Study Elementary Probability

Introduction to study elementary probability:

The elementary probability theory condition for all probability mass function (pmf) there is functions which give the probability in a divide random variable which is accurately corresponding to some of the value recognized. A pmf differs as of a common probability density function (pdf) in the values of a pdf, defined for the permanent random variables and not the probabilities as such desired.


How to study elementary probability:


The study of elementary probability function always defines as known in the relationship among the two such variables in a probability distribution function which is named as the "Probability Function". An elementary probability function assumes that the "variable" which indicates the values within the given range of such a random variable at its independent variable and "probability" defined as the dependent variable.

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Study the representation of probability function for elementary type:


An elementary probability function which relates the break up random variable is recognized as the "Probability Mass Function". In the sequence of a random variable the value of "X" assumes that the distinct set of values `x_1, x_2, x_3, ... x_n,` then the function "f" is defined by the f(xi) = P(X = xi) and that is called as the "Probability Function" or "Probability Mass Function". The pmf assigns a value for the required probability [P(X = xi)] in each of the possible values [xi] of the variable.

That gives the study of elementary probability value and the variable which represents the sequence of the distinct random variable which equals to some value. A study of discrete probability function f(x) defines the following properties.

f (xi) ≥ 0 is given by the probability for the variable to carryover a particular value which is always a positive real number.

Σ f (xi) = 1 [i = 1, 2, 3, ... ∞] The sum of the corresponding probabilities of all the given possible values and that suits the variable which represents the range of all the discrete random variable which may carry the value equal to One.

Elementary Probability

Introduction to elementary probability:

          The elementary probability started with gambling, The probability mostly used in the field of physical science, commerce, biology science, medical science, weather forecasting , etc., An experiment having more than one possible outcome is called statistical experiment. A statistical experiment is also known as a trial. Tossing a coin whether it results in a head or a tail is a trial. Certain statements implying more than one possible situation can also be termed as trials.


Elementary probability of an event

In a random experiment, let S be the sample space and E `sube` S. Then E is an event.

The elementary probability of occurrence of E is defined as

            P (E)= number of outcomes favorable to occurrence of E
                         number of all possible outcomes

                        = number of distinct element in E
                                number of distinct element in S

                        = n (E)
                           n (S)

                        Therefore P (E) = n(E)
                                                  n(S)


Elementary probability example problem

Example 1:

A coin is tossed once. Find the probability of getting a head?

Solution:

When a coin is tossed once, the sample space is given by S= {H, T},

Let E be the event of getting a head.

Then, E = {H}.

Therefore n(E)=1 and n(S)=2.

P (getting a head)= p(E)= n(E) / n(S) = 1 / 2.

Example 2:

Two coins are tossed once. Find the probability of

(i)Getting 2 heads

(ii)Getting a least 1 head

(iii)Getting no head

(iv)Getting 1 tail and 1 tail

Solution:

Where 2 coins are tossed once, the sample space is given by S= {HH, HT, TH, TT} and, therefore, n(S)=4.

(i)Getting 2 head

Let E1 = event of getting 2 heads. Then,

E1= {HH} and, therefore, n (E1) =1.

Therefore P (getting 2 heads) = P(E1) =  n(E1) / n(S) = 1 / 4

(ii)Getting at least 1 head

Let E2 = event of getting at least 1 head. Then,

E2= {HT, TH, HH} and, therefore, n (E2) =3.

Therefore P (getting at least 1 head) = P (E2) = n(E2) / n(S) = 3 / 4

(iii)Getting no head

Let E3 = event of getting no head. Then,

E3= {TT} and, therefore, n (E3) =1.

Therefore P (getting no head) = P (E3) = n(E3) / n(S) = 1 / 4

(iv)Getting 1 head and 1 tail

Let E4 = event of getting 1 head and 1 tail. Then,

E4= {HT, TH} and, therefore, n (E4) =2.

Therefore P (getting 1 head and 1 tail) = P (E4) = n(E4) / n(S) = 2 / 4 = 1 / 2.