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JAMB 2014 · UTME · Q50

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

Worked solution (try it first)
  1. From 10 to 30 inclusive there are 30−10+1=2130 - 10 + 1 = 21 numbers.
  2. The multiples of 3 are 12, 15, 18, 21, 24, 27, 30, which is 7 numbers.
  3. So the probability is 721=13\frac{7}{21} = \frac13, option D.

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