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

2 Answers

Oct 24, 2017

$P (E) = 0.6$ $P(E)=0.6$

Explanation:

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

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

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

Given that $P (! E) = 0.40$ $P(!E)=0.40$
This implies that
$XXX P (E) + 0.40 = 1.00$ $color(white)("XXX")P(E)+0.40=1.00$

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

Oct 29, 2017

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

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

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

or

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

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

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

$P (E) = 1 - P (E^{C})$ $"P"(E)=1-"P"(E^"C")$
$P (E) = 1 - 0.4$ $color(white)("P"(E))=1-0.4$
$P (E) = 0.6$ $color(white)("P"(E))=0.6$

which gives

$odds (E) = P (E) : P (E^{C})$ $"odds"(E)="P"(E):"P"(E^"C")$
$odds (E) = 0.6 : 0.4$ $color(white)("odds"(E))=0.6:0.4$

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

$odds (E) = 0.6 \times 5 : 0.4 \times 5$ $"odds"(E)=0.6xx5" ":" ""0.4xx5$
$odds (E) = 3 : 2$ $color(white)("odds"(E))=3:2$ .

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