Car parking related probability

probability combinatorics

Let $L$ denote the event that his left neighbor has left and $R$ the event that his right neighbor has left. Then:

$$P\left(L\cap R\right)=P\left(L\mid R\right)P\left(R\right)=\frac{9}{23}\frac{10}{24}$$

The second factor because of $24$ cars $10$ leave and all cars have equal chances to be one them. The first factor likewise, but now of $23$ cars $9$ are selected to leave.


Hint: Out of the $2$ neighbouring cars, both have to go, so you need a term of $2\choose 2$ in the numerator. Now, there are still $10-2=8$ cars to choose from the remaining $25-1-2=22$ cars. These are the favourable cases. All the possible cases are $24 \choose 10$. Do you know what to do next?

To compare

$$\frac{\dbinom{2}{2}\dbinom{22}{8}}{\dbinom{24}{10}}$$

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