Question

A sum of money is rounded off to the nearest rupee. The probability that the error occurred in rounding off is at least ten paise is
A. $\dfrac{{29}}{{100}}$
B. $\dfrac{{30}}{{100}}$
C. $\dfrac{{70}}{{100}}$
D. $\dfrac{{81}}{{100}}$

Answer 1

Hint: The formula for probability is equal to the division of favourable outcomes to the total outcomes. Total outcomes for this problem are all the possible errors in rounding off to the nearest rupee. Add all these numbers you will get the total outcomes. Favourable 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. Find the possible ways so that the error in possible outcomes is not more than 9 paise and note down the total of these numbers. Now solve by finding this probability of favourable outcomes and subtract this probability from 1 to get the required probability.

Complete step-by-step solution:
We already know that probability of an outcome is the division of favourable 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 where 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.
Favourable 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, favourable outcomes are 19. Now let’s find the probability where the round off error is not more than 9:
${\text{Probability}} = \dfrac{{{\text{Favorable outcomes}}}}{{{\text{Total outcomes}}}}$
Substituting favourable 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.
\[
  \therefore P(A\prime ) = \dfrac{{19}}{{100}} \\
   \Rightarrow 1 - P(A) = \dfrac{{19}}{{100}} = \dfrac{{100 - 19}}{{100}} \\
   \Rightarrow P(A) = \dfrac{{81}}{{100}}
 \]

So, option (D) is the correct answer.

Note: If total outcomes are favourable for a certain event then its probability is one as in the case of dice the probability of rolling a dice with a number six or lower is equal to one.
Conditional probability is having many applications in daily life such as in fields as diverse as calculus, insurance, and politics.
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. Here you may note that sometimes students tend to ignore the number 0 and sum the total of favourable outcomes as 18 so be careful while adding these numbers