Question

20 cards are numbered from 1 to 20. One card is then drawn at random. What is the probability that the number of the card drawn is

(i) A prime number ?

(ii) An odd number ?

(iii) A multiple of 5 ?

(iv) Not divisible by 3 ?

Answer 1

Clearly, the sample space is given by

$S=\{1,2,3,4,5,\dots \dots ,19,20\}$ and, therefore, $n\left(S\right)=20$.

(i) Let $E_1=$ event of getting a prime number. Then,

$E_1=\{2,3,5,7,11,13,17,19\}$ and, therefore, $n\left(E_1\right)=8$

$\therefore$ P(getting a prime number) $=P\left(E_1\right)=\frac{n\left(E_1\right)}{n\left(S\right)}=\frac{8}{20}=\frac{2}{5}$

(ii) Let $E_2=$ event of getting an odd number. Then,

$E_2=\{1,3,5,7,9,11,13,15,17,19\}$ and, therefore, $n\left(E_2\right)=10$

$\therefore$ P(getting a prime number) $=P\left(E_2\right)=\frac{n\left(E_2\right)}{n\left(S\right)}=\frac{10}{20}=\frac{1}{2}$

(iii) Let $E_3=$ event of getting a multiple of 5. Then,

$E_3=\{5,10,15,20\}$ and, therefore, $n\left(E_3\right)=4$

$\therefore$ P(getting a multiple of 5) $=P\left(E_3\right)=\frac{n\left(E_3\right)}{n\left(S\right)}=\frac{4}{20}=\frac{1}{5}$

(iv) Let $E_4=$ event of getting a number which is not divisible by 3. Then,

$E_4=\{1,2,4,5,7,8,10,11,13,14,16,17,19,20\}$ and so, $n\left(E_4\right)=14$

$\therefore$ P(getting a number which is not divisible by 3) $=P\left(E_4\right)=\frac{n\left(E_4\right)}{n\left(S\right)}=\frac{14}{20}=\frac{7}{10}$