Question

A natural number is chosen at random from amongst first 500. What is the probability that the number so chosen is divisible by 3 or 5?

Answer 1

Let S be the sample space.
Then n(S) = 500
∴ Total number of elementary events = 500
Let A be the event where the number selected is divisible by 3 and B be the event where the number selected is divisible by 5.
Then A = {3, 6, 9, 12, 15, ...498}
and B = {5, 10, 15, 20, 25, ...500}
and (A ∩ B) = { 15, 30, 45, ...495}
We have:
$n\left(A\right)=\frac{498}{3}=166$

$n\left(B\right)=\frac{500}{5}=100$

$n\left(A\cap B\right)=\frac{495}{15}=33$ [∵ LCM of 3 and 5 is 15]

$\therefore P\left(A\right)=\frac{166}{500},P\left(B\right)=\frac{100}{500}\mathrm{and}\mathit{}P\left(A\mathit{\cap }B\right)=\frac{33}{500}$

Now, required probability = P(a number is divisible by 3 or 5)
= P (A ∪ B)
= P(A) + P(B) $-$ P(A ∩ B)
= $\frac{166}{500}+\frac{100}{500}-\frac{33}{500}=\frac{266-33}{500}=\frac{233}{500}$