Q. 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) $\frac{23}{90}$
Explanation
(d)
Total number=90
Number divisible by 6 are {6,12,18,24,30,36,42,48,54,60,66,72,78,84,90}
Numbers divisible by 8 are {8,16,24,32,40,48,56,64,72,80,88}
Numbers divisible by 6 and 8 are {24,48,72}
Total number of numbers divisible by 6 or 8
$= 15 + 11 - 3 = 23$
$\therefore$ Required probability$= \frac{23}{90}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.