Question

An integer is chosen at random between $1$ and $100$. Find the probability that it is :

(i) divisible by $8$.

(ii) not divisible by $8$.

Answer 1

Step 1 : Calculation of the probability of getting an integer that is divisible by $8$.

As it is given that the integer is chosen from the numbers between $1$ and $100$. So, $1$ and $100$ are not included.

Then, the total number of outcomes will be,

$100-2=98$

Now, the number is divisible by $8$, then those numbers can be $8,16,24,32,40,48,56,64,72,80,88,\text{and}96.$

Therefore, the number of favorable outcomes are $12$.

The probability is calculated by the formula,

$\text{P(E)}=\frac{\text{F(O)}}{\text{T(O)}}$, where $\text{P(E)}$ is the probability of an event, $\text{F(O)}$ is the number of favorable outcomes, and $\text{T(O)}$ is the total number of outcomes.

Therefore,

$\begin{array}{rcl}\text{P(integer is divisible by}8)& =& \frac{12}{98}\\ & =& \frac{6}{49}\end{array}$

Step 2 : Calculation of the probability of getting an integer that is not divisible by $8$.

We know that the sum of the probabilities of all the elementary events is one. Then,

$$\begin{gathered} \therefore \begin{array}{l}\text{Probability (integer is divisible by}8)\end{array}+\begin{array}{l}\text{Probability (integer is not divisible by}8)\end{array}=1 \\ \Rightarrow \frac{6}{49}+\begin{array}{rcl}& & \text{P (integer is not divisible by}8)\end{array}=1 \\ \Rightarrow \begin{array}{l}\text{P (integer is not divisible by}8)\end{array}=1-\frac{6}{49} \\ \Rightarrow \begin{array}{l}\text{P (integer is not divisible by}8)\end{array}=\frac{43}{49} \end{gathered}$$

Final Answer :

Hence, the probabilities of integer divisible and not divisible by $8$ are $\frac{\mathbf{6}}{\mathbf{49}}\text{and}\frac{\mathbf{43}}{\mathbf{49}}$ respectively.