Question

The probability of choosing a number divisible by $6$ or $8$ from among $1$ to $90$ is

• A. $ \frac{1}{6} $
• B. $ \frac{1}{90} $
• C. $ \frac{1}{30} $
• D. $ \frac{23}{90} $

Answer 1

Answer: (D)

Total numbers $ =90 $
Numbers divisible by $ 6=6,\,12,18,24,30,36. $
$ 42,48,54,60,66,72,78,84,90 $
Number divisible by $ 8=8,16,24,32,40,48, $
$ 56,64,72,80,88 $
Total number of numbers divisible by 6 or 8
$ =15+11-3=23 $
$ \therefore $ Required probability $ =\frac{23}{90} $