Hazard game

In the Sportka hazard game, 6 numbers out of 49 are drawn. What is the probability that we will win:

a) second prize (we guess 5 numbers correctly)
b) the third prize (we guess 4 numbers correctly)?

Correct result:

p₁ =  0.0018 %
p₂ =  0.0969 %

Solution:

$$\begin{gathered} C_5(6)=\left({\displaystyle \genfrac{}{}{0px}{}{6}{5}}\right)={\displaystyle \frac{6!}{5!(6-5)!}}={\displaystyle \frac{6}{1}}=6 \\ {C}_1(43)=\left({\displaystyle \genfrac{}{}{0px}{}{43}{1}}\right)={\displaystyle \frac{43!}{1!(43-1)!}}={\displaystyle \frac{43}{1}}=43 \\ {C}_6(49)=\left({\displaystyle \genfrac{}{}{0px}{}{49}{6}}\right)={\displaystyle \frac{49!}{6!(49-6)!}}={\displaystyle \frac{49\cdot 48\cdot 47\cdot 46\cdot 45\cdot 44}{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}}=13983816 \\ n=49 \\ o=6 \\ s=n-o=49-6=43 \\ {p}_1=100\cdot {\displaystyle \frac{\left(\genfrac{}{}{0px}{}{o}{5}\right)\cdot \left(\genfrac{}{}{0px}{}{s}{1}\right)}{\left(\genfrac{}{}{0px}{}{n}{o}\right)}}=100\cdot {\displaystyle \frac{6\cdot 43}{13983816}}=0.0018\mathrm{\%} \end{gathered}$$

$$\begin{gathered} C_4(6)=\left({\displaystyle \genfrac{}{}{0px}{}{6}{4}}\right)={\displaystyle \frac{6!}{4!(6-4)!}}={\displaystyle \frac{6\cdot 5}{2\cdot 1}}=15 \\ {C}_2(43)=\left({\displaystyle \genfrac{}{}{0px}{}{43}{2}}\right)={\displaystyle \frac{43!}{2!(43-2)!}}={\displaystyle \frac{43\cdot 42}{2\cdot 1}}=903 \\ {p}_2=100\cdot {\displaystyle \frac{\left(\genfrac{}{}{0px}{}{o}{4}\right)\cdot \left(\genfrac{}{}{0px}{}{s}{2}\right)}{\left(\genfrac{}{}{0px}{}{n}{o}\right)}}=100\cdot {\displaystyle \frac{15\cdot 903}{13983816}}=0.0969\mathrm{\%} \end{gathered}$$

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