Question

An integer is chosen at random from the first two hundred numbers. then, the probability that the integer chosen is divisible by $6or8$, is

• A. $1/4$
• B. $2/4$
• C. $3/4$
• D. $noneofthese$

Answer 1

The correct option is A

$1/4$

Explanation for the correct option:

Step1. Finding integer divisible by 6

Let $A$ be the integer divisible by $6$

$\begin{array}{rcl}A& =& 6,12,18,\dots 198\\ 198& =& 6+(n-1)6\\ n& =& 33\end{array}$

So,

$P\left(A\right)=33/200$

Step2. Finding integer divisible by 8

$B$ be the integer divisible by $8$

$\begin{array}{rcl}B& =& 8,16,24,\dots 200\\ 200& =& 8+(n-1)8\\ n& =& 25\end{array}$

So,

$P\left(B\right)=25/200$

Step3. Finding integer divisible by $6or8$

Both are divisible by $6$ or $8$ both then

$\begin{array}{rcl}A\cap B& =& 24,48,\dots 192\\ 192& =& 24+(n-1)24\\ n& =& 8\end{array}$

So,

$P(A\cap B)=\frac{8}{200}$

Step4. Finding probability integer divisible by $6or8$

$\begin{array}{rcl}\mathit{P}\mathbf{(}\mathit{A}\mathbf{\cap }\mathit{B}\mathbf{)}& =& \mathit{P}\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{+}\mathit{P}\mathbf{\left(}\mathit{B}\mathbf{\right)}\mathbf{-}\mathit{P}\mathbf{(}\mathit{A}\mathbf{\cup }\mathit{B}\mathbf{)}\mathbf{}\mathbf{}\\ & =& \frac{33}{200}+\frac{25}{200}-\frac{8}{200}\\ & =& \frac{1}{4}\end{array}$

Hence, option $\left(A\right)$ is correct.