# I need to understand the difference between these two approaches for calculating probability

There are five empty boxes. Balls are placed independently one after another in randomly selected boxes. Find the probability that the fourth ball is the first to be placed in an occupied box.

I can think of one way of solving this problem as:

Placing the first ball in a box will always be unoccupied (hence the probability 1)

Secondly, the ball should be placed in an unoccupied box, so $$\frac 4 5$$ as there are only 4 box left.

Similarly, the third ball should be placed in an unoccupied box which means we have 3 boxes left, hence probability $$\frac{3}{5}$$

Now, so far we have occupied 3 boxes so, the fourth ball should be placed in one of the occupied 3 boxes, and the probability for this is $$\frac{3}{5}$$

All above events must occur, so the final probability is 4/5∗3/5∗3/5=36/125.

I got this answer and this matches with my booklet.

Another approach for solving this problem

I have in total $$5$$ boxes. I know that so far I have placed 4 balls in these 5 boxes. So, the total number of ways to do that is $$5^4$$.

I need to have one of the box contain 2 balls. So, remaining, 4 boxes must contain other 2 which can happen in $$4\choose2$$ $$4^2$$ ways. Hence, the final probability is: $$\frac{{4\choose2}4^2}{5^4}$$.

My question is that why is this answer not consistent with the previous approach. Is there anything that I am missing here?

• Thanks for reminding me. Feb 22 '21 at 9:17

In your second approach, you mix ordered and unordered (i.e. distinguishable and indistinguishable) things (balls, boxes).

I have in total $$5$$ boxes. I know that so far I have placed 4 balls in these 5 boxes. So, the total number of ways to do that is $$5^4$$.

Sure, but it means that you supposed the balls and boxes are labeled (ordered, distinguishable):

and — as an example — the configuration $$1 \mapsto \color{red}A\\ 2 \mapsto \color{blue}B\\ 3 \mapsto C\\ 4 \mapsto C\\$$ is for you different from

$$1 \mapsto \color{blue}B\\ 2 \mapsto \color{red}A\\ 3 \mapsto C\\ 4 \mapsto C\\$$

I need to have one of the box contain 2 balls.

Sure, but which one of the 5 labeled boxes? And which of the first 3 numbered balls you put to it as the first ball? There are $$\color{red}{(5\times 3)}$$ possibilities!

... So, remaining, 4 boxes must contain other 2 which can happen in $$4\choose2$$ $$4^2$$ ways.

Sure, only your calculation is somewhat strange:

• $$4\choose2$$ is OK (all possible pairs of remaining boxes), but

• the second multiplier is simply 2,

because there are exactly $$2$$ possibilities how to put the other 2 balls individually into a particular pair of boxes:

• ball $$x$$ into box $$X$$ and ball $$y$$ into box $$Y$$, or
• ball $$x$$ into box $$Y$$ and ball $$y$$ into box $$X$$.

As a result you have $$\color{blue}{\left({4\choose 2} \times 2\right)}$$ choices for the remaining 4 boxes and the remaining 2 balls.

Now your result will be correct, as the numerator of your final fraction is $$\color{red}{(5 \times 3)} \times \color{blue}{\left({4\choose 2} \times 2\right)}$$.

• You explained it really well. Thank you so much. Feb 22 '21 at 9:20
• @userNoOne, you're welcome. Feb 22 '21 at 11:05

In the second approach, possibilities include not only the fourth ball in the occupied box but also the others, e.g. balls 2 and 3. An correct example way of calculating by counting is:

• First choose three boxes, and put the balls into them in some order: $${5\choose 3}3!$$
• Pick one of the occupied boxes and put the fourth one: $${3 \choose 1}$$

That makes $$\frac{5\times 4 \times 3\times 3}{5^4}=\frac{36}{125}$$

• I understand @gunes, Thank you so much Feb 22 '21 at 9:18