First slide

Introduction to probability

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. The probability that the places neighboring his car are empty is

Moderate

A

$\frac{(r - 1)!}{(N - 1)!}$
B

$\frac{(r - 1)! (N - r)!}{(N - 1)!}$
C

$\frac{(N - r) (N - r - 1)}{(N + 1) (N + 2)}$
D

$\frac{^{N - r} C_{2}}{{}^{N - 1}C_{2}}$

Solution

Since there are r cars in l/ places, total number of selection of places out of N - 1 places for r - 1 cars (except the owner's car) is
$^{N - 1} C_{r - 1} = \frac{(N - 1)!}{(r - 1)! (N - r)!}$
If neighboring places are empty, then r - 1 cars must be parked in N - 3 places. So, the favorable number of cases is
$^{N - 3} C_{r - 1} = \frac{(N - 3)!}{(r - 1)! (N - r - 2)!}$
Therefore, the required probability is
$\begin{array}{l} = \frac{(N - 3)!}{(r - 1)! (N - r - 2)!} \times \frac{(r - 1)! (N - r)!}{(N - 1)!} \\ = \frac{(N - r) (N - r - 1)}{(N - 1) (N - 2)} = \frac{^{N - r} C_{2}}{^{N - 1} C_{2}} \end{array}$

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