Family

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

Final Answer:

p =  36.3054 %

Step-by-step explanation:

$$\begin{gathered} p_1=\frac{94}{94+100}=\frac{47}{97}\doteq 0.4845 \\ {p}_2=1-p_1=1-\frac{47}{97}=\frac{1\cdot 97}{97}-\frac{47}{97}=\frac{97}{97}-\frac{47}{97}=\frac{97-47}{97}=\frac{50}{97}\doteq 0.5155 \\ {C}_2(3)=\left(\genfrac{}{}{0ex}{}{3}{2}\right)=\frac{3!}{2!(3-2)!}=\frac{3}{1}=3 \\ p=100\cdot \left(\genfrac{}{}{0ex}{}{3}{2}\right)\cdot p_1^2\cdot p_2^1=100\cdot 3\cdot {\frac{47}{97}}^2\cdot {\frac{50}{97}}^1=36.3054\% \end{gathered}$$

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