1 Answer

Ask a Question

Gooner · Answer 1 · 2023-05-30T10:37:45+0000

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.

0 Comments

Please log in or register to add a comment.

Please log in or register to answer this question.

1 Answer

0 Comments

Please log in or register to add a comment.

Other Questions