Consider an even money game that pays 4:3 if event A occurs. What is the probability of event A?
I am thinking that here,
probability = denominator of odds / (denominator of odds + numerator of odds)
but I find other sources say
odds = probability / (1-p)
Not sure which to use
2 answers
How do I know which one to use
They are the same thing.
e.g.
suppose there are 2 possible outcomes to some experiment, A and B and
prob of event A is 5/19
then the prob of B is 1 - 5/19 = 14/19
odds in favour of event A = (5/19) ÷ (14/19)
= (5/19)(19/14) = 5/14 or 5 : 14
or
odds in favour of event A = (5/19) ÷ (1 - 5/19)
= (5/19) ÷ (14/19) , which is the same as above
So for yours, the odds in favour of event A = 4 : 3
working backwards:
the prob of A = 4/7
the prob of NOT A = 3/7
Notice I added the two terms of the given odds ratio to form my denominator
which what your first method suggests.
e.g.
suppose there are 2 possible outcomes to some experiment, A and B and
prob of event A is 5/19
then the prob of B is 1 - 5/19 = 14/19
odds in favour of event A = (5/19) ÷ (14/19)
= (5/19)(19/14) = 5/14 or 5 : 14
or
odds in favour of event A = (5/19) ÷ (1 - 5/19)
= (5/19) ÷ (14/19) , which is the same as above
So for yours, the odds in favour of event A = 4 : 3
working backwards:
the prob of A = 4/7
the prob of NOT A = 3/7
Notice I added the two terms of the given odds ratio to form my denominator
which what your first method suggests.
0 answers