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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?
Solution
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: \[n\left( A \right) = \frac{498}{3} = 166\]
\[n\left( B \right) = \frac{500}{5} = 100\]
\[\therefore P\left( A \right) = \frac{166}{500}, P\left( B \right) = \frac{100}{500} \text{ and } P\left( A \cap B \right) = \frac{33}{500}\]
Now, required probability = P(a number is divisible by 3 or 5)
= P (A ∪ B)
= P(A) + P(B) - P(A ∩ B)
= \[\frac{166}{500} + \frac{100}{500} - \frac{33}{500} = \frac{266 - 33}{500} = \frac{233}{500}\]
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