## Saturday, November 10, 2012

Consider the concept of  "theoretical probability".

You roll a fair 6 sided die. It is equally likely that any one of the 6 sides, denoted {1, 2, 3, 4, 5, 6} will appear face up.But the actual side that lands face up is not certain. If we roll the die 5 times, the possible outcomes are called the sample space and any occurrence in which the outcome is not known is called an experiment. Rolling a die is an example of an experiment. Therefore, for a single roll of the die, the sample space, denoted by S would be S ={1, 2, 3, 4, 5, 6}.

Suppose the result on a single roll of a die is 2. The next roll of the die is a 4. Each of these results are a subset of the sample space. A subset is called an event and denoted by E. Therefore, the subsets noted above are E ={2} and E = {4}.

The way in which we calculate theoretical probabilities is we divide the number of outcomes in the event divided by the total number of outcomes in the sample space.

What is the probability that a single roll of a fair 6 sided die lands 1 face up?
P(E) = (number of outcomes that result in 1 lands face up)/ (total number of possible outcomes) = 1/6

Example: A coin is flipped 3 times.
a. What is the probability that 2 of the 3 coins land heads up?
First we will figure out the sample space and the number of outcomes in the event.
We'll define the event E = { (H, H, T), (H, T, H), (T, H, H) }
The sample space is S = { (H, H, H), (H, H, T), (H, T, H), (H, T, T), (T, H, H), (T, T, H), (T, H, T), (T, T, T) }
There are 3 events and 8 outcomes in the sample space, therefore the probability that 2 of the 3 coins land heads up is 3/8.

• Note that an easy way to determine how large the sample space is in this problem is we take the number of
possibilities per coin toss and raise it to the power of the number of tosses (23 = 8).

b. What is the probability that at least 2 coins land tails up?
In this case, we define the event E = { (H, T, T), (T, T, H), (T, H, T), (T, T, T) }
The sample space is the same as in part a, therefore the probability that at least 2 coins land heads up is 4/8 = ½.

Notice the events that 0, 1, 2 and 3 heads appear are listed below with their associated probabilities.
0 heads: E = { (T, T, T) } P(E) = 1/8
1 head: E = { (H, T, T), (T, H, T), (T, T, H) } P(E) = 3/8
2 heads: E = { (H, H, T), (H, T, H), (T, H, H) } P(E) = 3/8
3 heads: E = { (H, H, H) } P(E) = 1/8

Notice if we add the probabilities of all the events, we get 1. The sum of theoretical probabilities of all possible outcomes in a sample space equals 1.

Example: Two fair 6 sided dice as rolled. What is the probability of getting a sum of 6, 7 or 8 ?
Since each die has equally likely outcomes, there are 6 ∙6 = 36 possible outcomes in the sample space.

Notice all of the possible outcomes below.
S = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) }

The outcomes in S that give us a sum of 6, 7 or 8, which we denote as event E are
E = { (1, 5), (5, 1), (2, 4), (4, 2), (3, 3), (3, 4), (4, 3), (1, 6), (6, 1), (2, 5), (5, 2), (2, 6), (6, 2), (3, 5), (5, 3), (4, 4) }

There are 16 outcomes in E and 36 outcomes in S. Therefore, the probability of getting a sum of 6, 7 or 8 is
16/36 =4/9