Tickets 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?
$\Big(\dfrac{1}{2}\Big)$
$\Big(\dfrac{2}{5}\Big)$
$\Big(\dfrac{3}{10}\Big)$
$\Big(\dfrac{3}{20}\Big)$

Question

Tickets 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? 1. 2. 3. 4.

KnowledgeBoat · Accepted Answer

The tickets are numbered from 1 to 20.

Total number of outcomes = 20

Let E be the event of getting a number multiple of 3, then

E = {3, 6, 9, 12, 15, 18}

The number of favorable outcomes to the event E = 6

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{6}{20} = \dfrac{3}{10}$

Hence, option 3 is the correct option.

Mathematics

Tickets 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?
$\Big(\dfrac{1}{2}\Big)$
$\Big(\dfrac{2}{5}\Big)$
$\Big(\dfrac{3}{10}\Big)$
$\Big(\dfrac{3}{20}\Big)$

Probability

Answer

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