#### notes

Let us think of a simple experiment. A bag contains 4 balls of the same size. Three of them are white and the fourth is black. You are supposed to pick one ball at random without seeing it. Then obviously, possibility of getting a white ball is more.

In Mathematical language, when possibility of an expected event is expressed in number, it is called ‘Probability’ . It is expressed as a fraction or percentage using the following formula.

For a random experiment, if sample space is ‘S’and ‘A’ is an expected event then probability of ‘A’ is P(A). It is given by following formula.

`P(A)="Number of sample points in event A"/"Number of sample points in sample spaces"="n(A)"/"n(S)"`

In the above experiment, getting a white ball is event A. As there are three white balls n(A) = 3, As the number of balls is 4, n(S) = 4

probability of getting a white ball is, `P(A)="n(A)"/"n(S)"=3/4`

Similarly, if getting black ball is event B, then n(B) = 1

`therefore P(B)="n(B)"/"n(S)"=1/4`

**Ex.** Find the probability of the following, when one coin is tossed.

(i) getting head (ii) getting tail

**Solution** : Let ‘S’ be the sample space.

S = {H, T} n(S) = 2

(i) Let event A be getting head

A = {H} ∴ n(A) = 1

`P(A)="n(A)"/"n(S)"=1/2`

(ii) Let event B be getting tail

B={T} ∴n(B)=1

`P(B)="n(B)"/"n(S)"=1/2`