## Designing probability functions for the experiments for which all the outcomes are equally likely

In the last article, we discussed the probability function which takes into account the probability value as the relative frequency. In this article, we discuss the probability function for equally likely outcomes. Let’s get started:

We all have seen experiments, where the outcomes are equally likely for example when an un-biased coin is tossed, both the outcomes are equally likely. The sample space just contains two outcomes i.e Heads, Tails

Since both, the outcomes are equally likely the probability of their occurrence would be the same

Now that we know the probability of heads and tails, we can compute the probability of all the possible subsets of the sample space:

The point being made over here is that once we know the probability of individual outcomes, we can leverage additivity action for computing the probability of any event/subset of the sample space.

Similarly, we can consider the case of rolling a dice, here 6 outcomes are possible and all of the outcomes are equally likely

Let’s look at some subsets from this sample space

Say we are interested in the event of getting an even number as the outcome, then this event would consist of the cases when the outcome is either 2 or 4 or 6 and since these three possibilities form the disjoint outcomes, we can sum up their individual probabilities to get the probability of the event that the outcome is even:

Similarly, if we have an event that the outcome is odd(O) then we’ll have the numbers {1, 3, 5} as part of the interest, and since these three outcomes are disjoint, we can compute the probability of the event that the outcome is odd as the sum of the probability of individual elements i.e

P(O) = P(1) + P(3) + P(5)

So, given the probability of individual events, we can then compute the probability of all the subsets of the sample space.

In the case of equally likely outcomes, the probability of any outcome is given as ‘1/n’ where ‘n’ is the total number of outcomes.

When we talk of the axioms of probability we talk about the events so we need a formula for P(A) where ‘A’ could be any event, right now we have P(A) only for those case which corresponds to the outcomes so only for singleton events

We can express ‘E’ as the union of ‘k’ singleton events and since these are disjoint events we can sum the probability of events individually to get the probability of ‘E’:

Now that we have a probability function for the case of equally likely outcomes, the next question is “does this function satisfies the axioms of probability?”

As pointed in the above image, this function satisfies the first two axioms.

The third axiom is as follows:

To prove this one, let’s assume we have ‘k₁’ elements in ‘A₁’ and ‘k₂’ elements in ‘A₂’, then using the function’s equation, we have the probability of the union as below:

Let’s look at some examples:

Say we are interested in knowing the probability of getting a black card from a deck of 52 cards and there are 26 black cards in a deck of 52 cards, the probability of drawing a black card would be given as:

Suppose we are dealing with 3 hands and we are interested in knowing the probability of getting 3 aces:

In this case, the total of possible outcomes is given by ⁵²C₃

And the number of ways of choosing 3 aces out of a total of 4 aces is given as ⁴C₃ which would be 4

Let’s take one more example:

R’ is the radius of the outer circle and ‘r’ is the radius of the inner circle

We are talking about an average person throwing a dart at a dartboard so we are saying that the person has an equally likely chance of hitting any point on the dartboard, with that assumption the probability would be given as the ratio of favorable outcomes which lies in the red circle to the total number of outcomes which lies in the outer circle

Now there are infinite points in both the circles, what we can do is to take the ratio of the areas of the two circles

In this article, we discussed the probability function for the experiments for which all the outcomes are equally likely.