# Cards

The player gets 8 cards of 32. What is the probability that it gets
a) all 4 aces
b) at least 1 ace

Result

a =  0.195 %
b =  70.451 %

#### Solution:

$C_{{ 4}}(4)=\dbinom{ 4}{ 4}=\dfrac{ 4! }{ 4!(4-4)!}=\dfrac{ 1 } { 1 }=1 \ \\ \ \\ C_{{ 4}}(28)=\dbinom{ 28}{ 4}=\dfrac{ 28! }{ 4!(28-4)!}=\dfrac{ 28 \cdot 27 \cdot 26 \cdot 25 } { 4 \cdot 3 \cdot 2 \cdot 1 }=20475 \ \\ \ \\ C_{{ 8}}(32)=\dbinom{ 32}{ 8}=\dfrac{ 32! }{ 8!(32-8)!}=\dfrac{ 32 \cdot 31 \cdot 30 \cdot 29 \cdot 28 \cdot 27 \cdot 26 \cdot 25 } { 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 }=10518300 \ \\ \ \\ a=100 \cdot \ { { 4 } \choose 4 } \cdot \ \dfrac{ { { 28 } \choose 4 } }{ { { 32 } \choose 8 } }=0.195 \%$
$C_{{ 8}}(28) = \dbinom{ 28}{ 8} = \dfrac{ 28! }{ 8!(28-8)!} = \dfrac{ 28 \cdot 27 \cdot 26 \cdot 25 \cdot 24 \cdot 23 \cdot 22 \cdot 21 } { 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 } = 3108105 \ \\ b=100 \cdot \ (1- \dfrac{ { { 32-4 } \choose 8 } }{ { { 32 } \choose 8 } } )=70.451 \%$

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