Whitney Garner wrote:

sud21 wrote:

Hi sud21!

I would be very leery of questions posted from unknown sources. Unfortunately, there is no way to answer this question in the form given. We might think this is as simple as multiplying the Pr(NOT A) times Pr(NOT B) = 1/5, but this will not always be the case. We need to know more about the events in terms of their dependence. Here are 2 examples of what I mean.

For the following 2 examples, let's pretend that we have a 10 sided die with the numbers 1-10 on the sides (to keep the probabilities the same.

Example 1

Event A = rolling an even number (2, 4, 6, 8, 10 are successes) = 0.5 probability
Event B = rolling a number >=5 (5, 6, 7, 8, 9, 10 are successes) = 0.6 probability

Now, if we use simple formulas:
Pr(A&B) = (0.5)(0.6) = 0.3
Pr(A only) = Pr(A & NOT B) = (0.5)(1-0.6) = (0.5)(0.4) = 0.2
Pr(B only) = Pr(B & NOT A) = (0.6)(1-0.5) = (0.6)(0.5) = 0.3
Pr(Neither) = Pr(NOT A & NOT B) = (1-0.5)(1-0.6) = (0.5)(0.4) = 0.2

Let's check this by looking at the actual outcomes for each probability:
Pr(A&B) = (6, 8, 10) = 3/10 = 0.3
Pr(A only) = (2, 4) = 2/10 = 0.2
Pr(B only) = (5, 7, 9) = 3/10 = 0.3
Pr(Neither) = (1, 3) = 2/10 = 0.2

These all check out, so it would seem as if the basic formulas work...BUT, what if we change the definition of event A...

Example 2
Event A = rolling a number >=6 = (6, 7, 8, 9, 10) = 0.5 probability
Event B = rolling a number >=5 = (5, 6, 7, 8, 9, 10) = 0.6 probability

The probabilities we calculated above would still be the same mathematically...but when we actually look at the results, we see that they SHOULD be different:
Pr(A&B) = (6, 7, 8, 9, 10) = 5/10 = 0.5
Pr(A only) = (empty) = 0.0
Pr(B only) = (5) = 1/10 = 0.1
Pr(Neither) = (1, 2, 3, 4) = 4/10 = 0.4

So you can see, there is no way to answer this question unless we know that the events are "independent"!


Whit

Whitney Garner wrote:Hi Pemdas!

I'm sorry if the language in my earlier reply wasn't 100% clear. You are exactly right that we need to be concerned about the Joint probability, because without it, there is no way to answer this question. You assumed that the joint probability must be the simple overlap (0.1), but from the 2 examples I provided, just because one event has a probability of 0.6 and the other has a probability of 0.5, we cannot draw any conclusions on their "overlap" (or joint likelihood). In both examples, the events I provided had the same probabilities as the original question, but in example 1 the joint probability was 0.3 (some overlap) while in example 2 it was 0.5 (perfect overlap). This results in 2 very different answers for the probability of Neither.

I also understand that my examples might have caused confusion because they are not as simply stated as the original. I wanted to use real world example events to indicate the various levels of "overlap" that might occur. This is always something we should consider. If the original had told us that each occurred with a probability of 0.5, we CANNOT assume that they are mutually exclusive just because they sum to 1. The point is that to find the probability of NEITHER, we would need more information, this problem is Incomplete.

I hope this clears up the confusion from my earlier response!

Whit