Question

A bag contains $5$ white balls, $7$ red balls, $4$ back balls and $2$ blue balls. One ball is drawn at random from the bag. What is the probability that the ball drawn is
$\left(i\right)$ White or blue
$\left(ii\right)$ Red or black
$\left(iii\right)$ Not white
$\left(iv\right)$ Neither white nor black

• A. $\left(i\right)\frac{1}{9}\left(ii\right)\frac{11}{18}\left(iii\right)\frac{17}{19}\left(iv\right)\frac{1}{2}$
• B. $\left(i\right)\frac{3}{14}\left(ii\right)\frac{11}{18}\left(iii\right)\frac{14}{23}\left(iv\right)\frac{1}{2}$
• C. $\left(i\right)\frac{7}{18}\left(ii\right)\frac{11}{18}\left(iii\right)\frac{13}{18}\left(iv\right)\frac{1}{2}$
• D. $\left(i\right)\frac{11}{13}\left(ii\right)\frac{11}{18}\left(iii\right)\frac{11}{17}\left(iv\right)\frac{1}{2}$

Answer 1

Answer: (C)

$\left(i\right)\frac{7}{18}\left(ii\right)\frac{11}{18}\left(iii\right)\frac{13}{18}\left(iv\right)\frac{1}{2}$

Since, total no. of balls = 5+7+4+2 = 18
(i) No. of balls that are white or blue = 5+2 = 7
$\therefore$ Required probability = $\frac{7}{18}$
(ii) No. of balls that are red or black = 7+4 = 11
$\therefore$ Required probability = $\frac{11}{18}$
(iii) No. of balls that are not white = 7+4+2 = 13
$\therefore$ Required probability = $\frac{13}{18}$
(iv) No. of balls that are neither white nor black = 7+2 = 9
$\therefore$ Required probability = $\frac{9}{18}=\frac{1}{2}$