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This problem is a very easy and cute problem of probability from ISI MStat PSA 2019 Problem 18.

Draw one observation \(N\) at random from the set \(\{1,2, \ldots, 100\}\). What is the probability that the last digit of \(N^{2}\) is \(1\)?

- \(\frac{1}{20}\)
- \(\frac{1}{50}\)
- \(\frac{1}{10}\)
- \(\frac{1}{5}\)

Last Digit of Natural Numbers

Basic Probability Theory

Combinatorics

But try the problem first...

Answer: is \(\frac{1}{5}\)

Source

Suggested Reading

ISI MStat 2019 PSA Problem Number 18

A First Course in Probability by Sheldon Ross

First hint

Try to formulate the sample space. Observe that the sample space is not dependent on the number itself rather only on the last digits of the number \(N\).

Also, observe that the number of integers in \(\{1,2, \ldots, 100\}\) is uniformly distributed over the last digits. So the sample space can be taken as \(\{0,1,2, \ldots, 9\}\). So, the number of elements in the sample space is \(10\).

**See the Food for Thought**!

Second Hint

This step is easy.

Find out the cases for which \(N^2\) gives 1 as the last digit. Use the reduced last digit sample space.

- 1 x 1
~~3 x 7~~(Since \(N^2\) and they must have the same last digit)~~7 x 3~~(Since \(N^2\) and they must have the same last digit)- 9 x 9

So, there are 2 possible cases out of 10.

Third Step

Therefore the probability = \( \frac{2}{10} = \frac{1}{5}\).

Food For Thought

- Observe that there is a little bit of handwaving in the First Step. Please make it more precise using the ideas of Probability that it is okay to use the sample space as the reduced version rather than \(\{1,2, \ldots, 100\}\).

- Generalize the problem for \(\{1,2, \ldots, n\}\).

- Generalize the problems for \(N^k\) for selecting an observation from \(\{1,2, \ldots, n\}\).

- Generalize the problems for \(N^k\) for selecting an observation from \(\{1,2, \ldots, n\}\) for each of the digits from \(\{0,1,2, \ldots, 9\}\).

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