Answer to Question #326687 in Statistics and Probability for Minnie

Question

If 4 cards are dealt from a standard deckof 52 playing cards, what is the probability that 

3 will be diamonds?

Using Hypergeometric Distribution

Answer 1

Suppose that $X$ is a random variable that has a hypergeometric distribution with parameters $N$, $K$. $N$ is a size of population, $K$ is a number of objects of a certain type. $n$ is a number of draws and $k$ is a number of successes(the objects of a certain type). We have: $P(X=k)=C_K^k\frac{C_{N-K}^{n-k}}{C_N^n},$ where $C_i^j$ denotes a binomial coefficient. In our case we have: $N=52$, $K=13$, $n=4$, $k=3.$ Thus, $P(X=k)=C_{13}^3\frac{C_{39}^{1}}{C_{52}^4}=\frac{13!}{3!10!}\frac{39}{\frac{52!}{4!48!}}\approx0.0412$