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

Given:

Tickets numbered from 1 to 20 are mixed up and a ticket is drawn at random.

To do:

We have to find the probability that the ticket drawn has a number which is a multiple of 3 or 7.

Solution:

Number of tickets $=20$

This implies,

The total number of possible outcomes $n=20$.

Multiples of 3 from 1 to 20 are 3, 6, 9, 12, 15 and 18.

Number of multiples of 3 from 1 to 20 $=6$

Multiples of 7 from 1 to 20 are 7 and 14.

Number of multiples of 7 from 1 to 20 $=2$

Total number of favourable outcomes(drawing a ticket that has a number which is a multiple of 3 or 7) $=6+2=8$.

We know that,

Probability of an event $=\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ possible\ outcomes}$

Therefore,

Probability that the ticket drawn has a number which is a multiple of 3 or 7 $=\frac{8}{20}$

$=\frac{2}{5}$

The probability that the ticket drawn has a number which is a multiple of 3 or 7 is $\frac{2}{5}$.   

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Updated on: 10-Oct-2022

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