# 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 =  0.002 %
p2 =  0.097 %

#### Solution:

$C_{{ 5}}(6) = \dbinom{ 6}{ 5} = \dfrac{ 6! }{ 5!(6-5)!} = \dfrac{ 6 } { 1 } = 6 \ \\ C_{{ 1}}(43) = \dbinom{ 43}{ 1} = \dfrac{ 43! }{ 1!(43-1)!} = \dfrac{ 43 } { 1 } = 43 \ \\ C_{{ 6}}(49) = \dbinom{ 49}{ 6} = \dfrac{ 49! }{ 6!(49-6)!} = \dfrac{ 49 \cdot 48 \cdot 47 \cdot 46 \cdot 45 \cdot 44 } { 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 } = 13983816 \ \\ n=49 \ \\ o=6 \ \\ s=n-o=49-6=43 \ \\ \ \\ p_{1}=100 \cdot \ \dfrac{ { { o } \choose 5 } \cdot \ { { s } \choose 1 } }{ { { n } \choose o } }=100 \cdot \ \dfrac{ 6 \cdot \ 43 }{ 13983816 }=0.002 \%$
$C_{{ 4}}(6) = \dbinom{ 6}{ 4} = \dfrac{ 6! }{ 4!(6-4)!} = \dfrac{ 6 \cdot 5 } { 2 \cdot 1 } = 15 \ \\ C_{{ 2}}(43) = \dbinom{ 43}{ 2} = \dfrac{ 43! }{ 2!(43-2)!} = \dfrac{ 43 \cdot 42 } { 2 \cdot 1 } = 903 \ \\ p_{2}=100 \cdot \ \dfrac{ { { o } \choose 4 } \cdot \ { { s } \choose 2 } }{ { { n } \choose o } }=100 \cdot \ \dfrac{ 15 \cdot \ 903 }{ 13983816 }=0.097 \%$

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