Three reds

What is the probability that when choosing 3 carats from seven carats, all 3 reds will be red?

Correct answer:

p =  1.129 %

Step-by-step explanation:

$$\begin{gathered} C_3(8)=\left({\displaystyle \genfrac{}{}{0px}{}{8}{3}}\right)={\displaystyle \frac{8!}{3!(8-3)!}}={\displaystyle \frac{8\cdot 7\cdot 6}{3\cdot 2\cdot 1}}=56 \\ {C}_3(32)=\left({\displaystyle \genfrac{}{}{0px}{}{32}{3}}\right)={\displaystyle \frac{32!}{3!(32-3)!}}={\displaystyle \frac{32\cdot 31\cdot 30}{3\cdot 2\cdot 1}}=4960 \\ p=100\cdot \left(\genfrac{}{}{0px}{}{8}{3}\right)\mathrm{/}\left(\genfrac{}{}{0px}{}{32}{3}\right)=100\cdot 56\mathrm{/}4960={\displaystyle \frac{35}{31}}\doteq 1.129=1.129\mathrm{\%} \\ \text{ Verifying Solution: } \\ {p}_1=100\cdot {\displaystyle \frac{8\cdot 7\cdot 6}{32\cdot 31\cdot 30}}={\displaystyle \frac{35}{31}}\doteq 1.129\mathrm{\%} \end{gathered}$$

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