Three colors

Find the probability that 3 balls of the same color will be drawn from fate with 10 white, 10 red, and 10 blue balls.

Correct answer:

p =  8.867 %

Step-by-step explanation:

$$\begin{gathered} C_3(30)=\left({\displaystyle \genfrac{}{}{0px}{}{30}{3}}\right)={\displaystyle \frac{30!}{3!(30-3)!}}={\displaystyle \frac{30\cdot 29\cdot 28}{3\cdot 2\cdot 1}}=4060 \\ n=10+10+10=30 \\ {p}_1={\displaystyle \frac{n}{n}}\cdot {\displaystyle \frac{9}{n-1}}\cdot {\displaystyle \frac{8}{n-2}}={\displaystyle \frac{30}{30}}\cdot {\displaystyle \frac{9}{30-1}}\cdot {\displaystyle \frac{8}{30-2}}={\displaystyle \frac{18}{203}}\doteq 0.0887 \\ p=100\cdot p_1=100\cdot 0.0887={\displaystyle \frac{1800}{203}}\doteq 8.867\mathrm{\%} \\ \text{ Verifying Solution: } \\ {C}_3(10)=\left({\displaystyle \genfrac{}{}{0px}{}{10}{3}}\right)={\displaystyle \frac{10!}{3!(10-3)!}}={\displaystyle \frac{10\cdot 9\cdot 8}{3\cdot 2\cdot 1}}=120 \\ {p}_2=3\cdot \left(\genfrac{}{}{0px}{}{10}{3}\right)\mathrm{/}\left(\genfrac{}{}{0px}{}{n}{3}\right)=3\cdot 120\mathrm{/}4060={\displaystyle \frac{18}{203}}\doteq 0.0887 \\ {p}_2=p_1 \end{gathered}$$

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