A number is chosen at random from the whole numbers 1 to 90, both included. What is the probability that the number is divisible by 6 or by 9 (or by both)?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\frac{2}{9}\)
Numbers divisible by both 6 and 9 are the multiples of their LCM, 18. They appear in both lists, so they are counted only once: n(A or B) = n(A) + n(B) − n(both).
Multiples of 6 from 1 to 90: \(90\div6=15\)
Multiples of 9 from 1 to 90: \(90\div9=10\)
Multiples of 18 from 1 to 90: \(90\div18=5\) (18, 36, 54, 72, 90)
Divisible by 6 or 9 \(=15+10-5=20\)
Probability \(\displaystyle =\frac{20}{90}=\frac{2}{9}\)