An urn contains 3 red balls and 3 green balls. A ball is drawn randomly. If red, kept out. If green, returned to urn. Second ball is drawn, repeat. What is the probability that the urn contains exactly 4 balls after 3 are drawn?
This is exactly as asking what's the probbly of drawing exactly 2 red balls in three intents. Let us denote $\displaystyle B_i$ the ball drawn in the i-th intent.
We have the following drawings that'll give 4 balls after 3 drawings:
$\displaystyle B_1=Red , B_2=Red, B_3=Green$
$\displaystyle B_1=Red , B_2=Green , B_3=Red$
$\displaystyle B_1=Green , B_2=Red , B_3=Red$
Now evaluate each event's probability. For example, in the first event the probability is $\displaystyle \frac{1}{2}\cdot\frac{2}{5}\cdot\frac{3}{4}$...can you see why?
Tonio