Question 901356
Q1) Case 1: 2 Red Balls are transfered from Urn1 to Urn 2.
   Case 2: 1 Red Ball and 1 White Ball are transfered from Urn1 to Urn 2.
   Case 3: 2 Red White Balls are transfered from Urn1 to Urn 2.

P(Case 1)= P (First Draw Red & Second Draw Red)=P(First Draw Red) *(Second Draw Red/First Draw Red) = 4/6 * 3/5 = 2/5.

P(Case 2)= P ((First Draw Red & Second Draw white) or (First draw White & Second draw Red))= P(First Draw red)*P(Second Draw White/First Draw Red) + P(First Draw White)*P(Second Draw Red/First Draw White) =4/6 * 2/5 + 2/6 * 4/5 = 4/15 + 4/15 = 8/15.

P(Case 3)= P (First Draw White & Second Draw White)=P(First Draw White) *(Second Draw White/First Draw White) = 2/6 * 1/5 = 1/15.

Note That P(Case 1)+ P(Case 2) + P(Case 3) = 1. (obviously).

So, 1)P(Draw Red Urn 2) = P(Draw Red Urn 2/Case 1) * (P(Case 1) + P(Draw Red Urn 2/Case 2) * (P(Case 2) + P(Draw Red Urn 2/Case 3) * (P(Case 3) = 5/8 * 2/5 + 4/8 * 8/15 + 3/8 * 1/15 = 1/4 + 4/16 + 1/40 = 21/40.
    2)a)P(Case 3/Draw Red Urn 2) = (P(Draw Red Urn 2 /Case 3)* (P(Case 3))/ P(Draw Red Urn 2) = (1/40)/(21/40) =1/21
    2 b) P(Case 2/Draw Red Urn 2) = (P(Draw Red Urn 2 /Case 2)* (P(Case 2))/ P(Draw Red Urn 2) = (4/16)/(21/40) =10/21.
    2 c)  P(Case 1/Draw Red Urn 2) = 1 - P(Case 3/Draw Red Urn 2) - P(Case 2/Draw Red Urn 2) = 1 - 1/21 -10/21 =10/21.

You can use Question 1 to solve question 2.