A car is parked by an owner amongst 25 cars in a row, not at either end. On his return he finds that exactly 15

 places are still occupied. The probability that both the neighboring places are empty  is

t is given that 15 places are occupied. This includes the owner's car also, and, hence 14 other cars are parked.

 There are 24 places (excluding places at the two ends) out of which 14 places can be chosen in ${ }^{22}{\mathrm{C}}_{14}$   ways.

 Excluding 15 (c) 92 (d) none of these the neighboring places there are 22 places in which 14 cars can be

 parked in${ }^{22}{\mathrm{C}}_{14}$  ways 

Hence, required probability $=\frac{{ }^{22}{C}_{14}}{{ }^{24}{C}_{14}}=\frac{15}{92}$