One green

In the container are 45 white and 15 balls. We randomly select 5 balls. What is the probability that it will be a maximum one green?

Result

p =  0.633

Solution:

C_{{ 5}}(60) = \dbinom{ 60}{ 5} = \dfrac{ 60! }{ 5!(60-5)!} = \dfrac{ 60 \cdot 59 \cdot 58 \cdot 57 \cdot 56 } { 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 } = 5461512 \ \\ C_{{ 0}}(15) = \dbinom{ 15}{ 0} = \dfrac{ 15! }{ 0!(15-0)!} = \dfrac{ 1 } { 1 } = 1 \ \\ C_{{ 1}}(15) = \dbinom{ 15}{ 1} = \dfrac{ 15! }{ 1!(15-1)!} = \dfrac{ 15 } { 1 } = 15 \ \\ C_{{ 5}}(45) = \dbinom{ 45}{ 5} = \dfrac{ 45! }{ 5!(45-5)!} = \dfrac{ 45 \cdot 44 \cdot 43 \cdot 42 \cdot 41 } { 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 } = 1221759 \ \\ C_{{ 4}}(45) = \dbinom{ 45}{ 4} = \dfrac{ 45! }{ 4!(45-4)!} = \dfrac{ 45 \cdot 44 \cdot 43 \cdot 42 } { 4 \cdot 3 \cdot 2 \cdot 1 } = 148995 \ \\ \ \\ p = \dfrac\dbinom{ {15}{ 0}\dbinom{ 45}{ 5} + \dbinom{ 15}{ 1}\dbinom{ 45}{ 4} }\dbinom{ { 45+15}{ 5} } = 0.633



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