# Balls

The urn is 8 white and 6 black balls. We pull 4 randomly balls. What is the probability that among them will be two white?

Result

p =  0.42

#### Solution:

$C_{{ 4}}(14) = \dbinom{ 14}{ 4} = \dfrac{ 14! }{ 4!(14-4)!} = \dfrac{ 14 \cdot 13 \cdot 12 \cdot 11 } { 4 \cdot 3 \cdot 2 \cdot 1 } = 1001 \ \\ C_{{ 2}}(8)=\dbinom{ 8}{ 2}=\dfrac{ 8! }{ 2!(8-2)!}=\dfrac{ 8 \cdot 7 } { 2 \cdot 1 }=28 \ \\ \ \\ C_{{ 2}}(6)=\dbinom{ 6}{ 2}=\dfrac{ 6! }{ 2!(6-2)!}=\dfrac{ 6 \cdot 5 } { 2 \cdot 1 }=15 \ \\ \ \\ p=\dfrac{ { { 8 } \choose 2 } \cdot \ { { 6 } \choose 2 } }{ { { 8+6 } \choose 4 } }=\dfrac{ 28 \cdot \ 15 }{ { { 8+6 } \choose 4 } } \doteq \dfrac{ 60 }{ 143 } \doteq 0.4196 \doteq 0.42$

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