marble probabilities A bucket has 3 marbles numbered 1, 2, and 3. Choose one and put it back along with another marble with the same number. Now choose a second and put it back along with a marble with the sum of your first two picks. Choose one more marble.
Find P(the first number was a 3 | third was a 3).
So what I did was draw out the entire tree of possibilities and then used the equation P(1st =3 AND 3rd =3)/P(3rd =3) and got 4/9 as the answer. Can anyone confirm this is correct? Also, is there an easier was to do this without drawing out the whole tree?
Thanks!
 A: Your answer is correct.
Your task can be somewhat simplified if you notice that you really aren't concerned about any other possible third draws than $3$, so there's no need to go through a third stage of branching. That eliminates $24$ branches.
Another thing that can simplify the task is the observation that if your first marble was $3$, then regardless of what you draw second, there will be two $3$ marbles out of a total of five marbles when you make your third draw. That lets you immediately conclude that $$P(\text{first and third draws were }3)=\frac13\cdot\frac25=\frac2{15}.$$ Then you add those with the probabilities of the other six ways that $3$ will be your third draw, which you can calculate by using the tree.
A: One way to avoid drawing the tree is counting the number of final leaves that would be in your tree. Format one triplet of draws as abc and denote an arbitrary draw by _:
3_3 has $2 \times 4 = 8$ leaves, since the second draw is arbitrary and by the third draw there are two 3-marbles.
x_3, where the x is not a 3, has $2 \times 4 = 8$ leaves
213 has 1 leaf on top of the one counted by the second case (since we have an extra 3)
123 has 1 leaf by similar argument
Result: $$\frac{8}{8 + 8 + 1 + 1} = \frac{8}{18} = \frac{4}{9}$$
