Question

A car is parked among $N$ cars standing in a row, but not at either end. On his return, the owner finds that exactly $r$ of the $N$ places are still occupied. What is the probability that both the places neighboring his car are empty?

• A. Required probability $=\frac{(N-r)(N-r-2)}{(N-1)(N-2)}$
• B. Required probability $=\frac{(N-r)(N-r-1)}{(N-1)(N-2)}$
• C. Required probability $=\frac{(N-r-2)(N-r-1)}{(N-1)(N-2)}$
• D. none of these

Answer 1

The correct option is B Required probability $=\frac{(N-r)(N-r-1)}{(N-1)(N-2)}$

On his return, the owner finds that exactly $r$ of $N$ places are still filled. This includes the space occupied as well
Therefore it means the remaining $(r-1)$ cars are parked in $(N-1)$ places.
Total number of ways ${=}^{N-1}{C}_{r-1}$
Since the neighboring place to car are empty, it means $(r-1)$ cars are parked among $(N-3)$ places
Number of favorable ways ${=}^{N-3}{C}_{r-1}$
$\therefore P=\frac{{}^{N-3}{C}_{r-1}}{{=}^{N-1}{C}_{r-1}}=\frac{(N-3)!(N-r)!}{(N-1)!(N-r-2)!}=\frac{(N-r)(N-r-1)}{(N-1)(N-2)}$