The aim of this question is to learn how to **calculate the probability** of simple experiments such as **rolling a die**.

The **probability of a particular event A** is given by:

\[ P( \ A \ ) \ = \ \dfrac{ n( \ A \ ) }{ n( \ S \ ) } \ = \ \dfrac{ \text{ No. of all possible outcomes for event A } }{ \text{ No. of all possible outcomes } } \]

Also, the probability of **complement of A** is given by:

\[ P( \ A_c \ ) \ = \ 1 \ – \ P( \ A \ ) \]

## Expert Answer

**All the possible outcomes while rolling a six-sided die are listed below:**

\[ S \ = \ \{ \ 1, \ 2, \ 3, \ 4, \ 5, \ 6 \ \} \]

And:

\[ \text{ No. of all possible outcomes } \ = \ n( \ S \ ) \ = \ 6 \]

Since:

\[ A \ = \ \{ \text{ all the possible outcomes smaller than 2 } \} \]

\[ \Rightarrow \ A \ = \ \{ \ 1 \ \} \]

And:

\[ \text{ No. of all possible outcomes for event A } \ = \ n( \ A \ ) \ = \ 1 \]

So:

\[ P( \ A \ ) \ = \ \dfrac{ n( \ A \ ) }{ n( \ S \ ) } \ = \ \dfrac{ 1 }{ 6 } \]

Since:

\[ A_c \ = \ \{ \text{ all the possible outcomes not smaller than 2 } \} \]

\[ \Rightarrow \ A \ = \ \{ \ 2, \ 3, \ 4, \ 5, \ 6 \ \} \]

And:

\[ \text{ No. of all possible outcomes for event } A_c \ = \ n( \ A_c \ ) \ = \ 5 \]

So:

\[ P( \ A_c \ ) \ = \ \dfrac{ n( \ A_c \ ) }{ n( \ S \ ) } \ = \ \dfrac{ 5 }{ 6 } \]

**The same problem can also be solved using the following formula:**

\[ P( \ A_c \ ) \ = \ 1 \ – \ P( \ A \ ) \]

\[ \Rightarrow P( \ A_c \ ) \ = \ 1 \ – \ \dfrac{ 1 }{ 6 } \]

\[ \Rightarrow P( \ A_c \ ) \ = \ \dfrac{ 5 \ – \ 1 }{ 6 } \]

\[ \Rightarrow P( \ A_c \ ) \ = \ \dfrac{ 5 }{ 6 } \]

## Numerical Result

\[ P( \ A \ ) \ = \ \dfrac{ 1 }{ 6 } \]

\[ P( \ A_c \ ) \ = \ \dfrac{ 5 }{ 6 } \]

## Example

Let’s say we roll a six-sided die and let $ A \ = $ get a number **smaller than 4**. Calculate P(Ac).

**All the possible outcomes while rolling a six-sided die are listed below:**

\[ S \ = \ \{ \ 1, \ 2, \ 3, \ 4, \ 5, \ 6 \ \} \]

And:

\[ \text{ No. of all possible outcomes } \ = \ n( \ S \ ) \ = \ 6 \]

Since:

\[ A \ = \ \{ \text{ all the possible outcomes smaller than 4 } \} \]

\[ \Rightarrow \ A \ = \ \{ \ 1, \ 2, \ 3 \ \} \]

And:

\[ \text{ No. of all possible outcomes for event A } \ = \ n( \ A \ ) \ = \ 3 \]

So:

\[ P( \ A \ ) \ = \ \dfrac{ n( \ A \ ) }{ n( \ S \ ) } \ = \ \dfrac{ 3 }{ 6 } \ = \ \dfrac{ 1 }{ 2 }\]

Since:

\[ P( \ A_c \ ) \ = \ 1 \ – \ P( \ A \ ) \]

\[ \Rightarrow P( \ A_c \ ) \ = \ 1 \ – \ \dfrac{ 1 }{ 2 } \ = \ \dfrac{ 2 \ – \ 1 }{ 2 } \ = \ \dfrac{ 1 }{ 2 }\]