# Suppose that you are rolling a six sided dice. Let A = get a number smaller than 2. What is P(Ac)?

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:

$P( \ A_c \ ) \ = \ 1 \ – \ P( \ A \ )$

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:

$P( \ A_c \ ) \ = \ 1 \ – \ P( \ A \ )$

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