# Understanding Sample Space and Sample Points In tossing a fair coin, there are only two possible outcomes, a Head (H) and a Tail (T).  If we let S be the set of all possible outcomes of this event, then, we write the set of possible outcomes as S = {H,T}.

If two fair coins are tossed, then the outcomes can be both heads {H,H} or both tails {T,T}. It can also be a head first then a tail {H,T}, or a tail first and then a head {T,H}. So, in tossing two coins, we have the set of possible outcomes S = {{T,T}, {T,H}, {H,T}, {H,H}}.

As the number of tosses increases, listing gets more difficult. One of the strategies that can be used to remedy this problem is by creating a tree diagram. The following problem is solved using a tree diagram. Notice that it is a three-coin tossing problem in disguise (try replacing B with H and G with T).

Problem: In a family of 3 children, what is the probability are three girls? What is the probability the first two children are boys? In probability problems, it is important to know, and sometimes list, all the possible outcomes of an event or series of events. The set of all possible outcomes of an event is called the sample space.

In the problem above, the set {G, G, B} stands for a family with a girl as the eldest, a girl as the middle child, and a boy as the youngest. This particular set is called a sample point. A sample point is a possible outcome of an event. In the problem above, the sample space S has 8 sample points, and there is only 1 sample point having three girls. Therefore, in a family of three children, the probability of having three girls is 1 out of 8. It is also clear that the probability that the first two children are boys is 2 out of 8.

Tree diagrams are only effective in solving problems about a single event with binary outcomes, or a series of such events.  In other cases, different problem solving strategies are used. For instance, in solving 2-die problems, a table is needed. As shown below, in rolling two standard cubical dice (numbered 1 through 6), there are 36 sample points. From the table, the probability of having a sum of 12 is 1 out of 36, and the probability of having a sum of 7 is 6 out of 36. From the discussion above, it is clear that the probability of an event (such as tossing a coin), is equal to the number of favorable outcomes (like having two heads) over the sample space.

If we let the probability of an event be P(E), n  be the number of favorable outcomes, and S be the sample space, then $P(E) = \frac{n}{S}$.