Tickets numbered 1 to 20 are mixed up and the | vedclass

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(D) Total number of tickets $= 20$.Multiples of $3$ between $1$ and $20$ are ${3, 6, 9, 12, 15, 18}$. There are $6$ such tickets.Multiples of $5$ betw

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$9/20$

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(D) Total number of tickets $= 20$.Multiples of $3$ between $1$ and $20$ are ${3, 6, 9, 12, 15, 18}$. There are $6$ such tickets.Multiples of $5$ between $1$ and $20$ are ${5, 10, 15, 20}$. There are $4$ such tickets.The number $15$ is common to both sets.Using the inclusion-exclusion principle,the number of favorable outcomes $= 6 + 4 - 1 = 9$.Therefore,the required probability $= \frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}} = \frac{9}{20}$.

Tickets numbered $1$ to $20$ are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of $3$ or $5$?

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