Question

A sum of money is rounded off to the nearest rupee. The probability that round off error is at least ten paise is

• A. $\frac{81}{100}$
• B. $\frac{82}{100}$
• C. $\frac{19}{100}$
• D. $\frac{19}{101}$

Answer 1

The correct option is A $\frac{81}{100}$

The sample space is $-0.50,-0.49,-0.48,...,-0.01,0.00,0.01,...0.49$
Let $A$ be the event that the round off error is at least $10$ paise.
Then $A^{\prime }$ is the event that the round off error is at most a paise.
Sample space for $A$ is given by
$-0.09,-0.08,...,-0.01,0.00,0.01,...0.09$
$\therefore P\left(A^{\prime }\right)=\frac{19}{100}\Rightarrow 1-P\left(A\right)=\frac{19}{100}\Rightarrow P\left(A\right)=\frac{81}{100}$