Skip to main content
A probability experiment is a process that leads to well-defined results called outcomes.
percobaan probabilitas adalah proses yang menghasilkan hasil yang terdefinisi dengan baik yang disebut sebagai outcome.
An outcome is the result of a single trial of a probability experiment.
outcome adalah hasil dari satu percobaan probabilitas.
A tree diagram can be used as a systematic way to find all possible outcomes of a probability experiment.Diagram pohon dapat digunakan sebagai cara sistematis untuk menemukan semua hasil yang mungkin dari suatu percobaan probabilitas.

​
Tree Diagram for Tossing Two Coins

Tree Diagram for Tossing Two Coins

​
Sample Spaces - Examples

ExperimentSample Space
Toss one coinH, T
Roll a die1, 2, 3, 4, 5, 6
Answer a true-false questionTrue, False
Toss two coinsHH, HT, TH, TT

​
Formula for Classical Probability

Classical probability assumes that all outcomes in the sample space are equally likely to occur.
Probabilitas klasik mengasumsikan bahwa semua hasil dalam ruang sampel memiliki kemungkinan yang sama untuk terjadi.
That is, equally likely events are events that have the same probability of occurring.
Artinya, kejadian yang sama kemungkinannya adalah kejadian yang memiliki probabilitas yang sama untuk terjadi.

The probability of any event EE is
number of outcomes in Etotal number of outcomes in the sample space\dfrac{\text{number of outcomes in E}}{\text{total number of outcomes in the sample space}}
This probability is denoted by          P(E)=n(E)n(S)~~~~~~~~~~ P(E) = \dfrac{n(E)}{n(S)} This probability is called classical probability, and it uses the sample space SS

​
Classical Probability - Examples

For a card drawn from an ordinary deck, find the probability of getting (a) a queen (b) a 6 of clubs (c) a 3 or a diamond.
Solution:(a) Since there are 4 queens and 52 cards, P(queen)=452=113 P(queen) = \dfrac{4}{52} = \dfrac{1}{13}
(b) Since there is only one 6 of clubs, then, P(6 of clubs)=152 P(6~of~clubs) = \dfrac{1} {52}
(c) There are four 3s and 13 diamonds, but the 3 of diamonds is counted twice in the listing. Hence there are only 16 possibilities of drawing a 3 or a diamond, thus P(3 or diamond)=1652=413 P(3~or~diamond) = \dfrac{16} {52} = \dfrac{4} {13}

When a single die is rolled, find the probability of getting a 9.
Solution:Since the sample space is 1, 2, 3, 4, 5, and 6, it is impossible to get a 9. Hence, P(9)=06=0 P(9) = \dfrac{0}{6} = 0
The sum of the probabilities of all outcomes in a sample space is one.

​
Complement of an Event

The complement of an event E is the set of outcomes in the sample space that are not included in the outcomes of event E. The complement of E is denoted by E‾\overline{E} (E bar). Complement of an Event

​
Complement of an Event - Examples

Find the complement of each event.
Solution: Getting a 1, 2, 3, 5, or 6.
Solution: Getting a consonant (assume y is a consonant).
Solution: Getting Saturday or Sunday.
Solution: Getting a girl.

​
Rule for Complementary Event

P(E‾)=1−P(E)P(\overline{E}) = 1 - P(E), or
P(E)=1−P(E‾)P(E) = 1 - P(\overline{E}), or
P(E)+P(E‾)=1P(E) + P(\overline{E}) = 1

​
Empirical Probability

The difference between classical and empirical probability is that classical probability assumes that certain outcomes are equally likely while empirical probability relies on actual experience to determine the probability of an outcome.
Perbedaan antara probabilitas klasik dan probabilitas empiris adalah bahwa probabilitas klasik mengasumsikan bahwa hasil tertentu sama kemungkinannya sementara probabilitas empiris bergantung pada pengalaman nyata untuk menentukan probabilitas suatu hasil.

​
Formula for Empirical Probability

Given a frequency distribution the probability of an event being in a given class is
P(E)=frequency for the classtotal frequencies in the distributionP(E) = \dfrac{\text{frequency for the class}}{\text{total frequencies in the distribution}}
          =fn ~~~~~~~~~~= \dfrac{f}{n}.
This probability is called the empirical probability and is based on observation.

​
Empirical Probability - Example

In a sample of 50 people, 21 had type O blood, 22 had type A blood, 5 had type B blood, and 2 had AB blood. Set up a frequency distribution.
Blood TypeFrequency
O21
A22
B5
AB2
50 = nn

Find the following probabilities for the previous example.
P(O)=2150.P(O) = \dfrac{21}{50}.
P(A or B)=2250+550=2750.P(A~or~B) = \dfrac{22}{50} + \dfrac{5}{50} = \dfrac{27}{50}.