Question

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

Answer 1

`P(E)=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
`color(white)("XXX")P(E)+0.40=1.00`

`color(white)("XXX")P(E)=0.60`

Answer 2

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

`"number of favourable outcomes"/"number of unfavourable outcomes"`

or

`"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")`
`color(white)("P"(E))=1-0.4`
`color(white)("P"(E))=0.6`

which gives

`"odds"(E)="P"(E):"P"(E^"C")`
`color(white)("odds"(E))=0.6:0.4`

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

`"odds"(E)=0.6xx5" ":" ""0.4xx5`
`color(white)("odds"(E))=3:2`.