(a) Let the events are

${R_1} = A$ red ball is drawn from urn $A$ and placed in $B$

${B_1} = A$ black ball is drawn from urn $A$ and placed in $B$

${R_2} = A$ red ball is drawn from urn $B$ and placed in $A$

${B_2} = A$ black ball is drawn from urn $B$ and placed in $A$

$R = A$ red ball is drawn in the second attempt from $A$

Then the required probability

$ = P({R_1}{R_2}R) + ({R_1}{B_2}R) + P({B_1}{R_2}R) + P({B_1}{B_2}R)$

$ = P({R_1})P({R_2})P(R) + P({R_1})P({B_2})P(R) + P({B_1})P({R_2})P(R) + $

$P({B_1})P({B_2})P(R)$

$ = \frac{6}{{10}} \times \frac{5}{{11}} \times \frac{6}{{10}} + \frac{6}{{10}} \times \frac{6}{{11}} \times \frac{5}{{10}} + \frac{4}{{10}} \times \frac{4}{{11}} \times \frac{7}{{10}} + \frac{4}{{10}} \times \frac{7}{{11}} \times \frac{6}{{10}}$

$ = \frac{{32}}{{55}}$.