Set up an appropriate sample space for each of Problems 1.1 to 1.10 and use itto solve the problem. Use either a uniform or non-uniform sample space or try both.

An integer is chosen at random with $1⩽\mathbf{N}⩽100$. What is the probability that is divisible by 11? That$\mathbf{N}>90$ ? That $\mathbf{N}⩽3$? That is a perfect square?

The numbers that do not include any fraction are said to be the integers. The integer is a whole number. It can be negative, positive and zero. But the numbers that include decimals are not considered as integers.

The integers are selected from 1 and 100 both inclusive, this implies that the total number of outcomes will be 100, so, the sample space is expressed as follows,

$\left(1 ,2 ,3 ,\cdots ,100\right)$

Each point of the obtained sample space has an equal probability of $\frac{1}{100}$.

A number between 1 and 100 which is divisible by 11 are 9 namely.

$\left(11 , 22 , 33 , 44 , 55 , 66 , 77 , 88 , 99\right)$

Find the probability that integer is divisible by 11 by adding the probabilities of each possible outcomes, that is 9 times$\frac{1}{100}$.

$\begin{array}{rcl}p& =& 9\times \frac{1}{100}\\ & =& \frac{9}{100}\\ & & \end{array}$

Thus, the probability that integer is divisible by 11 is$\frac{9}{100}$

A number between 1 and 100 which is more than 90 are10 namely.

$\left(91 , 92 , 93 , 94 , 95 , 96 , 97 , 98 , 99 , 100\right)$

Find the probability that integer selected is greater than 90 by adding the probabilities of each possible outcomes, that is 10 times $\frac{1}{100}$

$\begin{array}{rcl}p& =& 10\times \frac{1}{100}\\ & =& \frac{1}{10}\\ & & \end{array}$

Thus, the probability that integer selected is greater than 90 is$\frac{1}{10}$.

A number between 1 and 100 which is less than or equal to 3 are 3 namely$\left(1,2,3\right)$

This implies that the number of outcomes favourable are 3 and total number of outcomes are 100.

Find the probability that integer selected is less than or equal to 3 by adding the probabilities of each possible outcomes.

$\begin{array}{rcl}p& =& \frac{1}{100}+\frac{1}{100}+\frac{1}{100}\\ & =& \frac{3}{100}\\ & & \end{array}$

Thus,the probability that integer selected is less than or equal to 3 is$\frac{3}{100}$

A number between 1 and 100 which is a perfect square are 10 namely.

$\left(1 , 4 , 9 , 16 , 25 , 36 , 49 , 64 , 81 , 100\right)$

Find the probability that integer selected is a perfect square by adding the probabilities of each possible outcomes.

$\begin{array}{rcl}p& =& \frac{10}{100}\\ & =& \frac{1}{10}\\ & & \end{array}$

Thus, the probability that integer selected is a perfect square is$\frac{1}{10}$.