If the P(A) = 1/36, then probability of non-A =
1 - 1/36 = 35/36.
Given that P(A)= 1/36, what are the odds against A occuring?
I found this formula in my chapter
odds against A= 1-P(a)/P(a)
Would it be 1-1/36 over 1/36? Or how do I go about it?
2 answers
To carry that a bit further
odds = probability of what you want / probability of what you do not want
you want odds against A
so probability of what you want is 35/36
and the probability of what you do not want is A or 1/36
so odds against A are (35/36) / (1/36) = 35 to 1
which is [1-p(a)]/p(a)
odds = probability of what you want / probability of what you do not want
you want odds against A
so probability of what you want is 35/36
and the probability of what you do not want is A or 1/36
so odds against A are (35/36) / (1/36) = 35 to 1
which is [1-p(a)]/p(a)
2 answers