Hazard game

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

a) second prize (we guess five numbers correctly)
b) the third prize (we guess four numbers correctly)?

Final Answer:

p₁ =  0.0018 %
p₂ =  0.0969 %

Step-by-step explanation:

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

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

Try another example