Two Definitions of Probability

Classical

The classical definition of probability is only applicable when we are considering events that are equally likely.

\[P(E) = \frac{\text{number of outcomes favorable to E}}{\text{total number of outcomes}}\]

Example 1:

Roll a die. Find the probability of rolling

  • an even number
  • a number that is less than or equal to 4
  • a 7
  • a 1 or 2 or 3 or 4 or 5 or 6

  • there are 3 outcomes favorable to rolling an even number and there are 6 possible outcomes so \(P(\text{even number}) = \frac{3}{6}\)
  • there are 4 outcomes favorable to rolling a number that is less than or equal to 4 and there are 6 possible outcomes so \(P(\text{number} \le 4)=\frac{4}{6}\)
  • there are 0 outcomes favorable to rolling a 7 and there are 6 possible outcomes so \(P(\text{rolling a 7})=\frac{0}{6}=0\)
  • there are 6 outcomes favorable to rolling a 1 or 2 or 3 or 4 or 5 or 6 and there are 6 possible outcomes so \(P(\text{1 or 2 or 3 or 4 or 5 or 6})=\frac{6}{6}=1\) i.e. \(P(\text{sample space})=1\)

Relative Frequency

The relative frequency definition of probability is only applicable when we are considering events that can be repeated. Note that this can also include events that are equally likely so in that case the classical definition and the relative frequency definition will overlap.

\[P(E) = \frac{\text{number of times E occurs}}{\text{total number of trials where the number of trials is large}}\] Flip a coin. The probability of getting a head is equal to the long run proportion of heads in many many flips. In the trials below, you can see in the first flip we got a tail so the proportion of heads is 0/1. In the second flip, we got another tail so the proportion of heads is 0/2. In the third flip, we got a head so the proportion of heads is 1/3. We can see up to the eighth trial in this example.

Below we have about 500 trials and we can see that the proportion of heads settles down to about 0.5 which is the probability of getting heads when tossing a fair coin.

Try flipping your own coin here

Two Basic Properties of Probability

  1. For any event \(E\), \(0 \le P(E) \le 1\)
  2. \(P(\text{sample space}) = 1\)

The Complement Rule

\[P(A^c) = 1 - P(A)\]

Example 2:

Select one card at random from a standard deck of cards. Let A be the event that the chosen card is an Ace. Find \(P(A^c)\)

\(P(A) = \frac{4}{52}\) and \(P(A^c) = 1 - P(A) = \frac{48}{52}\)

The Addition Rule

\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]
Example 3: Select one card at random from a standard deck of cards. Let A be the event that the chosen card is an Ace. Let B be the event that the chosen card is red. Find the probability that the chosen card is an Ace or is red.

Using the addition rule, \(P(A \cup B) = P(A)+P(B)-P(A \cap B) = \frac{4}{52} + \frac{26}{52} - \frac{2}{52} = \frac{28}{52}\) Notice that we double counted the red Aces so that’s why we had to subtract the last term.

Another Example

Example 4:

Sixty people were asked for their gender and political party.

Democrat Republican Independent Total
Male 10 5 5 20
Female 15 15 10 40
Total 25 20 15 60

Find the probability that a randomly selected person is

  • female
  • female and democrat
  • female or democrat
  • not democrat

  • there are 40 females so \(P(F) = \frac{40}{60}\)
  • there are 15 female democrats so \(P(F \cap D) = \frac{15}{60}\)
  • \(P(F \cup D) = P(F) + P(D) - P(F \cap D) = \frac{40}{60}+\frac{25}{60}-\frac{15}{60}=\frac{50}{60}\)
  • \(P(D^c) = 1 - P(D) = 1 - \frac{25}{60} = \frac{35}{60}\)