Question

A box contains $20$ balls bearing numbers $1,2,3,...,20$ respectively. A ball is drawn at random from the box what is the probability that the number on the ball is
$\left(i\right)$ An odd number
$\left(ii\right)$ Divisible by $2$ or $3$
$\left(iii\right)$ A prime number
$\left(iv\right)$ Not divisible by $10$

• A. $\left(i\right)\frac{1}{3}\left(ii\right)\frac{12}{19}\left(iii\right)\frac{2}{5}\left(iv\right)\frac{9}{10}$
• B. $\left(i\right)\frac{1}{2}\left(ii\right)\frac{13}{20}\left(iii\right)\frac{2}{5}\left(iv\right)\frac{9}{10}$
• C. $\left(i\right)\frac{1}{3}\left(ii\right)\frac{18}{37}\left(iii\right)\frac{2}{5}\left(iv\right)\frac{2}{5}$
• D. $\left(i\right)\frac{1}{2}\left(ii\right)\frac{15}{29}\left(iii\right)\frac{2}{5}\left(iv\right)\frac{2}{5}$

Answer 1

The correct option is B $\left(i\right)\frac{1}{2}\left(ii\right)\frac{13}{20}\left(iii\right)\frac{2}{5}\left(iv\right)\frac{9}{10}$

Since, total balls = 20
(i) odd numbers from 1 to 20 are:
1,3,5,7,9,11,13,15,17,19
no. of balls bearing odd numbers = 10
$\therefore$ favorable outcomes= 10
$\therefore$ probability = $\frac{10}{20}=\frac{1}{2}$
(ii) number divisible by 2 or 3 from 1 to 20 are:
2,3,4,6,8,9,10,12,14,15,16,18,20
no. of balls bearing number divisible by 2 or 3 = 13
$\therefore$ favorable outcomes= 13
$\therefore$ probability = $\frac{13}{20}$
(iii) prime numbers from 1 to 20 are:
2,3,5,7,,11,13,17,19
no. of balls bearing prime number = 8
$\therefore$ favorable outcomes= 8
$\therefore$ probability = $\frac{8}{20}=\frac{2}{5}$
(iv) No. of balls bearing number not divisible by 10 = 18
$\therefore$ favorable outcomes= 18
$\therefore$ probability = $\frac{18}{20}=\frac{9}{10}$