Question

Bag $A$ contains three red and four white Balls; bag $B$ contains two red and three white balls. If one ball is drawn from bag $A$ and two balls from bag $B$, find the probability that:
(i) One ball is red and two balls are white;
(ii) All the three balls are of the same colour.

• A. i) $\frac{33}{70}$ and ii) $\frac{3}{14}$
• B. i) $\frac{22}{31}$ and ii) $\frac{1}{3}$
• C. i) $\frac{1}{2}$ and ii) $\frac{1}{2}$
• D. i) $\frac{8}{25}$ and ii) $\frac{2}{7}$

Answer 1

The correct option is A i) $\frac{33}{70}$ and ii) $\frac{3}{14}$

Possible selection are as follows
(i) Red from bag $A,2$ white from bag $B$
$1$ White from bag $A,1$ white from bag $B$
$1$ Red from $B$
$\therefore$ $P($one ball is red and two balls are white$)$
$=\frac{{}^3{C}_1}{{}^7{C}_1}\times \frac{{}^3{C}_2}{{}^5{C}_2}+\frac{{}^4{C}_1}{{}^7{C}_1}\times \frac{{}^3{C}_1{\times }^2{C}_1}{{}^5{C}_2}$

$=\frac{3}{7}\times \frac{3}{10}+\frac{4}{7}\times \frac{3\times 2}{10}$

$=\frac{9}{10}+\frac{24}{70}=\frac{33}{70}$

(ii) Possible selections are as follows
$1$ Red from bag $A,2$ Red from bag $B$
$1$ White from bag $A,2$ White from bag $B$
$\therefore$ $P($All the three balls are the same colour$)$

$=\frac{{}^3{C}_1}{{}^7{C}_1}\times \frac{{}^2{C}_2}{{}^5{C}_2}+\frac{{}^4{C}_1}{{}^7{C}_1}\times \frac{{}^3{C}_2}{{}^5{C}_2}$

$=\frac{3}{7}\times \frac{1}{10}+\frac{4}{7}\times \frac{3}{10}$

$=\frac{3}{70}+\frac{12}{70}=\frac{15}{70}=\frac{3}{14}$