A box contains 25 cards numbered from 1 to 25. A card is drawn at random from the bag.

Total number of outcomes = 25

(i) Let E1 be the event of getting a card divisible by 2 or 3.

Out of the given numbers, numbers divisible by 2 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22 and 24.
Out of the given numbers, numbers divisible by 3 are 3, 6, 9, 12, 15, 18, 21 and 24.
Out of the given numbers, numbers divisible by both 2 and 3 are 6, 12, 18 and 24.

Number of favourable outcomes = 16

$\therefore \mathrm{P}$ (getting a card divisible by 2 or 3$)=\mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{\text { Number of outcomes favourable to } \mathrm{E}_{1}}{\text { Number of all possible outcomes }}$

$=\frac{16}{25}$

Thus, the probability that the number on the drawn card is divisible by 2 or 3 is $\frac{16}{25}$.

(ii) Let E2 be the event of getting a prime number.

Out of the given numbers, prime numbers are 2, 3, 5, 7, 11, 13, 17, 19 and 23.

Number of favourable outcomes = 9

$\therefore \mathrm{P}$ (getting a prime number) $=\mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{\text { Number of outcomes favourable to } \mathrm{E}_{2}}{\text { Number of all possible outcomes }}$

$=\frac{9}{25}$

Thus, the probability that the number on the drawn card is a prime number is $\frac{9}{25}$.