# Left handed

It is known that 25% of the population is left-handed. What is the probability that there is a maximum of three left-handers at a seminar where there are 30 participants?

Result

p =  3.745 %

#### Solution:

$q=25/100=\dfrac{ 1 }{ 4 }=0.25 \ \\ r=1-q=1-0.25=\dfrac{ 3 }{ 4 }=0.75 \ \\ C_{{ 0}}(30)=\dbinom{ 30}{ 0}=\dfrac{ 30! }{ 0!(30-0)!}=\dfrac{ 1 } { 1 }=1 \ \\ \ \\ p_{0}={ { 30 } \choose 0 } \cdot \ q^0 \cdot \ r^{ 30-0 }=1 \cdot \ 0.25^0 \cdot \ 0.75^{ 30-0 } \doteq 0.0002 \ \\ C_{{ 1}}(30)=\dbinom{ 30}{ 1}=\dfrac{ 30! }{ 1!(30-1)!}=\dfrac{ 30 } { 1 }=30 \ \\ \ \\ p_{1}={ { 30 } \choose 1 } \cdot \ q^1 \cdot \ r^{ 30-1 }=30 \cdot \ 0.25^1 \cdot \ 0.75^{ 30-1 } \doteq 0.0018 \ \\ C_{{ 2}}(30)=\dbinom{ 30}{ 2}=\dfrac{ 30! }{ 2!(30-2)!}=\dfrac{ 30 \cdot 29 } { 2 \cdot 1 }=435 \ \\ \ \\ p_{2}={ { 30 } \choose 2 } \cdot \ q^2 \cdot \ r^{ 30-2 }=435 \cdot \ 0.25^2 \cdot \ 0.75^{ 30-2 } \doteq 0.0086 \ \\ C_{{ 3}}(30)=\dbinom{ 30}{ 3}=\dfrac{ 30! }{ 3!(30-3)!}=\dfrac{ 30 \cdot 29 \cdot 28 } { 3 \cdot 2 \cdot 1 }=4060 \ \\ \ \\ p_{3}={ { 30 } \choose 3 } \cdot \ q^3 \cdot \ r^{ 30-3 }=4060 \cdot \ 0.25^3 \cdot \ 0.75^{ 30-3 } \doteq 0.0269 \ \\ \ \\ p=100 \cdot \ (p_{0}+p_{1}+p_{2}+p_{3})=100 \cdot \ (0.0002+0.0018+0.0086+0.0269) \doteq 3.7449 \doteq 3.745 \%$

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