There are 10 men, 10 women, and 10 rooms. Each person randomly goes into a room.

What is the expected number of rooms with at least one man and woman?

Our prof. gave us the following solution however, I'm confused about the probability portion of the answer (especially the $(\frac{9}{10})^{10}$ part):

$$10 \times \left(\!1 - \left(\! \frac{9}{10} \!\right)^{\!\!10}\hspace{1mu}\right)^2$$

Solution 1:

The chance that a particular room does not have a specific man is $\frac 9{10}$. The chance that it does not have any men is $\left (\frac 9{10}\right )^{10}$ The chance that it has at least one man is $1-\left (\frac 9{10}\right )^{10}$ The chance that it has at least one man and at least one woman is $\left(1-\left (\frac 9{10}\right )^{10}\right )^2$ For the expected number of rooms with at least one man and at least one woman, multiply by $10$, getting $$10\left(1-\left (\frac 9{10}\right )^{10}\right )^2$$

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