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Question
Tickets numbered from 1 to 20 are mixed up together and then a ticket is drawn at random. What is the probability that the ticket has a number which is a multiple of 3 or 7?
Solution
Total number of elementary events, n(S) = 20C1 = 20
Multiples of 3 or 7 = 3, 6, 9, 12, 15, 18, 7, 14
Thus, favourable number of events, n(E) = 8C1 = 8
Hence, required probability = \[\frac{n\left( E \right)}{n\left( S \right)} = \frac{8}{20} = \frac{2}{5}\]
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