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

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Solution

Let S be the sample space. Then, clearly n(S) = 500.

Let $E_{1}$ = event of getting a number divisible by 3,

and $E_{2}$ = event of getting a number divisible by 5. Then,

$(E_{1} \cap E_{2})$ = event of getting a number divisible by both 3 and 5 = event of getting a number divisible by 15.

$∴ E_{1} =$ {3, 6, 9,..., 495, 498}, $E_{2} =$ {5, 10, 15,..., 495, 500}

and $(E_{1} \cap E_{2}) =$ {15, 30, 45,...,495}.

$∴ n (E_{1}) = (\frac{498}{3}) = 166, n (E_{2}) = (\frac{500}{5}) = 100$

and $n (E_{1} \cap E_{2}) = (\frac{495}{15}) = 33.$

$∴ P (E_{1}) = \frac{n (E_{1})}{n (S)} = \frac{166}{500} = \frac{83}{250}, P (E_{2}) = \frac{n (E_{2})}{n (S)} = \frac{100}{500} = \frac{1}{5}$

and $P (E_{1} \cap E_{2}) = \frac{n (E_{1} \cap E_{2})}{n (S)} = \frac{33}{500} .$

$∴$ P(the chosen number is divisible by 3 or 5)

$= P (E_{1} o r E_{2}) = P (E_{1} \cup E_{2})$

$= P (E_{1}) + P (E_{2}) - P (E_{1} \cap E_{2})$

$= \frac{83}{250} + \frac{1}{5} - \frac{33}{500}) = \frac{233}{500} .$

Hence, the required probability is $\frac{233}{500} .$

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