Why does he multiply the combinatorics by the probabilities here but only compute the combinatorics in the example posted in the comments?
1 Like
I solved the problem in the first picture using the approach he uses here and expected to get 1/11. i.e.
RRRR = 4! / (0!4!) = 1
RRRN = 4! / (1!3!) = 4
RRNN = 4! / (2!2!) = 6
You should multiply by the probabilities. The reason it isn’t needed in the second example is that the probability of the event happening and not happening is the same, so the probability cancels out (in other words, Greg is making use of a shortcut). This is not the case in the first case (where the probability of the event happening and not happening is \frac{1}{3} and \frac{2}{3} respectively).
1 Like
Thank you