There are 25 tickets numbered as 1, 2, 3, 4,..., 25 respectively. One ticket is drawn at random. What is the probability that the number on the ticket is a multiple of 3 or 5?
(a) $\frac{2}{5}$ (b) $\frac{11}{25}$ (c) $\frac{12}{25}$ (d) $\frac{13}{25}$

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There are 13 tickets numbered 1 to 13. If a ticket is randomly selected, find the probability that the outcome is a divisor of 12.

The correct option is D 613Sample space=(1,2,3,4,5,6,7,8,9,10,11,12,13) Numbers divisor of 12 from 1 to 13=6P(A)=no. of possible outcomesNo. of total outcomes =613

There are $13$ tickets numbered $1$ to $13$ . If a ticket is randomly selected, find the probability that the outcome is a divisor of $12$ .

The correct option is D $\frac{6}{13}$

Sample space $= (1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13)$

Numbers divisor of 12 from $1 t o 13 = 6$

$P (A)$ $= \frac{no. of possible outcomes}{No. of total outcomes}$

$= \frac{6}{13}$