Starting a new term in university comes with 4 brand new topics to learn. One of these is Probability. I’d like to share a little bit of (simple) maths that I learnt in my lecture last week on calculating probabilities when there are equally likely outcomes.

I will not define what a probability measure or space is rigorously, as it is quite irrelevant for the simple examples I’m considering today. But, I will introduce the following notation:

Let Ω be a set and *F* be a set of all subsets of Ω. The elements of Ω are called **outcomes** and the elements of* F* are called **events**. As we are using this as a probability model, Ω will be an abstraction of a real set of outcomes, and *F* will model observable events.

Then let us write, **P(A)** as the probability of the event A occurring. Note that:

P(A) = |A|/|Ω|.

Now, let us consider a few examples of how we would use this. In the following examples we use symmetry and randomness to argue that there are equally likely outcomes.

### Throwing a Die

The throw of a die has six possible outcomes, and so

Ω = {1, 2, 3, 4, 5, 6} and P(A) = |A|/6 for A ⊆ Ω.

Therefore, for example P({2, 4, 6}) = **1/2**.

### Balls from a Bag

A bag contains *r* balls. Say we draw *k* balls from this bag without looking. If we consider that the balls are labelled by {1, . . . , n}, we have selected a subset of {1, . . . , n} of size *k*.

Therefore, Ω is the set of subsets of {1, . . . , n} of size *k*, so

with the probability of a individual outcome being 1/|Ω|.

### Pack of Cards

In an idealised well-shuffled pack of cards, every possible order is equally likely, so it is modelled by the set Ω of permutations of {1, . . . , 52}.

Because of this, we can see that |Ω| = 52!

**What is the probability of the event A that the first two cards are aces?**

There are 4 choices for the first ace, 3 choices for the second, and the rest can come in any order. So,

|A| = 4 × 3 × 50!

and

P(A) = |A|/|Ω| = 12/(52 × 51) = **1/221**.

### Distribution of Largest Digit

Suppose that *r* digits are chosen from a table of random numbers. **What is the probability that ***k* is the greatest digit drawn?

Firstly, model the set of possible outcomes by

Ω = {0, 1, . . . , 9}^{n}

Consider the event Ak = [no digit exceeds *k*], and the event Bk = [the largest digit is *k*] So:

|Ω| = 10^{n} ;

|Ak| = (k + 1)^{n} ;

|Bk| = |Ak \ Ak−1| = (k + 1)^{n} − k^{n}.

Therefore, P(Bk) = **[(k + 1)**^{n} − k^{n}]/10^{n}.

Hope you enjoyed today’s post. M x