Cards marked with numbers 1, 3, 5, ..., 101 are placed in a bag and mixed thoroughly.

​Given number 1, 3, 5, ..., 101 form an AP with a = 1 and d = 2.
Let Tn = 101. Then,
1 + (n − 1)2 = 101
⇒ 1 + 2n  − 2 = 101
⇒ 2n = 102
⇒ n = 51

Thus, total number of outcomes = 51.

(i) Let E1 be the event of getting a number less than 19.

Out of these numbers, numbers less than 19 are 1, 3, 5, ... , 17.
Given number 1, 3, 5, .... , 17 form an AP with a = 1 and d = 2.
Let Tn = 17. Then,
1 + (n − 1)2 = 17
⇒ 1 + 2n  − 2 = 17
⇒ 2n = 18
⇒ n = 9

Thus, number of favourable outcomes = 9.

$\therefore P($ getting a number less than 19$)=P\left(E_{1}\right)=\frac{\text { Number of outcomes favourable to } E_{1}}{\text { Number of all possible outcomes }}$

$=\frac{9}{51}=\frac{3}{17}$

Thus, the probability that the number on the drawn card is less than 19 is $\frac{3}{17}$.

(ii) Let E2 be the event of getting a prime number less than 20.

Out of these numbers, prime numbers less than 20 are 3, 5, 7, 11, 13, 17 and 19.

Thus, number of favourable outcomes = 7.

$\therefore P($ getting a prime number less than 20$)=P\left(E_{2}\right)=\frac{\text { Number of outcomes favourable to } E_{2}}{\text { Number of all possible outcomes }}$

$=\frac{7}{51}$

Thus, the probability that the number on the drawn card is a prime number less than 20 is $\frac{7}{51}$.