2.3  Interpretations of Probability
2.3  Interpretations of ProbabilityInterpretations of Probability
Before discussing the specific rules of probability, it's important to recognize there are three main interpretations of probability.
 Classical Interpretation of Probability

The probability that event E occurs is denoted by P(E). When all outcomes are equally likely, then:
 \[P(E) = \frac{number\ of\ outcomes\ in\ E}{number\ of\ possible\ outcomes}\]
 Subjective Probability

Subjective probability reflects personal belief which involves personal judgment, information, intuition, etc.
For example, what is P (you will get an A in a certain course)? Each student may have a different answer to the question.
 Relative Frequency Concept of Probability (Empirical Approach)

If a particular outcome happens over a large number of events then the percentage of that outcome is close to the true probability.
For example, if we flip the given coin 10,000 times and observe 4555 heads and 5445 tails, then for that coin, P(H)=0.4555.
 \[P(E) \approx \frac{number\ of\ outcomes\ in\ E}{number\ of\ possible\ outcomes}\]
Example 25
 Find the probability that exactly one head appears in two flips of a fair coin.
Answer:
The possible outcomes are listed as: {(H, H), (H, T), (T, H), (T, T)}.
Note that the four outcomes are of equal probability since the coin is fair.
\[P(getting\ exactly\ one\ H\ in\ two\ flips\ of\ a\ fair\ coin)=P\{(H,T),(T,H)\}=\frac{2}{4}=\frac{1}{2}\]
 Find the probability that two heads appear in two flips of a fair coin.
Answer:
\(P(getting\ two\ H\ in\ two\ flips\ of\ a\ fair\ coin)=P\{(H,H)\}=\frac{1}{4}\)
 Find the probability that the sum of two faces is greater than or equal to 10 when one rolls a pair of fair dice.
Answer:
The possible outcomes of the experiment are:
{ (1,1) (2,1) (3,1) (4,1) (5,1) (6,1) (1,2) (2,2) (3,2) (4,2) (5,2) (6,2) (1,3) (2,3) (3,3) (4,3) (5,3) (6,3) (1,4) (2,4) (3,4) (4,4) (5,4) (6,4) (1,5) (2,5) (3,5) (4,5) (5,5) (6,5) (1,6) (2,6) (3,6) (4,6) (5,6) (6,6) }
If S denotes the sum of the points in the two faces:
\[ \begin {align} P(S\ greater\ than\ or\ equal\ to\ 10)& =P(S=10)+P(S=11)+P(S=12)\\
& = P\{(4,6),(5,5),(6,4)\}+P\{(5,6),(6,5)\}+P\{(6,6)\} \\
& = \frac{3}{36}+\frac{2}{36}+\frac{1}{36} \\
& =\frac{1}{6} \\
\end {align} \]
Try It! Rolling Dice
Directions: Try each of these problems and click the buttons to reveal the correct answers.
Again using the example where we flip two fair sixsided dice, find the following:
 \(A={(3, 5)}\)
 \(B=\text {a 4 is rolled on the first die}\)
 \(C=\text {a 5 is rolled on the second die}\)
 \(D=\text {the sum of the dice is 7}\)
 \(E={(7, 4)}\)

\(P(B\cap D)\) and \(P(B\cup D)\).\(P(B\cap D)=\frac{1}{36}\) and \(P(B\cup D)=\frac{11}{36}\).

\(P(D\cap C)\) and \(P(D\cup C)\).\(P(D\cap C)=\frac{1}{36}\) and \(P(D\cup C)=\frac{11}{36}\)

\(P(A\cap D)\) and \(P(A\cup D)\)\(P(A\cap D)=0\) and \(P(A\cup D)=\frac{7}{36}\)

\(P(B\cap C)\) and \(P(B\cup C)\)\(P(B\cap C)=\frac{1}{36}\) and \(P(B\cup C)=\frac{11}{36}\)