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Ticket numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn bears a number which is a multiple of 3 or 7?

1/15
1/2
2/5
7/20

Correct Answer: 2/5

Explanation:

Clearly, n(S) = 20 and E = { 3, 6, 9, 12, 15, 18, 7, 14 } i.e., n(E) = 8
∴ P(E) = n(E)/ n(S) = 8/20 = 2/5

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