They can not be dependent. Oswald's position in each lineup can be any one of four. Each witness has a 1/4 probability of a random correct pick.
Nobody claimed that the lineup picks were random events.
Well duh, of course there was nothing random about picking Oswald, he was the man they all saw.
As you state it yes, they are four times worse.Like six persons rolling a four-sided dice looking for exactly one combination.
https://mathflight.files.wordpress.com/2014/11/4-4-multiplication-rule-basics.pdfPlease view page 4, example "homicide case".
If that's a legitimate source then I can't argue with that, it certainly fits all the parameters.Ironically I first wrote down 1 in 4096 because that was the amount of options and then because I'm a bit rusty I thought I'd better check and came up with many examples similar to my post above which seem to make sense?Q. There are four friends ? Adam, Bella, Christopher and Drew. All of them are asked to choose any number in their mind. Now what is the probability that every one of them has the same number in mind? The chosen number must be in between 1 to 5 (including 1 and 5).A1. Without loss of generality, let's say Adam picks first. The probability that the other 3 choose the exact same number as Adam is 1/5∗1/5∗1/5, which gives the result of 1/125.Note: this answer presumes that the numbers are chosen at random.JohnM