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)

$\frac{23}{90}$

Answer 2

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 $=\frac{23}{90}$