odds for one event happening: 1/118 odds for second event happening: 1/76 what is odds of either event happeni

Question

odds for one event happening: 1/118

odds for second event happening: 1/76

what is odds of either event happening

FIRST ANSWER 10 POINTS (correct answer)

one engineering and one computer science and we're too lazy to figure this out

Answer 1

If I understand you correctly, you have two events, A and B, and you say that

P(A) = 1/118, and P(B) = 1/76.

Now, the probability of either A or B happening is given by

P(A or B) = P(A) + P(B) - P(A and B),

i.e. the sum of the individual probabilities, minus the probability that both happen. You don't give P(A and B), but if the events are independent (i.e. the occurrence of one event does not affect the probability that the other event occurs), then

P(A and B) = P(A) x P(B),

and so,

P(A or B) = P(A) + P(B) - P(A) P(B).

If you put the numbers in,

P(A or B) = (1/118) + (1/76) - [ (1/118)x(1/76) ]
P(A or B) = 0.0215209634,

i.e. there is about a 2.15% chance of either event A or event B occurring (but not both). This is the usual definition of "or" in probability.

Answer 2

I agree - just sum the two odds if it is either or both, if is either but NOT both (an exclusive OR) then you have add the odds then substract the product.

1/118 + 1/76 - 1/118/76 = 0.021520963....

Answer 3

1/118 + 1/76 would be the odds for at least one event happening if you don't mind that both can occur also.

If you demand that only one occurs but not both, then the odds would be:

(75/76 * 1/118) + (117/118 * 1/76).... (I think)

Answer 4

Depends on if both outcomes are feasible.

If not, the probability of either happening is just the sum of your two probabilities minus the joint prob. of the two events.

P(a or b,NOT a and b) = P(a) + P(b) - P(a and b)

If you allow both to succeed, it is the sum of your two probabilities .

P(a or b, a and b) = P(a) + P(b)

If you are confused by this, look up Venn Diagram on the web. It is a graphical way to depict the probabilities in question. I find it useful for keeping track of when and when not to subtract out the joint probability P(a and b).

By the way, if P(a) and P(b) are not correlated (dependent on each other), you can write P(a and b) = P(a)P(b).

Answer 5

Approximately 1:48.5