ISI MStat PSA 2019 Problem 18 | Probability and Digits

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

Probability and Digits - ISI MStat Year 2019 PSA 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 \(1\)?

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

Prerequisites


Last Digit of Natural Numbers

Basic Probability Theory

Combinatorics

Check the Answer


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

ISI MStat 2019 PSA Problem Number 18

A First Course in Probability by Sheldon Ross

Try with Hints


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!

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.

So, there are 2 possible cases out of 10.

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

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