Question

A box contains $50$ discs which are numbered from $1$ to $50$. If one disc is drawn at random from the box, find the probability that is bears
(i) A two digit number less than $30$
(ii) A prime number
(iii) A number divisible by $3$

• A. (i) $\frac{1}{5}$ (ii) $\frac{7}{12}$ (iii) $\frac{7}{35}$
• B. (i) $\frac{2}{5}$ (ii) $\frac{3}{10}$ (iii) $\frac{8}{25}$
• C. (i) $\frac{1}{5}$ (ii) $\frac{13}{18}$ (iii) $\frac{11}{18}$
• D. None of these
  

Answer 1

The correct option is B (i) $\frac{2}{5}$ (ii) $\frac{3}{10}$ (iii) $\frac{8}{25}$

Total number of discs are $50$ which are numbered from $1$ to $50$.
(1) Two digit number less than $30$ are:- $10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29$
Here favorable discs $=20$
$\therefore$ probability $=$ $\frac{20}{50}=\frac{2}{5}$
(ii) prime numbers between $1$ to $50$ are:-
$2,3,5,7,11,13,17,19,23,29,31,37,41,43,47$
Here favorable discs $=15$
$\therefore$ probability $=$ $\frac{15}{50}=\frac{3}{10}$
(iii) Numbers divisible by $3$ between $1$ to $50$ are:-
$3,6,\mathrm{9..............48}$
So, total number of favorable discs $=16$.
$\therefore$ probability $=$ $\frac{16}{50}=\frac{8}{25}$