Balls

The urn is 8 white and 6 black balls. We pull 4 randomly balls. What is the probability that among them will be two white?

Correct result:

p =  0.4196

Solution:

$$\begin{gathered} C_4(14)=\left({\displaystyle \genfrac{}{}{0px}{}{14}{4}}\right)={\displaystyle \frac{14!}{4!(14-4)!}}={\displaystyle \frac{14\cdot 13\cdot 12\cdot 11}{4\cdot 3\cdot 2\cdot 1}}=1001 \\ {C}_2(8)=\left({\displaystyle \genfrac{}{}{0px}{}{8}{2}}\right)={\displaystyle \frac{8!}{2!(8-2)!}}={\displaystyle \frac{8\cdot 7}{2\cdot 1}}=28 \\ {C}_2(6)=\left({\displaystyle \genfrac{}{}{0px}{}{6}{2}}\right)={\displaystyle \frac{6!}{2!(6-2)!}}={\displaystyle \frac{6\cdot 5}{2\cdot 1}}=15 \\ p={\displaystyle \frac{\left(\genfrac{}{}{0px}{}{8}{2}\right)\cdot \left(\genfrac{}{}{0px}{}{6}{2}\right)}{\left(\genfrac{}{}{0px}{}{8+6}{4}\right)}}={\displaystyle \frac{28\cdot 15}{\left(\genfrac{}{}{0px}{}{8+6}{4}\right)}}={\displaystyle \frac{60}{143}}=0.4196 \end{gathered}$$

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