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?
If the P(A) = 1/36, then probability of non-A =
1 - 1/36 = 35/36.
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)
To find the odds against event A occurring, you can use the formula:
Odds against A = (1 - P(A)) / P(A)
In your case, P(A) = 1/36. Plugging this value into the formula, we have:
Odds against A = (1 - 1/36) / (1/36)
To simplify this expression, we can first find the common denominator:
Odds against A = (36/36 - 1/36) / (1/36)
Next, let's perform the subtraction:
Odds against A = (35/36) / (1/36)
To divide two fractions, we can multiply the first fraction by the reciprocal of the second fraction:
Odds against A = (35/36) * (36/1)
We can cancel out the common factor of 36:
Odds against A = 35/1
Therefore, the odds against event A occurring are 35 to 1. This means that for every 35 outcomes where A does not occur, there is 1 outcome where A occurs.