I'm trying to solve a problem two different ways, and I can'd seem to figure out where I'm going wrong.
I have 4 buckets (A,B,C,D), and 4 identical marbles. Each marble has an equal chance of being put in any of the 4 buckets, and each is placed independently (each bucket can have 0-4 marbles placed in it, with a total of 4 across all buckets)
I need to calculate the probability of bucket A being empty after the four marbles are placed.
My intuition says I can calculate the probability of the marble being placed in to any bucket except A as $\frac{3}{4}$. Then I can multiply the probabilities together since they are independent. So I can do $\left(\frac{3}{4}\right)^ 4$, which seems right.
But I think I should also be able to get the answer by calculating the number of ways that leaves A empty divided by the total number of ways the marbles can be placed in the 4 buckets. When I do that I get:
total ways:
$\left(\dbinom{4}{4}\right)$ $\longrightarrow$ $\dbinom{7}{4}$ $\longrightarrow$ 35
ways with A empty: $\left(\dbinom{3}{4}\right)$ $\longrightarrow$ $\dbinom{6}{4}$ $\longrightarrow$ 15
But that gives me $\frac{15}{35}$, which is not the same as $\left(\frac{3}{4}\right)^ 4$
I'm guessing I am over counting or something else dumb, but I'm really stumped. Thanks for the help!
Best Answer
This is a Bernoulli process, a stochastic process which has two outcomes, success ("the marble is in $A$!") and failure ("the marble is not in $A$."). Let $s$ denote the probability of success and $f$ denote the probability of failure.
The formula for the probability of $k$ successes in $n$ trials is $$Pr[k\mbox{ successes in }n\mbox{ trials }] = \binom{n}{k}s^kf^{n-k}.$$
Where did this come from? There are $\binom{n}{k}$ different ways of arranging those $k$ successes among the $n$ tries. By independence, each of those different ways has probability $s^kf^{n-k}$ --- there are $k$ successes and the rest, all $n-k$ of them, are failures. Since we're not interested in the order of the successes, just the total number, we can add up the probabilities of all the different ways of getting $k$ successes in $n$ trials to arrive at the formula.
In your case, $s = \frac{1}{4}$, $f = \frac{3}{4}$, and $n=4$. Now we see that your first idea is correct: $$Pr[0\mbox{ successes }] = \binom{4}{0}\bigg(\frac{1}{4}\bigg)^0\bigg(\frac{3}{4}\bigg)^4 = \bigg(\frac{3}{4}\bigg)^4.$$
What about your second idea? Let's think this through. For the denominator, we want to count the number of ways of distributing $4$ marbles among $4$ buckets. Let's think of this as a process: first we'll toss the first marble, then we'll toss the second marble, and so forth. How many options does the first marble have? $4$. The second? Again, $4$. So on. By the independence of the marble dropping, there are $4^4$ possible outcomes for this experiment. Repeating the argument, there are $3^4$ outcomes in which the $A$ bucket does not receive any marbles. Hence, $\frac{3^4}{4^4} = \big(\frac{3}{4}\big)^4$, just as we computed above.
Edit: After reading Andre Nicolas' answer, I want to add the following. One can only use counting to compute probability if all outcomes are equally likely. But as he notes, not all the multisets are equally likely! Some may be arrived at via multiple routes through the experiment's tree; for example, $\{A,A,A,A\}$ can only happen one way, while $\{A,B,C,D\}$ can happen $4!$ ways, since, e.g., $\{A,B,C,D\}=\{D,C,B,A\}$. So if you want to think in terms ofmultisets, you need to modify your counting. Take the different types of multisets (all one bucket, two buckets, three buckets, all four buckets), count the number of multisets in each type and multiply by the number of ways each multiset can occur, then add everything together.