Question

A sum of money is rounded off to the nearest rupee. The probability that round off error is at least 10 paise is:
(a) $ \dfrac{81}{100} $
(b) $ \dfrac{82}{100} $
(c) $ \dfrac{19}{100} $
(d) $ \dfrac{19}{101} $

Answer 1

Hint: The formula for probability is equal to the division of favorable outcomes to the total outcomes. Total outcomes for this problem are all the possible errors in rounding off to the nearest rupee are -0.50, -0.49, -0.48………-0.01, 0.00, 0.01, 0.02, 0.03……….0.49. Add all these numbers you will get the total outcomes. Favorable outcomes are the outcomes in which error is at least 10 paise which we are going to find by subtracting the probability of getting at most 9 paise from 1. The possible ways so that the error in possible outcomes is not more than 9 paise are -0.09, -0.08, -0.07……… -0.01, 0.00, 0.01, 0.02 …... 0.09. Note down the total of these numbers and find this probability of favorable outcomes and subtract this probability from 1 to get the required probability.

Complete step-by-step answer:
We know that probability of an outcome is the division of favorable outcomes to the total outcomes which we have shown below:
 $ \text{Probability}=\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}} $
Total outcomes are the possible number of errors in paise that could occur in rounding off to the nearest rupee.
 -0.50, -0.49, -0.48………-0.01, 0.00, 0.01, 0.02, 0.03……….0.49
All the above numbers are in rupee. Addition of the above numbers is 100 because from -0.50 to -0.01 50 numbers are there then from 0.01 to 0.49 49 numbers are there and one more number 0.00 so adding 50, 49 and 1 we get the total number as 100.
From the above, we have got the total outcomes as 100.
We have to find the probability in which the round off error is at least 10 paise which we are going to find by subtracting the probability in which the round off error is not more than 9 paise from 1.
Favorable outcomes in finding the probability where the round off error is not more than 9 paise are the possible ways in which the round off error is not more than 9 paise.
-0.09, -0.08, -0.07……., -0.01, 0.00, 0.01, 0.02….., 0.09
All the numbers are in rupee. Addition of all the above numbers is 19.
Hence, favorable outcomes are 19.
Probability in which the round off error is not more than 9 is equal to:
 $ \text{Probability}=\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}} $
Substituting favorable outcomes as 19 and total outcomes as 100 in the above equation we get,
 $ \text{Probability}=\dfrac{19}{100} $
Subtracting the above probability from 1 we get the probability in which the round off error is at least 10 paise.
 $ \begin{align}
  & 1-\dfrac{19}{100} \\
 & =\dfrac{100-19}{100} \\
 & =\dfrac{81}{100} \\
\end{align} $
Hence, the correct option is (a).

Note: You might have thought why we have taken negative values in the possible ways like -0.50, -0.49 because the error is always in positive and negative like the round off error to nearest rupee is $ \pm 0.20 $ .
One more thing to be noted is that you might make mistake in counting the number of possible errors like in the following:
-0.09, -0.08, -0.07……., -0.01, 0.00, 0.01, 0.02….., 0.09
You might ignore the number 0 and write the total as 18 so be careful while adding these numbers.