Question

A box contains 4 Green, 5 Blue and 3 Red balls. A ball is drawn at random. Find the probability that the selected ball is
(i) Red in colour (ii) not Green in colour.

• A. (i)$\frac{1}{4}$ (ii) $\frac{2}{3}$
• B. (i)$\frac{4}{1}$ (ii) $\frac{2}{11}$
• C. (i)$\frac{5}{6}$ (ii) $\frac{5}{6}$
• D. (i)$\frac{13}{6}$ (ii) $\frac{2}{9}$

Answer 1

Answer: (A)

(i)$\frac{1}{4}$ (ii) $\frac{2}{3}$

$4$- Green $5$- Blue $3$- Red

$\left(i\right)$ Red in colour $\to$ total = $4+5+3=12$

$\Rightarrow \frac{3}{12}=\frac{1}{4}$

$\left(ii\right)$ not Green in colour:- means balls maybe blue or Red.

Blue + Red = $5+3=8$ and total = $12$

So : Answer is $\frac{8}{12}=\frac{2}{3}$