Expand this (A+B)^3.
After you are done, ponder your expression with A = 0.96 and B = 0.04
Ok, I have a scenario I can't get my head around.
I have a system which captures information about vehicles going past a camera. 96% of vehicles will be captured correctly, so 4% will not be captured correctly.
Any particular vehicle has a 96% chance of being captured correctly.
If I have 3 capture points, the probability of any particular vehicle being MISSED is 0.04x0.04x0.04 = 0.000064 , so the chance of it being read correctly on at least one point is the inverse of that = 99.9936%
Hopefully that reasoning is correct.
What I can't grasp is how to work out the probability of it matching on any two of the three points.
Let's say we take point A to point B - 96% on each point means the probability of matching both is 92% (0.96 x 0.96). And matching at A AND B AND C is 0.96*0.96*0.96 = 88%
But what if I care about matching it at A-B OR A-C - the percentage is obviously higher than 92%, but I can't work out how to...um...work it out. My degree was an awfully long time ago.
Thanks!
guess it has to do something along these lines... if p(A) , p(B) and p(C) denote the chances that the vehicle being read at the capture points A, B and C, the answer might have to do with
p(ABC') + p(AB'C) + p(A'BC) where p(A') , p(B') and p(C') denote the chances that the vehicle being missed at the capture points A, B and C... wish it works...