Family

94 boys are born per 100 girls. Determine the probability that there are two boys in a randomly selected family with three children.

Result

p =  36.305 %

Solution:

p1=9494+100=47970.4845 p2=1p1=10.484550970.5155  C2(3)=(32)=3!2!(32)!=31=3  p=100 (32) p12 p21=100 3 0.48452 0.5155136.305436.305p_{ 1 }=\dfrac{ 94 }{ 94+100 }=\dfrac{ 47 }{ 97 } \doteq 0.4845 \ \\ p_{ 2 }=1 - p_{ 1 }=1 - 0.4845 \doteq \dfrac{ 50 }{ 97 } \doteq 0.5155 \ \\ \ \\ C_{{ 2}}(3)=\dbinom{ 3}{ 2}=\dfrac{ 3! }{ 2!(3-2)!}=\dfrac{ 3 } { 1 }=3 \ \\ \ \\ p=100 \cdot \ { { 3 } \choose 2 } \cdot \ p_{ 1 }^2 \cdot \ p_{ 2 }^1=100 \cdot \ 3 \cdot \ 0.4845^2 \cdot \ 0.5155^1 \doteq 36.3054 \doteq 36.305 \\%



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