Firstly, balls are drawn with replacement.

First ball you draw can be any of the 3. So probability of drawing first ball = 3/3 = 1

For each of the 3 possible scenarios of drawing the first ball (either ball 1, 2 or 3),

the second ball drawn must be one of the other two balls (cannot be the same ball, if not the draw will stop)

So, probability of drawing second ball that is not the same as first ball = 2/3

For the third draw, the ball drawn has to be the ball that has not been drawn before yet, in order to proceed to the fourth draw.

So, probability of drawing third ball that is not the same as first or second ball = 1/3

For the fourth draw, drawing any ball will always result in the draw being stopped (since all 3 balls have been drawn once before already in the first three draws)

So probability = 3/3 = 1


And that is why you have
3/3 x 2/3 x 1/3 x 3/3

Another way to do this question.


Let's assume we want to draw ball 1 first.

Probability of getting ball 1 on first draw
= 1/3

For second draw, we can only pick ball 2 or 3.

Probability of getting ball 2 or 3 on second draw = 2/3

For third draw,

If ball 2 was gotten in the second draw then you can only draw ball 3 now.
If ball 3 was gotten in the second draw, you can only draw ball 2 now.

Either way, probability of getting (ball 2 if second draw was ball 3 or ball 3 if second draw was ball 2)

= 1/3

For fourth draw, any ball will do to end the draw

Probability of drawing a ball on fourth
draw = 3/3 = 1

Probability for this scenario
= 1/3 x 2/3 x 1/3 x 3/3
= 2/27


There are two other scenarios
(one whereby you draw ball 2 first, the other whereby you draw ball 3 first.)

So the probability = 3 x 2/27 =
= 6/27
= 2/9