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]

\[\therefore P\left( A \right) = \frac{166}{500}, P\left( B \right) = \frac{100}{500} \text{ and }  P\left( A \cap B \right) = \frac{33}{500}\]

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