Question

An integer is chosen at random between 1 to 100. Find the probability that it is
1) Divisible by 8
2) Not divisible by 8

Answer 1

Hint:
First we have to find the total number of integers between the given numbers
Then, we will find the probability by using formula, P(E) $ = \dfrac{{n(E)}}{{n(S)}}$
For the probability which is not divisible is given by \[P(E'){\text{ = 1 - P}}({\text{E}})\]

Complete step by step solution:
Here, we are given that the integer is chosen at random between 1 to 100.
We have to find the probability which is divisible by 8 or which is not.
Total number of integers between 1 to 100: 2, 3, 4, 5, 6, ………., 99
So, sample space U = 98
So, total number of outcomes, n(s) = 98
A) Now, total number which are divisible by 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96
Total number which are divisible by 8 = 12
So, number of possible events, n(E) = 12
Now, the probability which is divisible by 8, P(E)
P(E) $ = \dfrac{{n(E)}}{{n(S)}} = \dfrac{{12}}{{98}}$
$ = \dfrac{6}{{49}}$
$\therefore $ The probability which is divisible by $8 = \dfrac{6}{{49}}$

B) Now, the probability which is not divisible by 8, P(E’)
 $ = - \dfrac{6}{{49}}$
$ = \dfrac{{\left( {49 - 6} \right)}}{{49}}$
$ = \dfrac{{43}}{{49}}$
$\therefore $ probability which is not divisible by $8 = \dfrac{6}{{49}}$

Note:
Note:
The formula of probability is given by the ratio of number of favourable outcomes to the number of total outcomes.
Let X be any event. Now the probability of event X can be given by
$P\left( X \right) = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of favourable outcomes}}}}$.