Respuesta :
We will have the following:
First, we will remeber that:
[tex]P(X=x)=\mleft(\begin{array}{c}n \\ x\end{array}\mright)p^xq^{n-x}[/tex]Now, the probability of at least 5 girls will be:
[tex]P(X\ge5)=P(X=5)+P(X=6)[/tex][tex]\Rightarrow P(X\ge5)=\mleft(\begin{array}{c}7 \\ 5\end{array}\mright)(0.5)^5(0.5)^{7-5}+\mleft(\begin{array}{c}7 \\ 6\end{array}\mright)(0.5)^6(0.5)^{7-6}[/tex][tex]\Rightarrow P(X\ge5)=(\frac{7!}{(7-5)!5!})(0.5)^5(0.5)^2+(\frac{7!}{(7-6)!6!})(0.5)^6(0.5)^1[/tex][tex]\Rightarrow P(X\ge5)=(21)(\frac{1}{32})(\frac{1}{4})+(7)(\frac{1}{64})(0.5)\Rightarrow P(X\ge5)=\frac{21}{128}+\frac{7}{128}[/tex][tex]\Rightarrow P(X\ge5)=\frac{7}{32}\Rightarrow P(X\ge5)=0.21875[/tex]So, the probability of having at least 5 girls is 0.21875.
Now, the probability of having at most 4 girls will be:
[tex]P(X\le4)=\sum ^4_{x=0}P(x)\Rightarrow P(X\le4)=P(0)+P(1)+P(2)+P(3)+P(4)[/tex]So:
[tex]P(X\le4)=\mleft(\begin{array}{c}7 \\ 0\end{array}\mright)(0.5)^0(0.5)^{7-0}+(\begin{array}{c}7 \\ 1\end{array})(0.5)^1(0.5)^{7-1}+(\begin{array}{c}7 \\ 2\end{array})(0.5)^2(0.5)^{7-2}+(\begin{array}{c}7 \\ 3\end{array})(0.5)^3(0.5)^{7-3}+(\begin{array}{c}7 \\ 4\end{array})(0.5)^4(0.5)^{7-4}[/tex][tex]\Rightarrow P(X\le4)=(1)(1)(\frac{1}{128})+(7)(\frac{1}{2})(\frac{1}{64})+(21)(\frac{1}{4})(\frac{1}{32})+(35)(\frac{1}{8})(\frac{1}{16})+(35)(\frac{1}{16})(\frac{1}{8})[/tex][tex]\Rightarrow P(X\le4)=\frac{1}{128}+\frac{7}{128}+\frac{21}{128}+\frac{35}{128}+\frac{35}{128}\Rightarrow P(X\le4)=\frac{99}{128}[/tex][tex]P(X\le4)=0.7734375[/tex]So, the probability of having at most 4 girls is 0.7734375.
