Question

$\left(A\right)$ A box contains $3$ blue, $2$ white and $4$ red marbles. If a marble is drawn from the box. What is the probability that it will be
$\left(i\right)$ White $\left(ii\right)$ Blue $\left(iii\right)$ Red?
$\left(B\right)$ A bag contains $5$ red, $8$ green and $7$ white balls. One ball is drawn at random from the bag, find the probability of getting
$\left(i\right)$ A white ball or a green ball
$\left(ii\right)$ Neither a green ball nor a red ball.

• A. $\left(A\right)\left(i\right)\frac{1}{9}\left(ii\right)\frac{2}{3}\left(iii\right)\frac{5}{9}\left(B\right)\left(i\right)\frac{6}{13}\left(ii\right)\frac{11}{20}$
• B. $\left(A\right)\left(i\right)\frac{5}{9}\left(ii\right)\frac{1}{4}\left(iii\right)\frac{2}{11}\left(B\right)\left(i\right)\frac{1}{2}\left(ii\right)\frac{5}{19}$
• C. $\left(A\right)\left(i\right)\frac{2}{9}\left(ii\right)\frac{1}{3}\left(iii\right)\frac{4}{9}\left(B\right)\left(i\right)\frac{3}{4}\left(ii\right)\frac{7}{20}$
• D. None of these

Answer 1

The correct option is C $\left(A\right)\left(i\right)\frac{2}{9}\left(ii\right)\frac{1}{3}\left(iii\right)\frac{4}{9}\left(B\right)\left(i\right)\frac{3}{4}\left(ii\right)\frac{7}{20}$

(a) Total no. of marbles=3+2+4=9
(1) No. of white marbles=2
$\therefore$ probability of getting a white marble=$\frac{2}{9}$
(2) No. of blue marbles=3
$\therefore$ probability of getting a blue marble= $\frac{3}{9}=\frac{1}{3}$
(3) No. of red marbles=4
$\therefore$ probability of getting a red marble=$\frac{4}{9}$
(b)Since, Total no. of balls = 5+8+7=20
(1) No. of balls that are white or green = 8+7=15
$\therefore$ probability of getting a white or green ball = $\frac{15}{20}=\frac{3}{4}$
(2) No. of balls that are neither green nor red =7
$\therefore$ probability of getting a ball that are neither green nor red = $\frac{7}{20}$