Combination tells us the number of ways we can create a subset of \(k\) elements given a set of \(N\) elements.
Suppose we have a set of \(N\) elements. From this set we want to create a subset of \(k\) elements. It is given by the formula \({N \choose k} = \frac{N!}{K! (N-k)!}\). This is also called the binomial coefficient and it tells us how many \(k\) element subsets are there in a set containing \(N\) elements.
Example 1:
In an experiment, we toss a coin \(N\) number of times. Out of \(N\) tosses what is the probability of \(k\) heads.
Let the probability of heads be \(P(H) = p\).
Let the probability of tails be \(P(T) = (1-p)\).
Now let’s say that the result of an experiment where \(N=6\) and \(k=4\) be HHTTHTH.
\[\begin{align} P(HHTTHTH) &= p.p.(1-p).(1-p).p.(1-p).p \\ &= p^4 (1-p)2 \end{align}\]For any combination of \(4\) heads, the probability will be \(p^4(1-p)^2\)
Therefore
\[\begin{align} P(k=4) &= (\text{No. of k heads}) \times p^k(1-p)^{N-k} \\ &= {N \choose k}p^k(1-p)^{N-k} \end{align}\]Example 2:
Event B: \(3\) out of \(10\) tosses were heads.
Given that event \(B\) has occured, what is the conditional probability that event A: the first two tosses were heads has occured?
Number of elements in \(B\) = \({10 \choose 3}\)
Probability of \(B\) occuring is \(P(B) = {10 \choose 3}p^3(1-p)^{7}\)
In event \(A\), the first two tosses were heads. Therefore the number of elements in the event \(\mid A \cap B \mid\) is the number of choices for the third head since we already know that the first two tosses were heads. Since the third toss can be in any of the remaining \(8\) positions, \(\mid A \cap B \mid = 8\)
Therefore the probability that the first two tosses are heads in the event \(B\) is \(\frac{\mid A \cap B \mid}{\mid B \mid} = \frac{8}{10 \choose 3}\)