Question 1158866
I assume that if the number on the top of the die that was tossed is neither 3 nor 6, no marbles are drawn. Your turn is over. better luck next time.
 
Each of the 6 faces/numbers on a fair die has the same probability to appear on top : {{{1/6}}} .
The probability of getting a 3 is {{{1/6}}} and the probability of getting a 6 is {{{1/6}}}  for each draw, the probability of getting either a 3 or a 6 is {{{1/6+1/6=2/6=1/3}}} .
A teacher might write that as {{{P(3or6)=1/3}}} .

Let's think of what happens in the {{{1/3}}} of the cases when the game gets to the point when marbles can be drawn.

Urn B starts with {{{1+2=3}}} marbles.
After transferring one marble from urn B to urn A,
urn A will have 9 marbles, either 6red and 3 while or 5red and 4 white,
depending on the color of the marble drawn from urn B

The entire play sequence can be diagrammed as follows
{{{drawing(600,360,0,100,-60,0,
locate(1,-18.5,start),locate(21,-1,Die),
locate(39,-1,urn),locate(45,-1,B),
locate(57,-1,urn),locate(63,-1,A),
locate(75,-1,Probability),locate(75,-6,4/6=2/3),
line(9,-20,20,-10),line(9,-20,20,-30),
locate(20,-5,neither),locate(30,-5,3)
locate(21,-7,nor),locate(27,-7,6),
locate(23,-28,3),locate(21,-30,or),locate(25,-30,6),
locate(40,-21,red),locate(72,-14,(1/3)(red(1/3))(2/3)=highlight(2/27)),
line(28,-30,38,-23),line(28,-32,38,-40),
locate(39,-39,white),locate(72,-25,(1/3)(red(1/3))(1/3)=1/27),
line(45,-22,55,-17),line(45,-23,55,-28),
line(46,-41,55,-36),line(46,-41.5,55,-46.5),
locate(57,-15,red),locate(56,-26.5,white),
locate(57,-34,red),locate(56,-45.5,white),
locate(14,-15,"2/3"),locate(14,-22,"1/3"),
locate(33,-26,red("1/3")),locate(33,-33,green("2/3")),
locate(51,-19,"2/3"),locate(51,-24,"1/3"),
locate(51,-38,"5/9"),locate(51,-42,"4/9"),
locate(72,-33,(1/3)(green(2/3))(5/9)=10/81),locate(72,-44,(1/3)(green(2/3))(4/9)=highlight(8/81)))
)}}}
 
a. what is the probability that both marbles are red?
{{{(1/3)(red(1/3))(2/3)=highlight(2/27)}}}
b. what is the probability that both marbles are white?
{{{(1/3)(green(2/3))(4/9)=highlight(8/81)}}}
 
After the die shows a 3 or a 6, when drawing one of the 3 marbles from urn B, the probability of getting each one of them is {{{1/3}}} ,
so the probability of drawing the only one that is red is {{{1/3}}} .
A teacher could write that as {{{P("red|3 or 6")=red(1/3)}}} .
However, drawing that red marble only happens in  that happens in {{{red(1/3)}}} of the {{{1/3}}} of the plays when the die was a 3 or 6.
It happens in {{{(1/3)*(red(1/3))=1/9}}} of the plays.
A teacher could write that as {{{P("3 or 6",red)}}}={{{P(3 or 6)*P("red|3 or 6")=(1/3)*red(1/3)=1/9}}} .
What happens if the marble drawn from urn B is not red, and what is the probability of that?
If the marble drawn from urn B is not red, it must be one of the 2 white marbles.
The probability of that happening after getting a 3 or 6 on the die is {{{P("white|3 or 6")=green(2/3)}}},
and that will happen in {{{green(2/3)}}} of the {{{1/3}}} of plays when the die showed a 3 or a 6.
That happens in {{{(1/3)*green(2/3)=2/9}}} of the plays.
A teacher could write that as {{{P("3 or 6,white")=P("3 or 6")*P("white|3 or 6")=(1/3)*green(2/3)}}}={{{2/9}}} .
A teacher would say that red and white are "mutually exclusive" events, that it has to be one or the other, so if red happens {{{1/3}}} of the times , white has to happen the other {{{2/3}}} out of the {{{3/3=1}}} of the total of times.
 
What happens to urn A, after drawing a marble from urn B?
The game starts with {{{5+3=8}}} marbles in urn A.
After drawing a marble from urn B and putting it into urn A,
urn A will contain {{{8+1=9}}} marbles,
and the probability of drawing each of them is {{{1/9}}} .
 
We know that if the die shows a 3 or a 6 and the marble drawn from urn B (and added to urn A) is red, {{{5+1=6]}}} of the 9 marbles in urn A will be red and in that case, the probability of drawing one of them would be {{{6/9=2/3}}} .
The total probability of getting two red marbles is the product of probabilities of the desired outcome for all 3 steps (the die toss, urn B draw, and urn A draw):
{{{(1/3)(red(1/3))(2/3)=highlight(2/27)}}} .
 
If the marble drawn from urn B (and added to urn A) is white, {{{3+1=4}}} of the 9 marbles in urn A will be red and the probability of drawing one of them would be {{{4/9}}}
In that case, you got 2 white marbles, and the probability of that is the product of the probabilities for each step
{{{(1/3)(green(2/3))(4/9)=highlight(8/81)}}}