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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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The correct answer is:
$\frac{7}{15}$
No. of total events $=30$
Multiples of 3 are $3,6,9,12,15,18,21$,
$24,27,30$
Multiples of 5 are $5,10,15,20,25,30$
Then multiples of 3 or 5 are : $3,5,6,9,10,12,15$,
$18,20,21,24,25,27,30$
$\therefore$ No. of favourable events $=14$
The required probability
$=\frac{\text { No. of favourable events }}{\text { No. of total events }}=\frac{14}{30}=\frac{7}{15}$
Multiples of 3 are $3,6,9,12,15,18,21$,
$24,27,30$
Multiples of 5 are $5,10,15,20,25,30$
Then multiples of 3 or 5 are : $3,5,6,9,10,12,15$,
$18,20,21,24,25,27,30$
$\therefore$ No. of favourable events $=14$
The required probability
$=\frac{\text { No. of favourable events }}{\text { No. of total events }}=\frac{14}{30}=\frac{7}{15}$
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