Probabilities: Compound Events

In this post I solved three probability questions using the basic principle of discrete probability: find the ratio of favourable to total cases. The first 2 examples were very simple, but the third involved a few different steps.

The counting strategy works, and is probably the simplest way to solve that problem. But there are times when such direct calculation may not be possible or desirable. In those situations, it is useful to know a more general technique, that looks at the event “2 Red balls and 1 Black” as a composite of smaller and simpler events, and finds its probability by operating over the probabilities of the smaller ones. This is going to be a very algebraic post, and so I’ll use shorter symbols to describe the events.

To start us off, I will call the event “2 Red balls and 1 Black” simply E.

It is useful to think of events as Sets of states. I will use the same name for an event and its respective set. For a set S, the number of elements in it is |S|. If N is the total number of states, then the probability of event S  is the probability of choosing a state from S at random: |S|/N.

As I did in the previous post, I can divide the set E into three sets: RRB, RBR, BRR. Because all of these are disjoint sets (a state falls into exactly one such set), the probability of E is

This is a principle to keep in mind:

If an event is the union of several disjoint sub-events, its probability is the sum of the probabilities of its sub-events.

Now, look at each of these sub-events and observe that each of them requires three different conditions. Event RRB, for example, only happens when

• we draw a Red ball from Urn 1 AND
• we draw a Red ball from Urn 2 AND
• we draw a Black ball from Urn 3

Each of these conditions corresponds to a smaller event that corresponds only to one Urn, and is not affected by whatever the outcome of the other urns. We say that these events are independent. In set terms, this event will be satisfied for any red ball of urn 1, and any red ball of urn 2, and any black ball of urn 3. So,

This is the second principle to remember:

If an event is the intersection of independent sub-events, its probability is the product of the probabilities of each sub-event.

In this fashion, I can reduce the original event E to a condition on very simple events, each of them representing an experiment on a single urn. These are as simple as the first two challenges in here, and we are in familiar territory. We simply divide the favourable cases by the total number of cases, to find:

P[R1] = 4/7

P[B1] = 3/7

P[R2] = 5/9

P[B2] = 4/9

P[B3] = 4/8

P[R3] = 4/8

P[RRB] = P[R1] x P[R2] x P[B3] = 4/7 x 5/9 x 4/8 = 10/63

P[RBR] = P[R1] x P[B2] x P[R3] = 4/7 x 4/9 x 4/8 = 8/63

P[BRR] = P[B1] x P[R2] x P[R3] = 3/7 x 5/9 x 4/8 = 5/42

P[E] = P[RRB] + P[RBR] + P[BRR] = 10/63 + 8/63 + 5/42 = 17/42.