Given:
Bag \[1:3\] red balls,
Bag \[2:2\] red balls and \[1\] white ball
Bag \[3:3\] white balls
Now, let E1, E2 and E3 be the events of choosing Bag \[1\], Bag \[2\] and Bag \[3\] respectively and a ball is drawn from it.
And, we have
\[P({{E}_{1}})\text{ }=\text{ }1/6,\text{ }P({{E}_{2}})\text{ }=\text{ }2/6\]and \[P({{E}_{3}})\text{ }=\text{ }3/6\]
Therefore, the required probabilities are \[7/18\] and \[11/18\].