**Question:**

Cards bearing numbers 1, 3, 5, .... , 35 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card bearing

(i) a prime number less than 15,

(ii) a number divisible by 3 and 5.

**Solution:**

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

Let *Tn* = 35. Then,

1 + (*n* − 1)2 = 35

⇒ 1 + 2*n* − 2 = 35

⇒ 2*n* = 36

⇒ *n* = 18

Thus, total number of 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.

Number of favourable outcomes = 5.

$\therefore \mathrm{P}$ (getting a prime number less than 15$)=\mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{\text { Number of outcomes favourable to } \mathrm{E}_{1}}{\text { Number of all possible outcomes }}$

$=\frac{5}{18}$

Thus, the probability of getting a card bearing a prime number less than 15 is

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

Out of these numbers, number divisible by 3 and 5 means number divisible by 15 is 15.

Number of favourable outcomes = 1.

$\therefore \mathrm{P}$ (getting a number divisible by 3 and 5$)=\mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{\text { Number of outcomes favourable to } \mathrm{E}_{2}}{\text { Number of all possible outcomes }}$

$=\frac{1}{18}$

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