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A group includes 5 boys and 5 girls which are randomly placed in a row. What is the probability that both the first person and the last person in the row is a girl?

I tried to visualize it but I don't know how calculate the probability in this exercise.

Can someone give me a hint?

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First, pick two girls and sit them on the ends. Then complete with the remaining people. This can be done in $2\cdot {5 \choose 2}\cdot 8!$ ways. So, the probability would be $\frac{2\cdot{5 \choose 2}\cdot 8!}{10!} = \frac{2}{9}$.

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  • $\begingroup$ Of course. Thank you very much to both of you. $\endgroup$ – Anonymous196 Jun 1 '16 at 5:45
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The probability that there is a girl in the first row is $\frac{5}{10}=\frac{1}{2}$. Then the probability that there is a girl in the front and last row is $$\frac{1}{2} \times \frac{4}{9}= \frac{2}{9}$$

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