Question

If an integer is chosen at random from first $100$ positive integers, then the probability that the chosen number is a multiple of $4$ or $6$ is

• A. $\frac{41}{100}$
• B. $\frac{33}{100}$
• C. $\frac{1}{10}$
• D. None of these

Answer 1

The correct option is B

$\frac{33}{100}$

Explanation for the correct option:

Step 1. Find the required probability:

Total number of cases = 100

Let $A$ be the multiples of $4$ event and $B$ be the multiples of $6$ event.

Therefore,

The probability of getting a multiple of $4$, $P\left(A\right)=\frac{25}{100}$

The probability of getting a multiple of $6$, $P\left(B\right)=\frac{16}{100}$

The probability of getting neither multiple of $4$ nor multiple of $6$, $P(A\cap B)=\frac{8}{100}$

Step 2. The probability of chosen number be a multiple of 4 or 6, $\mathit{P}\mathbf{\left(}\mathit{A}\mathit{U}\mathit{B}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathit{P}\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathit{P}\mathbf{\left(}\mathit{B}\mathbf{\right)}\mathbf{}\mathbf{–}\mathbf{}\mathit{P}\mathbf{(}\mathit{A}\mathbf{\cap }\mathit{B}\mathbf{)}$

$\Rightarrow$ $P\left(AUB\right)=\frac{25}{100}+\frac{16}{100}–\frac{8}{100}$

$\Rightarrow$ $P\left(AUB\right)=\frac{(25+16-8)}{100}$

$\Rightarrow$ $\mathit{P}\mathbf{\left(}\mathit{A}\mathit{U}\mathit{B}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{33}}{\mathbf{100}}$

Hence, Option ‘B’ is Correct.