Given numbers 1,3,5.....35 form an AP with a = 1 and d = 2

Let T_n = 35 Then,

1 + (n - 1)2 = 35

⇒ 1 + 2n - 2 = 35

 ⇒ 2n = 36 

⇒ n = 18

Thus, Total number of favorable outcomes = 18

(i) Let E₁ be the event of getting a prime number less than 15.

out of these numbers, prime numbers less than 15 are 3,5,7,11 and 13.

The number of favorable outcomes = 5

Therefore P(getting a prime number less than 15) = P(E₁) = \(\frac{number\,of\,outcomes\,favorable\,to\,E_1}{number\,of\,all\,possible\,outcomes}\) = \(\frac{5}{18}\)

Thus, the probability of getting card bearing a prime number less than 5 is \(\frac{5}{18}\)

(ii) Let E₂ be the event of getting a number divisible by 3 and 5

out of these numbers, the number divisible by 3 and 5 means number divisible by 15 is 15

The number of favorable outcomes = 1

Therefore P(getting a number divisible by 3 and 5) = P(E₂) = \(\frac{number\,of\,outcomes\,favorable\,to\,E_2}{number\,of\,all\,possible\,outcomes}\)= \(\frac{1}{18}\)

Thus, the probability of getting card bearing a number divisible by 3 and 5 is \(\frac{1}{18}\)