_2(5)=\left(\genfrac{}{}{0ex}{}{5}{2}\right)=\frac{5!}{2!(5-2)!}=\frac{5\cdot 4}{2\cdot 1}=10 \\ {C}_1(9)=\left(\genfrac{}{}{0ex}{}{9}{1}\right)=\frac{9!}{1!(9-1)!}=\frac{9}{1}=9 \\ {C}_1(5)=\left(\genfrac{}{}{0ex}{}{5}{1}\right)=\frac{5!}{1!(5-1)!}=\frac{5}{1}=5 \\ {C}_2(9)=\left(\genfrac{}{}{0ex}{}{9}{2}\right)=\frac{9!}{2!(9-2)!}=\frac{9\cdot 8}{2\cdot 1}=36 \\ {p}_1=\frac{\left(\genfrac{}{}{0ex}{}{5}{2}\right)\cdot \left(\genfrac{}{}{0ex}{}{9}{1}\right)+\left(\genfrac{}{}{0ex}{}{5}{1}\right)\cdot \left(\genfrac{}{}{0ex}{}{9}{2}\right)}{\left(\genfrac{}{}{0ex}{}{5+9}{3}\right)}=\frac{10\cdot 9+5\cdot 36}{\left(\genfrac{}{}{0ex}{}{5+9}{3}\right)}=0.7418 \end{gathered}$$

$$\begin{gathered} C_0(5)=\left(\genfrac{}{}{0ex}{}{5}{0}\right)=\frac{5!}{0!(5-0)!}=\frac{1}{1}=1 \\ {C}_3(9)=\left(\genfrac{}{}{0ex}{}{9}{3}\right)=\frac{9!}{3!(9-3)!}=\frac{9\cdot 8\cdot 7}{3\cdot 2\cdot 1}=84 \\ {p}_2=\frac{\left(\genfrac{}{}{0ex}{}{5}{0}\right)\cdot \left(\genfrac{}{}{0ex}{}{9}{3}\right)+\left(\genfrac{}{}{0ex}{}{5}{1}\right)\cdot \left(\genfrac{}{}{0ex}{}{9}{2}\right)}{\left(\genfrac{}{}{0ex}{}{5+9}{3}\right)}=\frac{1\cdot 84+5\cdot 36}{\left(\genfrac{}{}{0ex}{}{5+9}{3}\right)}=0.7253 \end{gathered}$$

Try another example