Combining cups and coins probabilities problem

Question

We have 13 cups and 5 coins. We randomly hide the 5 coins under the caps. We can hide multiple coins under a single cup, so there can be 0 or 1 or 2 or 3 or 4 or 5 coins under any single cap. We then ask someone to turn over any 5 cups. What is the probability of him finding 0 or 1 or 2 or 3 or 4 or 5 coins under the caps?

Text says:

Probability of each coin being under a cap is 5/13.

Let P(n) be the probability of n coins being hidden under revealed cups:

P(5) = ((5 / 13) ^ 5) * ((8 / 13) ^ 0) = 0.00841653

P(4) = ((5 / 13) ^ 4) * ((8 / 13) ^ 1) * 5 = 0.06733227

P(3) = ((5 / 13) ^ 3) * ((8 / 13) ^ 2) * 10 = 0.21546326

P(2) = ((5 / 13) ^ 2) * ((8 / 13) ^ 3) * 10 = 0.34474122

P(1) = ((5 / 13) ^ 1) * ((8 / 13) ^ 4) * 5 = 0.27579297

P(0) = ((5 / 13) ^ 0) * ((8 / 13) ^ 5) = 0.08825375

S(P) = 1.00000000

But I have no idea how the above formulas derive.

What is the mathematical description of this problem? Is it "combined probabilities" of two events?

Can someone please explain how those probabilities are calculated?

Answer 1

As @whuber said, these are probabilities from the binomial distribution. But the example may be a bit confusing: five cups are turned, and five coins are placed.

Slightly different example

To make the example a bit less opaque, change it a bit, say six coins instead of five are hidden under the cups. Let us try to calculate the probability that in all 2 out of 6 coins then are found under the five cups.

Probability of finding 2 coins under prespecified cups

It does certainly not matter for the probability if the person who turns the cups has decided beforehand which cups to turn without telling. So say, the five leftmost cups will be turned. Now, randomly hide the coins under the cups.

• Coin number 1 has probability $\frac{5}{13}$ to be discovered and $\frac{8}{13}$ to remain hidden. The same holds for coins number 2, 3, 4, 5, 6.
• the probability that coins number 1 and number 2 are discovered, while the other four remain hidden, is $\frac{5}{13}\cdot\frac{5}{13}\cdot\frac{8}{13}\cdot\frac{8}{13}\cdot\frac{8}{13}\cdot\frac{8}{13}$, because we place coins under cups independently of each other.
• the probability that coins number 3 and 5 are discovered, and coins 1, 2, 4, 6 remain hidden, is also $\left(\frac{5}{13}\right)^2\cdot\left(\frac{8}{13}\right)^4$.
• there are in all $\binom{6}{2}=15$ possibilities to choose which two coins out of five are discovered, and all these choices have the same probability. So, summing up the probabilities for all choices of two out of five, we get $ P(2\text{ out of }6)=15\cdot \left(\frac{5}{13}\right)^2\cdot\left(\frac{8}{13}\right)^4. $

Modelling as a two step process

If you want to consider both the the distribution of coins under the cups and the selection of cups to turn as random, you would interpret the probability that we just have calculated as a conditional probability: $ P(2\text{ out of }6 \text{ coins are discovered}\mid \text{the first five cups are turned}). $ This probability is equal to the probability if an other set of five cups is turned, and all sets of five cups are equally likely to be turned. Therefore, you would get the same result if you consider this as a two step random process.

General case

When you have $n$ coins, and the probability of discovering a coin is $p$, then the probability of finding $k$ out of the $n$ coins is $ P(k\text{ out of }n)=\binom{n}{k}p^k (1-p)^{n-k}. $