Combinatorial Problems

As mentioned in the Introduction, if we assume that the elementary events of an experiment with finitely many possible outcomes are equally likely, then the assignment of probabilities is quite simple and straightforward. For example, if we want the probability of drawing an Ace, when the experiment consists of the drawing of a card under the assumption that any card is as likely to be drawn as any other, then we can say that $$\frac{1} {52}$$ is the probability of drawing any of the 52 cards, and $$\frac{4} {52} = \frac{1} {13}$$ is the probability of drawing an Ace, since there are four Aces in the deck. We obtain the probability by taking the number of outcomes making up the event that an Ace is drawn and dividing it by the total number of outcomes in the sample space. Thus the assignment of probabilities is based on the counting of numbers of outcomes, if these are equally likely. Now the counting was very simple in the above example, but in many others it can become quite involved. For example, the probability of drawing two Aces if we draw two cards at random (this means “with equal probabilities for all possible outcomes”) from our deck is $$\frac{4\cdot 3} {52\cdot 51} =.0045$$ , since, as we shall see in the next section, 4 ⋅ 3 = 12 is the number of ways in which two Aces can be drawn and 52 ⋅ 51 = 2652 is the total number of possible outcomes, that is, of possible pairs of cards.