A driver parks a car in a row of $25$ cars randomly at any place but not ends. After coming back he finds $10$ cars are gone so what is the probability that both the neighbouring cars have gone?
What I did $$\dfrac{{24\choose 8}}{{23\choose 1}{24\choose 10}}$$ $24C8$ as we want two cars to go so we want to select only $8$ cars. And driver can park in $23$ ways and cars can go in $24C10$ ways.
But that doesn't yield the answer what am I missing on? Please any hints using basic probability equations.
Best Answer
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.