Question

An integer is chosen at random from the first two hundred natural numbers. The probability of getting an integer divisible by 6 or 8 is

• A.
  
  $\frac{6}{13}$
• B.
  
  $\frac{2}{5}$
• C.
  $\frac{1}{13}$
• D.
  $\frac{1}{4}$

Answer 1

The correct option is D

$\frac{1}{4}$
The natural number less than or equal to 200 and divisible by 6 are 6, 12, …, 198.

Now, these numbers form an AP with a = 6 and d = 6.

Also, last term, $l=a+(n–1)d\Rightarrow 198=6+(n–1)\times 6\Rightarrow 192=(n–1)\times 6\Rightarrow n=33$

Therefore, total numbers that are divisible by 6 are 33.

Now, the natural numbers less than equal to 200 and divisible by 8 are 8, 16, …, 200.

Clearly, these numbers form an AP with $a=8andd=8.$

Also, last term, $l=a+(n–1)d\Rightarrow 200=8+(n–1)\times 8\Rightarrow 192=(n–1)\times 8\Rightarrow n=25$

Therefore, total numbers divisible by 8 are 25.

Now, the numbers divisible by both 6 and 8 are 24, 48, 72, 96, 120, 144, 168 and 192.

Therefore, total numbers that are divisible by both 6 and 8 is 8.

Number of favourable outcomes $=33+25–8=50$
$\therefore$ Required probability $=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{50}{200}=\frac{1}{4}$


Hence, the correct answer is option d.