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.