**Question:**

Cards numbered 1 to 30 are put in a bag. A card is drawn at random from the bag. Find the probability that the number on the drawn card is

(i) not divisible by 3,

(ii) a prime number greater than 7,

(iii) not a perfect square number.

**Solution:**

Total number of outcomes = 30.

(i) Let E1 be the event of getting a number not divisible by 3.

Out of these numbers, numbers divisible by 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27 and 30.

Number of favourable outcomes = 30 − 10 = 20

$\therefore \mathrm{P}($ getting a number not divisible by 3$)=\mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{\text { Number of outcomes favourable to } \mathrm{E}_{1}}{\text { Number of all possible outcomes }}$

$=\frac{20}{30}=\frac{2}{3}$

Thus, the probability that the number on the card is not divisible by 3 is $\frac{2}{3}$.

(ii) Let E2 be the event of getting a prime number greater than 7.

Out of these numbers, prime numbers greater than 7 are 11, 13, 17, 19, 23 and 29.

Number of favourable outcomes = 6

$\therefore \mathrm{P}($ getting a prime number greater than 7$)=\mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{\text { Number of outcomes favourable to } \mathrm{E}_{2}}{\text { Number of all possible outcomes }}$

$=\frac{6}{30}=\frac{1}{5}$

Thus, the probability that the number on the card is a prime number greater than 7 is $\frac{1}{5}$.

(iii) Let E3 be the event of getting a number which is not a perfect square number.

Out of these numbers, perfect square numbers are 1, 4, 9, 16 and 25.

Number of favourable outcomes = 30 − 5 = 25

$\therefore \mathrm{P}$ (getting non-perfect square number) $=\mathrm{P}\left(\mathrm{E}_{3}\right)=\frac{\text { Number of outcomes favourable to } \mathrm{E}_{3}}{\text { Number of all possible outcomes }}$

$=\frac{25}{30}=\frac{5}{6}$

Thus, the probability that the number on the card is not a perfect square number is $\frac{5}{6}$.