The **sample space** of an experiment is the set of all possible outcomes of the experiment.

For example, suppose we roll a dice one time. The sample space of possible outcomes includes:

Sample space = 1, 2, 3, 4, 5, 6

Using notation, we write the symbol for sample space as a cursive S and the outcomes in brackets as follows:

*S* = {1, 2, 3, 4, 5, 6}

**Examples of Sample Spaces**

Here are a few more examples of sample spaces:

**Example 1: Coin Toss**

Suppose we toss a coin one time. If we let *H* = the coin lands on heads and *T* = the coin lands on tails, then the sample space for this coin toss is:

*S* = {H, T}

**Example 2: Marbles in a Bag**

Suppose we randomly select one marble from a bag that contains three marbles: a red marble, a green marble, and a blue marble. If we let *R* = red, *G* = green, and *B* = blue, then the sample space is:

*S* = {R, G, B}

**Example 3: Coin Toss & Dice Roll**

Suppose we toss a coin and roll a dice at the same time. If we let *H1* represent the result of a “Head” and a “1” then the sample space for the results is:

*S* = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

**The Fundamental Counting Principle**

The **fundamental counting principle** is a way to calculate the total number of potential outcomes of an experiment.

This principle states that if event A has *n* distinct outcomes and event B has *m* distinct outcomes, then the total number of potential outcomes can be calculated as:

Total outcomes = *m* * *n*

**Example 1: Coin Toss & Dice Roll**

For example, if we toss a coin and roll a dice at the same time, then the total number of outcomes in the sample space can be calculated as:

Total outcomes = (2 ways a coin can land) * (6 ways a dice can land) = **12** possible outcomes.

We wrote out these 12 outcomes in the previous example:

*S* = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

**Example 2: Counting Outfit Combinations**

This principle can also be used to calculate the total outcomes in a sample space for more than two events.

For example, suppose a random drawer contains 3 different shirts, 4 different pants, and 2 different socks. If we randomly select one article of clothing each without looking, the total number of possible outfits would be calculated as:

Total outfits = 3 * 4 * 2 = **24** possible outfits

**Visualizing Sample Spaces with Tree Diagrams**

When the number of outcomes in a sample space is large, it can be helpful to construct a **tree diagram** to visualize the different combinations of outcomes.

For example, suppose a closet contains 2 different shirts, 2 different pants, and 2 different socks. If we randomly select one article of clothing each without looking, the total number of possible outfits could be visualized as:

This diagram helps us visualize the eight different potential outcomes in the sample space.

We can also use the fundamental counting principle to confirm that there should be eight different outcomes:

Total outcomes = 2 shirts * 2 pants * 2 socks = 8 possible outfits

**Calculating Probabilities of Outcomes in Sample Spaces**

Once we have identified the sample space of some experiment, we can calculate the probability of some event *A* occurring by using the following formula:

P(A) = (Sample Space of A) / (Total Sample Space)

For example, suppose we roll a dice one time. The sample space can be written as:

*S* = {1, 2, 3, 4, 5, 6}

If we define event A as the dice landing on the number “2”, then the sample space of event A can be written as:

*S* = {2}

Thus, the probability of event A occurring can be calculated as:

P(A) = 1/6

If we define event A as the dice landing on an even number, then the sample space of event A can be written as:

*S* = {2, 4, 6}

Thus, the probability of event A occurring can be calculated as:

P(A) = 3/6

OK, I am flipping two coins at the same time. The sample space consists of {(H,H), (H,T), (T,T)} because the two coins are indistinguishable. Obviously something seriously wrong is going on here because in reality the probability of {(H,T)} is NOT 1/3 as you might calculate from the above. The correct result seems to depend on knowing which coin is first and which is second so that there are four possible outcomes. How can we avoid such pitfalls.

Another example: rolling two dice at once where the dice are indistinguishable, so that there are only 21 possible outcomes instead of the 36 that are possible when there is a first die and a second die.