Question

A sum of money is rounded off to the nearest rupee. The probability that the round off error is at least ten paisa is $\frac{9k}{100}$.The value of $k$ is

Answer 1

The sample space is $S=\{-0.50,-0.49,-0.48,...,-0.01,0.00,0.01,...0.49\}$
Let $E$ be the event that the round off error is at least 10 paise,
then $E^{\prime }$ is the event that the round off error at most a paise
$\therefore E^{\prime }=\{-0.09,-0.08,...,-0.01,0.00,0.01,...,0.09\}$
Hence $n\left(E^{\prime }\right)=19$ and $n\left(S\right)=100$
$\therefore P\left(E^{\prime }\right)=\frac{n\left(E^{\prime }\right)}{n\left(S\right)}=\frac{19}{100}$
Hence required probability $P\left(E\right)=1-P\left(E^{\prime }\right)=1-\frac{19}{100}=\frac{81}{100}$
$\therefore k=9$