Question

A car is parked by an owner amongst $25$ cars in a row, not at either end. On the return, he finds that exactly $15$ places are still occupied. The probability that both the neighboring places are empty.

• A. $\frac{91}{276}$
• B. $\frac{15}{184}$
• C. $\frac{15}{92}$
• D. None of these

Answer 1

The correct option is C

$\frac{15}{92}$

Explanation for the correct option:

Step1. Find the total number of ways :

As it is given that when the owner returns then there are still $15$ places occupied. So, if we excluded the owner's cars we have $24$ places for $14$ cars.

The total number of ways $={}^{24}C_{14}$

Step2. Find the required probability :

Given the condition the neighboring places are empty. Then $14$ cars must be parked in [ $25-$($1$ owner car )$-$($2$ neighboring place)] $=22$ places

So, the favorable number of ways to park the $14$ cars that must be parked in $22$ places $={}^{22}C_{14}$

$\therefore$Required probability $=\frac{{}^{22}C_{14}}{{}^{24}C_{14}}$ $[\because$Probability = Favorable cases / Total cases$]$

$=\frac{{\displaystyle \frac{22!}{\left(22-14\right)!14!}}}{{\displaystyle \frac{24!}{\left(24-14\right)!14!}}}$ $\left[\mathbf{\because }{}^{\mathbf{n}}\mathit{C}_{\mathbf{r}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{n}\mathbf{!}}{\left(n-r\right)\mathbf{!}\mathbf{r}\mathbf{!}}\right]$

$$\begin{gathered} =\frac{22!10!14!}{24!8!14!} \\ =\frac{10\times 9\times 8!}{24\times 23\times 8!} \\ =\frac{90}{552} \\ =\frac{\mathbf{15}}{\mathbf{92}} \end{gathered}$$

Hence, the correct option is (C).