Respuesta :
We will have the following:
***Having at least 7 girls:
[tex]\begin{gathered} P(x\ge7)=\frac{9!}{7!(9-7)!}(0.5)^7(0.5)^{9-7}+\frac{9!}{8!(9-8!)}(0.5)^8(0.5)^{9-8} \\ \\ \Rightarrow P(x\ge7)=\frac{45}{512}\Rightarrow P(x\ge7)\approx0.088 \end{gathered}[/tex]The probability is 45/512, that is approximately 0.088.
***Having at most 8 girls:
[tex]\begin{gathered} P(x\leq8)=\left(\begin{array}{c}9 \\ 0\end{array}\right)(0.5)^0(0.5)^9+\left(\begin{array}{c}9 \\ 1\end{array}\right)(0.5)^1(0.5)^8+\left(\begin{array}{c}9 \\ 2\end{array}\right)(0.5)^2(0.5)^7+\left(\begin{array}{c}9 \\ 3\end{array}\right)(0.5)^3(0.5)^6+\left(\begin{array}{c}9 \\ 4\end{array}\right)(0.5)^4(0.5)^5+\left(\begin{array}{c}9 \\ 5\end{array}\right)(0.5)^5(0.5)^4+\left(\begin{array}{c}9 \\ 6\end{array}\right)(0.5)^6(0.5)^3+\left(\begin{array}{c}9 \\ 7\end{array}\right)(0.5)^7(0.5)^2+\left(\begin{array}{c}9 \\ 8\end{array}\right)(0.5)^8(0.5)^1 \\ \\ \Rightarrow P(x\leq8)=\frac{1}{512}+\frac{9}{512}+\frac{9}{128}+\frac{21}{128}+\frac{63}{256}+\frac{63}{256}+\frac{21}{128}+\frac{9}{128}+\frac{9}{512} \\ \\ \Rightarrow P(x\leq8)=\frac{511}{512}\Rightarrow P(x\leq8)\approx0.99 \end{gathered}[/tex]So, the probability of having at most 8 girls is 511/512, that is approximately 0.99.
