Probability 17023

In the draw pole, five numbers out of 35 are drawn. The third prize is paid for three guessed numbers. What is the probability of winning the third prize if we submit a ticket with one of five numbers?

Correct answer:

p =  0.0134

Step-by-step explanation:

$$\begin{gathered} C_2(30)=\left(\genfrac{}{}{0ex}{}{30}{2}\right)=\frac{30!}{2!(30-2)!}=\frac{30\cdot 29}{2\cdot 1}=435 \\ {C}_3(5)=\left(\genfrac{}{}{0ex}{}{5}{3}\right)=\frac{5!}{3!(5-3)!}=\frac{5\cdot 4}{2\cdot 1}=10 \\ {C}_5(35)=\left(\genfrac{}{}{0ex}{}{35}{5}\right)=\frac{35!}{5!(35-5)!}=\frac{35\cdot 34\cdot 33\cdot 32\cdot 31}{5\cdot 4\cdot 3\cdot 2\cdot 1}=324632 \\ p=\frac{\left(\genfrac{}{}{0ex}{}{5}{3}\right)\cdot \left(\genfrac{}{}{0ex}{}{35-5}{2}\right)}{\left(\genfrac{}{}{0ex}{}{35}{5}\right)}=\frac{10\cdot \left(\genfrac{}{}{0ex}{}{35-5}{2}\right)}{324632}=0.0134 \end{gathered}$$

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