# The probability of an event E not occurring is 0.4. What are the odds in favor of E occurring?

Oct 24, 2017

$P \left(E\right) = 0.6$

#### Explanation:

An event must either occur ($E$) or not occur (!E)

Therefore the sum of the probabilities of an event occurring and an event not occurring must be equal to 100%

That is P(E)+P(!E)=1.00

Given that P(!E)=0.40
This implies that
$\textcolor{w h i t e}{\text{XXX}} P \left(E\right) + 0.40 = 1.00$

$\textcolor{w h i t e}{\text{XXX}} P \left(E\right) = 0.60$

Oct 29, 2017

The odds in favour of $E$ occurring are $3 : 2$.

#### Explanation:

An odds in favour is a ratio of "how likely an event is to occur" to "how likely it is to NOT occur". This can be derived from

$\text{number of favourable outcomes"/"number of unfavourable outcomes}$

or

$\text{proability of event occuring"/"probability of event not occurring}$

and is usually expressed in colon notation as $n : m ,$ where $n$ and $m$ are whole numbers.

Given "P"(E^"C")=0.4, we can deduce that

"P"(E)=1-"P"(E^"C")
$\textcolor{w h i t e}{\text{P} \left(E\right)} = 1 - 0.4$
$\textcolor{w h i t e}{\text{P} \left(E\right)} = 0.6$

which gives

"odds"(E)="P"(E):"P"(E^"C")
$\textcolor{w h i t e}{\text{odds} \left(E\right)} = 0.6 : 0.4$

This can be scaled up by 5, so that both numbers in the odds are whole numbers:

$\text{odds"(E)=0.6xx5" ":" } 0.4 \times 5$
$\textcolor{w h i t e}{\text{odds} \left(E\right)} = 3 : 2$.