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The Smith family was one of the first to come to the U.S. They had 9 children. Assuming that the probability of a child being a girl is .5, find the probability that the Smith family had: at least 7 girls ? at most 8 girls?

User Vishal G
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4 votes

We will have the following:

***Having at least 7 girls:


\begin{gathered} P(x\ge7)=(9!)/(7!(9-7)!)(0.5)^7(0.5)^(9-7)+(9!)/(8!(9-8!))(0.5)^8(0.5)^(9-8) \\ \\ \Rightarrow P(x\ge7)=(45)/(512)\Rightarrow P(x\ge7)\approx0.088 \end{gathered}

The probability is 45/512, that is approximately 0.088.

***Having at most 8 girls:


\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)=(1)/(512)+(9)/(512)+(9)/(128)+(21)/(128)+(63)/(256)+(63)/(256)+(21)/(128)+(9)/(128)+(9)/(512) \\ \\ \Rightarrow P(x\leq8)=(511)/(512)\Rightarrow P(x\leq8)\approx0.99 \end{gathered}

So, the probability of having at most 8 girls is 511/512, that is approximately 0.99.

User Gooner
by
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