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(D) Total number of outcomes $= 30$. Multiples of $3$ in the range $1$ to $30$ are: $3, 6, 9, 12, 15, 18, 21, 24, 27, 30$ (Total $10$). Multiples of $

Vedclass · Accepted Answer

$\frac{7}{15}$

Vedclass · Answer

(D) Total number of outcomes $= 30$. Multiples of $3$ in the range $1$ to $30$ are: $3, 6, 9, 12, 15, 18, 21, 24, 27, 30$ (Total $10$). Multiples of $5$ in the range $1$ to $30$ are: $5, 10, 15, 20, 25, 30$ (Total $6$). Multiples of both $3$ and $5$ (i.e.,multiples of $15$) are: $15, 30$ (Total $2$). Using the inclusion-exclusion principle,the number of favorable outcomes (multiples of $3$ or $5$) is: $10 + 6 - 2 = 14$. Therefore,the required probability $= \frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}} = \frac{14}{30} = \frac{7}{15}$.

If one ticket is selected at random from $30$ tickets,each with a distinct number from $1$ to $30$,then the probability that the number on the selected ticket is a multiple of $3$ or $5$ is

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