A number is chosen at random from 10 to 30 both inclusive. What is the probability that the number is divisible by 3?

\(\frac{2}{15}\)

\(\frac{1}{10}\)

\(\frac{1}{3}\)

\(\frac{2}{5}\)

Correct answer is C

Sample space S = {10, 11, 12, ... 30}

Let E denote the event of choosing a number divisible by 3

Then E = {12, 15, 18, 21, 24, 27, 30} and n(E) = 7

Prob (E) = \(\frac{n(E)}{n(S)}\)

Prob (E) = \(\frac{7}{21}\)

Prob (E) = \(\frac{1}{3}\)

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A number is chosen at random from 10 to 30 both inclusive. What is the probability that the number is divisible by 3?

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