Cards bearing numbers 1, 3, 5, …. , 35 are kept in a bag. A card is drawn at random from the bag.
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 + 2n  − 2 = 35
⇒ 2n = 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 518518.


(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 $\frac{1}{18}$.

Administrator

Leave a comment

Please enter comment.
Please enter your name.